Solving for $x$ with $\operatorname{arcsinh}$

Can someone please help solve for $x$ in this expression below. If someone could write out the steps on how to solve the function it would be greatly appreciated. Thank you!

$$\frac{9\operatorname{arcsinh}\left(\dfrac{3x+25\pi}{3}\right)+(3x+25\pi)\sqrt{(3x+25\pi)^2+9}-9\operatorname{arcsinh}\left(\dfrac{25\pi}{3}\right)-25\pi\sqrt{625\pi^2+9}}{6\pi}=250.$$

• Why do you think it is possible to solve this? – GEdgar Jun 20 '18 at 17:36
• Highly doubtful you'll get an analytic closed-form expression. You'll probably have to solve numerically. – Adrian Keister Jun 20 '18 at 17:36
• someone else solved it doing it equal to 1000 and got a numerical value (but showed no work) so I assumed that it would be possible in someway – AHarrington Jun 20 '18 at 17:38
• Numerical value $x=8.5864874140498282084$. For example, you can use Wolfram Alpha. – GEdgar Jun 20 '18 at 17:41
• thank you! can you show your steps to how you solved it out? – AHarrington Jun 20 '18 at 17:43

You could first simplify the equation using $$\frac{3x+25 \pi}3=y\implies x=y-\frac{25 \pi }{3}$$ which makes the equation to be $$9 y\sqrt{y^2+1} +9 \sinh ^{-1}(y)-25 \pi \left(60+\sqrt{9+625 \pi ^2}\right)-9 \sinh ^{-1}\left(\frac{25 \pi }{3}\right)$$ that is to say $$y\sqrt{y^2+1} + \sinh ^{-1}(y)=k \qquad \text{with} \qquad k=\frac{25}9 \pi \left(60+\sqrt{9+625 \pi ^2}\right)+ \sinh ^{-1}\left(\frac{25 \pi }{3}\right)$$ and $k$ is quite large $(k\approx 1213.45)$. So, since $\sinh ^{-1}(y)$ does not vary very fast, you can have a first approximation solving $$y\sqrt{y^2+1}=k \implies y^2(y^2+1)=k^2$$ which is a quadratic in $y^2$. But, since $k$ is large, we can make it $y^4=k^2$ that is to say $y=\sqrt k$. In other words, a quick and dirty estimate is $y\approx 34.8346$.
So, now, consider that you look for the zero of function $$f(y)=y\sqrt{y^2+1} + \sinh ^{-1}(y)-k$$ $$f'(y)=2 \sqrt{y^2+1}$$ and use Newton method starting from the guess given above. The iterates would be $$\left( \begin{array}{cc} n & y_n \\ 0 & 34.834556018882314153 \\ 1 & 34.766492763616163489 \\ 2 & 34.766426194028451786 \\ 3 & 34.766426193964771862 \end{array} \right)$$ which is the solution for twenty significant figures. Now, go back to $x=y-\frac{25 \pi }{3}$ to get the solution for $x$.
Even for small values of $k$, we can have a quite good approximation of the solution of $$y\sqrt{y^2+1} + \sinh ^{-1}(y)-k=0$$ Let $t=\sqrt k$, $y=t(1+\epsilon)$ and expand as a series around $\epsilon=0$. Limited to first order, this should lead to $$\left(t\,\sqrt{t^2+1} +\sinh ^{-1}(t)-t^2\right)+\frac{2 \left(t^3+t\right) \epsilon }{\sqrt{t^2+1}}+O\left(\epsilon ^2\right)=0$$ then $\epsilon$ and $$y_{est}=\frac{t \left(\sqrt{t^2+1}+t\right)-\sinh ^{-1}(t)}{2 \sqrt{t^2+1}}$$
To numerically check its validity, give $y$ a value, compute $k$ then $t$ and then $y_{est}$. The table below shows some results $$\left( \begin{array}{ccc} y & k & y_{est} \\ 1 & 2.29559 & 1.058452457 \\ 2 & 5.91577 & 2.032535418 \\ 3 & 11.3053 & 3.017879758 \\ 4 & 18.5871 & 4.010649880 \\ 5 & 27.8075 & 5.006831348 \\ 10 & 103.497 & 10.00146264 \\ 15 & 228.902 & 15.00055180 \\ 20 & 404.189 & 20.00027069 \\ 25 & 629.412 & 25.00015440 \\ 30 & 904.594 & 30.00009712 \\ 35 & 1229.75 & 35.00006543 \\ 40 & 1604.88 & 40.00004638 \\ 45 & 2030.00 & 45.00003419 \\ 50 & 2505.11 & 50.00002600 \end{array} \right)$$
• We could have a much better approximation using the same approach but, instead of Taylor series, build the simplest [1,1] Padé approximant around $\epsilon=0$. The formula becomes much more complex (but perfectly legible). For the first value of the table, it would give $y_{est}=1.00489$ !!! – Claude Leibovici Jun 22 '18 at 9:31