Optimal. Leaf size=119 \[ -\frac{3 (-e)^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{3 (-e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{40 d^{5/2}}-\frac{\sqrt{-e} \sqrt{d+e x^2}}{20 d x^4}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5} \]
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Rubi [A] time = 0.0599371, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5151, 266, 51, 63, 208} \[ -\frac{3 (-e)^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{3 (-e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{40 d^{5/2}}-\frac{\sqrt{-e} \sqrt{d+e x^2}}{20 d x^4}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5} \]
Antiderivative was successfully verified.
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Rule 5151
Rule 266
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{x^6} \, dx &=-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{1}{5} \sqrt{-e} \int \frac{1}{x^5 \sqrt{d+e x^2}} \, dx\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{1}{10} \sqrt{-e} \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{d+e x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{-e} \sqrt{d+e x^2}}{20 d x^4}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{\left (3 (-e)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{d+e x}} \, dx,x,x^2\right )}{40 d}\\ &=-\frac{\sqrt{-e} \sqrt{d+e x^2}}{20 d x^4}-\frac{3 (-e)^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}+\frac{\left (3 (-e)^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{80 d^2}\\ &=-\frac{\sqrt{-e} \sqrt{d+e x^2}}{20 d x^4}-\frac{3 (-e)^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}-\frac{\left (3 (-e)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{40 d^2}\\ &=-\frac{\sqrt{-e} \sqrt{d+e x^2}}{20 d x^4}-\frac{3 (-e)^{3/2} \sqrt{d+e x^2}}{40 d^2 x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5}-\frac{3 (-e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{40 d^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.141426, size = 114, normalized size = 0.96 \[ -\frac{3 e^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{-e}}{\sqrt{e} \sqrt{d+e x^2}}\right )}{40 d^{5/2}}+\sqrt{-e} \left (\frac{3 e}{40 d^2 x^2}-\frac{1}{20 d x^4}\right ) \sqrt{d+e x^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{-e} x}{\sqrt{d+e x^2}}\right )}{5 x^5} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.04, size = 150, normalized size = 1.3 \begin{align*} -{\frac{1}{5\,{x}^{5}}\arctan \left ({x\sqrt{-e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{e}{10\,{d}^{2}{x}^{2}}\sqrt{-e}\sqrt{e{x}^{2}+d}}-{\frac{3\,{e}^{2}}{40}\sqrt{-e}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ){d}^{-{\frac{5}{2}}}}-{\frac{1}{20\,{d}^{2}{x}^{4}}\sqrt{-e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{e}{40\,{d}^{3}{x}^{2}}\sqrt{-e} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{2}}{40\,{d}^{3}}\sqrt{-e}\sqrt{e{x}^{2}+d}} \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{-d \sqrt{-e} x^{5} \int \frac{\sqrt{e x^{2} + d}}{e^{2} x^{9} + d e x^{7} -{\left (e x^{7} + d x^{5}\right )}{\left (e x^{2} + d\right )}}\,{d x} - \arctan \left (\sqrt{-e} x, \sqrt{e x^{2} + d}\right )}{5 \, x^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.84853, size = 529, normalized size = 4.45 \begin{align*} \left [\frac{3 \, e^{2} x^{5} \sqrt{-\frac{e}{d}} \log \left (-\frac{e^{2} x^{2} + 2 \, \sqrt{e x^{2} + d} d \sqrt{-e} \sqrt{-\frac{e}{d}} + 2 \, d e}{x^{2}}\right ) - 16 \, d^{2} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) + 2 \,{\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt{e x^{2} + d} \sqrt{-e}}{80 \, d^{2} x^{5}}, -\frac{3 \, e^{2} x^{5} \sqrt{\frac{e}{d}} \arctan \left (\frac{\sqrt{e x^{2} + d} d \sqrt{-e} \sqrt{\frac{e}{d}}}{e^{2} x^{2} + d e}\right ) + 8 \, d^{2} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) -{\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt{e x^{2} + d} \sqrt{-e}}{40 \, d^{2} x^{5}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 17.1436, size = 148, normalized size = 1.24 \begin{align*} - \frac{\operatorname{atan}{\left (\frac{x \sqrt{- e}}{\sqrt{d + e x^{2}}} \right )}}{5 x^{5}} - \frac{\sqrt{- e}}{20 \sqrt{e} x^{5} \sqrt{\frac{d}{e x^{2}} + 1}} + \frac{\sqrt{e} \sqrt{- e}}{40 d x^{3} \sqrt{\frac{d}{e x^{2}} + 1}} + \frac{3 e^{\frac{3}{2}} \sqrt{- e}}{40 d^{2} x \sqrt{\frac{d}{e x^{2}} + 1}} - \frac{3 e^{2} \sqrt{- e} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{e} x} \right )}}{40 d^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.18388, size = 143, normalized size = 1.2 \begin{align*} -\frac{1}{40} \,{\left (\frac{3 \, \arctan \left (\frac{\sqrt{-x^{2} e^{2} - d e} e^{\left (-\frac{1}{2}\right )}}{\sqrt{d}}\right ) e^{\left (-\frac{5}{2}\right )}}{d^{\frac{5}{2}}} + \frac{{\left (5 \, \sqrt{-x^{2} e^{2} - d e} d e + 3 \,{\left (-x^{2} e^{2} - d e\right )}^{\frac{3}{2}}\right )} e^{\left (-6\right )}}{d^{2} x^{4}}\right )} e^{5} - \frac{\arctan \left (\frac{x \sqrt{-e}}{\sqrt{x^{2} e + d}}\right )}{5 \, x^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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