I hope I can find closed form solution for two following definite integrals. Unfortunately I don't have Mathematica and I can't find similar integrals in Tables. Can any one help me please?

\begin{equation} \int_0^\infty \frac{e^{-x}}{x^{1/2}+a x^{3/2}} dx \end{equation}

\begin{equation} \int_0^\infty \frac{xe^{-x}}{x^{1/2}(1+bx)^2} dx \end{equation}


Here is a solution for the first integral \begin{align} \tag{a} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x &= 2\int\limits_{0}^{\infty} \frac{z\mathrm{e}^{-z^{2}}}{z+az^{3}} \mathrm{d}z \\ \tag{b} &= \frac{2}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}/a} \frac{1}{1+y^{2}} \mathrm{d}y \\ \tag{c} &= \frac{\pi}{\sqrt{a}} \mathrm{e}^{1/a} \mathrm{erfc}\left(\frac{1}{\sqrt{a}}\right) \end{align}


a. $x=z^{2}$

b. $y^{2}=az^{2}$

c. From DLMF, we have the following integral definition of the complementary error function \begin{equation} \mathrm{erfc}(z) = \frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-z^{2}t^{2}} \frac{1}{t^{2} + 1} \mathrm{d}t \end{equation}


For a proof of the above equation, see Show $\frac{2}{\pi} \mathrm{exp}(-z^{2}) \int_{0}^{\infty} \mathrm{exp}(-z^{2}x^{2}) \frac{1}{x^{2}+1} \mathrm{d}x = \mathrm{erfc}(z)$


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