# Help with unsolved indefinite integral ! $\int \!r\ln \left( r \right) \sqrt {ar+{b}^{2}+{r}^{2}}\,{\rm d}r$

I really need to solve this indefinite integral:

$$\int \!r\ln \left( r \right) \sqrt {ar+{b}^{2}+{r}^{2}}\,{\rm d}r$$

It seems much more complicated than it looks. I have found a integral table with a integral resembles my integral, and the solution involves Polylogs. The link of the intgral table is below:

http://www-elsa.physik.uni-bonn.de/~dieckman/IntegralsIndefinite/IndefInt.html

Anyone willing to take this challenge and help with this integral ?

Thank you very much in advance

• For what it is worth, the current stock version of Mathematica doesn't seem to be able to solve your integral given no additional information.
– Bill
Sep 23, 2019 at 7:38
• Please add solution for this integral? ,because I can't find on this website? Sep 23, 2019 at 16:41

Solution by derivative of AppellF1 function (using Mathematica):

solution = Limit[D[Integrate[r*r^t*Sqrt[a r + b^2 + r^2], r], t], t -> 0]

(* (r^2*Sqrt[b^2 + r*(a + r)]*(AppellF1[2, -1/2, -1/2, 3, (-2*r)/(a + Sqrt[a^2 -
4*b^2]),
(2*r)/(-a + Sqrt[a^2 - 4*b^2])]*(-1 + 2*Log[r]) +
2*(Derivative[0, 0, 0, 1, 0, 0][AppellF1][2, -1/2, -1/2, 3, (-2*r)/(a + Sqrt[a^2 -
4*b^2]),
(2*r)/(-a + Sqrt[a^2 - 4*b^2])] + Derivative[1, 0, 0, 0, 0, 0][AppellF1][2, -1/2,
-1/2, 3,
(-2*r)/(a + Sqrt[a^2 - 4*b^2]), (2*r)/(-a + Sqrt[a^2 - 4*b^2])])))/
(4*Sqrt[(-a + Sqrt[a^2 - 4*b^2] - 2*r)/(-a + Sqrt[a^2 - 4*b^2])]*
Sqrt[(a + Sqrt[a^2 - 4*b^2] + 2*r)/(a + Sqrt[a^2 - 4*b^2])])*)


Check:

 N[(solution /. r -> 2 /. a -> 1 /. b -> 1) - (solution /. r -> 1 /. a -> 1 /. b -> 1), 20]
(* 1.4951478165944420424 + 0.*10^-20 I *)

NIntegrate[r *Log[r]*Sqrt[a r + b^2 + r^2] /. a -> 1 /. b -> 1, {r, 1, 2}, WorkingPrecision -> 20]

(* 1.4951478165944420424  *)

• I will check your answer and mark as solution if correct, thank you very much for your time. Could you explain what technique you used to obtain a solution ? Sep 23, 2019 at 19:07
• Very simple technique: $\underset{t\to 0}{\text{lim}}\frac{\partial r^t}{\partial t}=\ln (r)$ Sep 23, 2019 at 19:47
• Thank you very much, your answer is correct!! If its not too much to ask, Is there a way to substitute the Appell functions derivatives for another function ? I am asking that because I am going to implement in fortran, and it seems there is no code for those derivatives. Sep 23, 2019 at 19:59
• @RomildoJuniorCarlaBeatriz get much simpler answers without the Appell function... See my answer Sep 24, 2019 at 7:09
• Maybe this helps:functions.wolfram.com/HypergeometricFunctions/AppellF1 Sep 24, 2019 at 14:42

The integral can be constructed from the following generic cases $$\int x \log(x)\sqrt{a^2+x^2} dx= \frac{1}{3} \left(-\frac{1}{3} \sqrt{a^2+x^2} \left(4 a^2+x^2\right)-a^3 \log(x)+\left(a^2+x^2\right)^{3/2} \log(x)+a^3 \log\left[a \left(a+\sqrt{a^2+x^2}\right)\right]\right)$$

$$\int x \log(x+b)\sqrt{a^2+x^2}dx =\frac{1}{6} \left(-\frac{1}{3} \sqrt{a^2+x^2} \left(8 a^2+6 b^2-3 b x+2 x^2\right)-2 \left(a^2+b^2\right)^{3/2} \log(b+x)+2 \left(a^2+x^2\right)^{3/2} \log(b+x)+b \left(3 a^2+2 b^2\right) \log\left[x+\sqrt{a^2+x^2}\right]+2 \left(a^2+b^2\right)^{3/2} \log\left[a^2-b x+\sqrt{a^2+b^2} \sqrt{a^2+x^2}\right]\right)$$

and $$\int x \log(c x+b)\sqrt{a^2+x^2}dx=\frac{1}{18 c^3}\left(-c \sqrt{a^2+x^2} \left(6 b^2-3 b c x+2 c^2 \left(4 a^2+x^2\right)\right)-6 \left(b^2+a^2 c^2\right)^{3/2} \log(b+cx)+6 c^3 \left(a^2+x^2\right)^{3/2} \log(b+c x)+3 b \left(2 b^2+3 a^2 c^2\right) \log\left[x+\sqrt{a^2+x^2}\right]+6 \left(b^2+a^2 c^2\right)^{3/2} \log\left[a^2 c-b x+\sqrt{b^2+a^2 c^2} \sqrt{a^2+x^2}\right]\right)$$

Finally, we consider a form that yields result in terms of dilogarithm function (setting $$y(x)=\sqrt{1+x^2}$$) $$\int \log(x)\sqrt{1+x^2}dx=\frac{1}{24} (-\pi^2-6 x y+12 x y \log(x)+6 \text{arcsinh}(x) \Big(\text{arcsinh}(x)+2 \log(x)-2 \log(1+x+y)-1-12 \text{Li}_2(-x-y)+12 \text{Li}_2(1-x-y)\Big)$$

• OP want's integral: $\int x \log (x) \sqrt{a x^2+b x+c} \, dx$ ? Sep 24, 2019 at 14:39
• @MariuszIwaniuk It is a simple matter to bring the integral to this form... Sep 24, 2019 at 14:40
• Then,what is the answer for: $\int r \ln (r) \sqrt{a r+b^2+r^2} \, dr$ ? Sep 24, 2019 at 14:44
• @MariuszIwaniuk If one insists on $a\neq0$ then the dilogarithm function appears :) Sep 24, 2019 at 15:00
• @yarchik thank you for your time, but as MariuszIwaniuk said, i need the answer for \int \!r\ln \left( r \right) \sqrt {ar+{b}^{2}+{r}^{2}}\,{\rm d}r , the parameter "a" cant be zero and the polynomial inside the square root must be complete Sep 25, 2019 at 18:43