Optimal. Leaf size=297 \[ -\frac{14 d^{9/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{135 e^{9/4} \sqrt{d+e x^2}}-\frac{28 d^2 \sqrt{x} \sqrt{d+e x^2}}{135 e^2 \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{28 d^{9/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{135 e^{9/4} \sqrt{d+e x^2}}+\frac{28 d x^{3/2} \sqrt{d+e x^2}}{405 e^{3/2}}-\frac{4 x^{7/2} \sqrt{d+e x^2}}{81 \sqrt{e}}+\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]
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Rubi [A] time = 0.187218, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{28 d^2 \sqrt{x} \sqrt{d+e x^2}}{135 e^2 \left (\sqrt{d}+\sqrt{e} x\right )}-\frac{14 d^{9/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{135 e^{9/4} \sqrt{d+e x^2}}+\frac{28 d^{9/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{135 e^{9/4} \sqrt{d+e x^2}}+\frac{28 d x^{3/2} \sqrt{d+e x^2}}{405 e^{3/2}}-\frac{4 x^{7/2} \sqrt{d+e x^2}}{81 \sqrt{e}}+\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]
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^{7/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \, dx &=\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{9} \left (2 \sqrt{e}\right ) \int \frac{x^{9/2}}{\sqrt{d+e x^2}} \, dx\\ &=-\frac{4 x^{7/2} \sqrt{d+e x^2}}{81 \sqrt{e}}+\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{(14 d) \int \frac{x^{5/2}}{\sqrt{d+e x^2}} \, dx}{81 \sqrt{e}}\\ &=\frac{28 d x^{3/2} \sqrt{d+e x^2}}{405 e^{3/2}}-\frac{4 x^{7/2} \sqrt{d+e x^2}}{81 \sqrt{e}}+\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (14 d^2\right ) \int \frac{\sqrt{x}}{\sqrt{d+e x^2}} \, dx}{135 e^{3/2}}\\ &=\frac{28 d x^{3/2} \sqrt{d+e x^2}}{405 e^{3/2}}-\frac{4 x^{7/2} \sqrt{d+e x^2}}{81 \sqrt{e}}+\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (28 d^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{135 e^{3/2}}\\ &=\frac{28 d x^{3/2} \sqrt{d+e x^2}}{405 e^{3/2}}-\frac{4 x^{7/2} \sqrt{d+e x^2}}{81 \sqrt{e}}+\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (28 d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{135 e^2}+\frac{\left (28 d^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{e} x^2}{\sqrt{d}}}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{135 e^2}\\ &=\frac{28 d x^{3/2} \sqrt{d+e x^2}}{405 e^{3/2}}-\frac{4 x^{7/2} \sqrt{d+e x^2}}{81 \sqrt{e}}-\frac{28 d^2 \sqrt{x} \sqrt{d+e x^2}}{135 e^2 \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{2}{9} x^{9/2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{28 d^{9/4} \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 )}{135 e^{9/4} \sqrt{d+e x^2}}-\frac{14 d^{9/4} \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 )}{135 e^{9/4} \sqrt{d+e x^2}}\\ \end{align*}
Mathematica [C] time = 0.105563, size = 124, normalized size = 0.42 \[ \frac{2 x^{3/2} \left (-14 d^2 \sqrt{\frac{e x^2}{d}+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-\frac{e x^2}{d}\right )+14 d^2+45 e^{3/2} x^3 \sqrt{d+e x^2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+4 d e x^2-10 e^2 x^4\right )}{405 e^{3/2} \sqrt{d+e x^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.936, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{7}{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}{9} \, x^{\frac{9}{2}} \log \left (\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - \frac{1}{9} \, x^{\frac{9}{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{7}{2} \, \log \left (x\right )\right )}}{9 \,{\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{7}{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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 4 \, d e^{\frac{1}{2}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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