Optimal. Leaf size=115 \[ b \text{Unintegrable}\left (\frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}},x\right )-\frac{5 a e}{2 d^3 \sqrt{d+e x^2}}-\frac{5 a e}{6 d^2 \left (d+e x^2\right )^{3/2}}+\frac{5 a e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 d^{7/2}}-\frac{a}{2 d x^2 \left (d+e x^2\right )^{3/2}} \]
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Rubi [A] time = 0.208684, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{a+b \tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin{align*} \int \frac{a+b \tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx &=a \int \frac{1}{x^3 \left (d+e x^2\right )^{5/2}} \, dx+b \int \frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx\\ &=\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x^2 (d+e x)^{5/2}} \, dx,x,x^2\right )+b \int \frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx\\ &=\frac{a}{3 d x^2 \left (d+e x^2\right )^{3/2}}+b \int \frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx+\frac{(5 a) \operatorname{Subst}\left (\int \frac{1}{x^2 (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 d}\\ &=\frac{a}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 a}{3 d^2 x^2 \sqrt{d+e x^2}}+b \int \frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx+\frac{(5 a) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{d+e x}} \, dx,x,x^2\right )}{2 d^2}\\ &=\frac{a}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 a}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 a \sqrt{d+e x^2}}{2 d^3 x^2}+b \int \frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx-\frac{(5 a e) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{4 d^3}\\ &=\frac{a}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 a}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 a \sqrt{d+e x^2}}{2 d^3 x^2}+b \int \frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx-\frac{(5 a) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{2 d^3}\\ &=\frac{a}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 a}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 a \sqrt{d+e x^2}}{2 d^3 x^2}+\frac{5 a e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 d^{7/2}}+b \int \frac{\tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx\\ \end{align*}
Mathematica [A] time = 53.7482, size = 0, normalized size = 0. \[ \int \frac{a+b \tan ^{-1}(c x)}{x^3 \left (d+e x^2\right )^{5/2}} \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.595, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\arctan \left ( cx \right ) }{{x}^{3}} \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e x^{2} + d}{\left (b \arctan \left (c x\right ) + a\right )}}{e^{3} x^{9} + 3 \, d e^{2} x^{7} + 3 \, d^{2} e x^{5} + d^{3} x^{3}}, 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 [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arctan \left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac{5}{2}} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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