Optimal. Leaf size=302 \[ -\frac{6 e^{7/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right ),\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{12 e^2 \sqrt{x} \sqrt{d+e x^2}}{35 d^2 \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{12 e^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}+\frac{12 e^{7/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{4 \sqrt{e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}} \]
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Rubi [A] time = 0.181199, antiderivative size = 302, 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, 325, 329, 305, 220, 1196} \[ -\frac{12 e^2 \sqrt{x} \sqrt{d+e x^2}}{35 d^2 \left (\sqrt{d}+\sqrt{e} x\right )}+\frac{12 e^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{6 e^{7/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}+\frac{12 e^{7/4} \left (\sqrt{d}+\sqrt{e} x\right ) \sqrt{\frac{d+e x^2}{\left (\sqrt{d}+\sqrt{e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{e} \sqrt{x}}{\sqrt [4]{d}}\right )|\frac{1}{2}\right )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{4 \sqrt{e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}} \]
Antiderivative was successfully verified.
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Rule 6221
Rule 325
Rule 329
Rule 305
Rule 220
Rule 1196
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{x^{9/2}} \, dx &=-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}+\frac{1}{7} \left (2 \sqrt{e}\right ) \int \frac{1}{x^{7/2} \sqrt{d+e x^2}} \, dx\\ &=-\frac{4 \sqrt{e} \sqrt{d+e x^2}}{35 d x^{5/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}-\frac{\left (6 e^{3/2}\right ) \int \frac{1}{x^{3/2} \sqrt{d+e x^2}} \, dx}{35 d}\\ &=-\frac{4 \sqrt{e} \sqrt{d+e x^2}}{35 d x^{5/2}}+\frac{12 e^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}-\frac{\left (6 e^{5/2}\right ) \int \frac{\sqrt{x}}{\sqrt{d+e x^2}} \, dx}{35 d^2}\\ &=-\frac{4 \sqrt{e} \sqrt{d+e x^2}}{35 d x^{5/2}}+\frac{12 e^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}-\frac{\left (12 e^{5/2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{35 d^2}\\ &=-\frac{4 \sqrt{e} \sqrt{d+e x^2}}{35 d x^{5/2}}+\frac{12 e^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}-\frac{\left (12 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+e x^4}} \, dx,x,\sqrt{x}\right )}{35 d^{3/2}}+\frac{\left (12 e^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 )}{35 d^{3/2}}\\ &=-\frac{4 \sqrt{e} \sqrt{d+e x^2}}{35 d x^{5/2}}+\frac{12 e^{3/2} \sqrt{d+e x^2}}{35 d^2 \sqrt{x}}-\frac{12 e^2 \sqrt{x} \sqrt{d+e x^2}}{35 d^2 \left (\sqrt{d}+\sqrt{e} x\right )}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{7 x^{7/2}}+\frac{12 e^{7/4} \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 )}{35 d^{7/4} \sqrt{d+e x^2}}-\frac{6 e^{7/4} \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 )}{35 d^{7/4} \sqrt{d+e x^2}}\\ \end{align*}
Mathematica [C] time = 0.0995408, size = 131, normalized size = 0.43 \[ \frac{-4 e^{5/2} x^5 \sqrt{\frac{e x^2}{d}+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-\frac{e x^2}{d}\right )+4 \sqrt{e} x \left (-d^2+2 d e x^2+3 e^2 x^4\right )-10 d^2 \sqrt{d+e x^2} \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{35 d^2 x^{7/2} \sqrt{d+e x^2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.843, size = 0, normalized size = 0. \begin{align*} \int{{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ){x}^{-{\frac{9}{2}}}}\, 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*} 2 \, d \sqrt{e} \int -\frac{\sqrt{e x^{2} + d} x}{7 \,{\left ({\left (e^{2} x^{4} + d e x^{2}\right )} x^{\frac{9}{2}} -{\left (e x^{2} + d\right )} e^{\left (\log \left (e x^{2} + d\right ) + \frac{9}{2} \, \log \left (x\right )\right )}\right )}}\,{d x} - \frac{\log \left (\sqrt{e} x + \sqrt{e x^{2} + d}\right )}{7 \, x^{\frac{7}{2}}} + \frac{\log \left (-\sqrt{e} x + \sqrt{e x^{2} + d}\right )}{7 \, x^{\frac{7}{2}}} \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 (\frac{\operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right )}{x^{\frac{9}{2}}}, 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*} \int \frac{\operatorname{artanh}\left (\frac{\sqrt{e} x}{\sqrt{e x^{2} + d}}\right )}{x^{\frac{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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