Optimal. Leaf size=269 \[ \frac{6 d^{5/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{25 e^{5/4} \sqrt{d+e x^2}}-\frac{12 d^{5/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{5/4} \sqrt{d+e x^2}}-\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{e}}+\frac{12 d \sqrt{x} \sqrt{d+e x^2}}{25 e \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{2}{5} x^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]
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Rubi [A] time = 0.157495, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {6221, 321, 329, 305, 220, 1196} \[ \frac{6 d^{5/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{5/4} \sqrt{d+e x^2}}-\frac{12 d^{5/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{5/4} \sqrt{d+e x^2}}-\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{e}}+\frac{12 d \sqrt{x} \sqrt{d+e x^2}}{25 e \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{2}{5} x^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]
Antiderivative was successfully verified.
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Rule 6221
Rule 321
Rule 329
Rule 305
Rule 220
Rule 1196
Rubi steps
\begin{align*} \int x^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \, dx &=\frac{2}{5} x^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{5} \left (2 \sqrt{e}\right ) \int \frac{x^{5/2}}{\sqrt{d+e x^2}} \, dx\\ &=-\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{e}}+\frac{2}{5} x^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{(6 d) \int \frac{\sqrt{x}}{\sqrt{d+e x^2}} \, dx}{25 \sqrt{e}}\\ &=-\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{e}}+\frac{2}{5} x^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{(12 d) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{25 \sqrt{e}}\\ &=-\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{e}}+\frac{2}{5} x^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{\left (12 d^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{25 e}-\frac{\left (12 d^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{e} x^2}{\sqrt{d}}}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{25 e}\\ &=-\frac{4 x^{3/2} \sqrt{d+e x^2}}{25 \sqrt{e}}+\frac{12 d \sqrt{x} \sqrt{d+e x^2}}{25 e \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{2}{5} x^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{12 d^{5/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{5/4} \sqrt{d+e x^2}}+\frac{6 d^{5/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{25 e^{5/4} \sqrt{d+e x^2}}\\ \end{align*}
Mathematica [C] time = 0.0904489, size = 109, normalized size = 0.41 \[ \frac{2 x^{3/2} \left (2 d \sqrt{\frac{e x^2}{d}+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-\frac{e x^2}{d}\right )-2 \left (d+e x^2\right )+5 \sqrt{e} x \sqrt{d+e x^2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )\right )}{25 \sqrt{e} \sqrt{d+e x^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.848, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{3}{2}}}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{5} \, x^{\frac{5}{2}} \log \left (\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - \frac{1}{5} \, x^{\frac{5}{2}} \log \left (-\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - 2 \, d \sqrt{e} \int -\frac{x e^{\left (\frac{1}{2} \, \log \left (e x^{2} + d\right ) + \frac{3}{2} \, \log \left (x\right )\right )}}{5 \,{\left (e^{2} x^{4} + d e x^{2} -{\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{\frac{3}{2}} \operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{3}{2}} \operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.29049, size = 1, normalized size = 0. \begin{align*} +\infty \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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