Optimal. Leaf size=125 \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}+\frac{b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e} \]
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Rubi [A] time = 0.106275, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2338, 266, 50, 63, 208} \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}+\frac{b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e} \]
Antiderivative was successfully verified.
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Rule 2338
Rule 266
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{(b n) \int \frac{\left (d+e x^2\right )^{5/2}}{x} \, dx}{5 e}\\ &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{(b n) \operatorname{Subst}\left (\int \frac{(d+e x)^{5/2}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{(b d n) \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{\left (b d^2 n\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{\left (b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d+e x}} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac{\left (b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{5 e^2}\\ &=-\frac{b d^2 n \sqrt{d+e x^2}}{5 e}-\frac{b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac{b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{5 e}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}\\ \end{align*}
Mathematica [A] time = 0.138715, size = 181, normalized size = 1.45 \[ \sqrt{d+e x^2} \left (\frac{d^2 \left (15 a+15 b \left (\log \left (c x^n\right )-n \log (x)\right )-23 b n\right )}{75 e}+\frac{1}{75} d x^2 \left (30 a+30 b \left (\log \left (c x^n\right )-n \log (x)\right )-11 b n\right )+\frac{1}{25} e x^4 \left (5 a+5 b \left (\log \left (c x^n\right )-n \log (x)\right )-b n\right )\right )+\frac{b d^{5/2} n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )}{5 e}-\frac{b d^{5/2} n \log (x)}{5 e}+\frac{b n \log (x) \left (d+e x^2\right )^{5/2}}{5 e} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.459, size = 0, normalized size = 0. \begin{align*} \int x \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \, 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 = 1.58339, size = 734, normalized size = 5.87 \begin{align*} \left [\frac{15 \, b d^{\frac{5}{2}} n \log \left (-\frac{e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) - 2 \,{\left (3 \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 23 \, b d^{2} n - 15 \, a d^{2} +{\left (11 \, b d e n - 30 \, a d e\right )} x^{2} - 15 \,{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{150 \, e}, -\frac{15 \, b \sqrt{-d} d^{2} n \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) +{\left (3 \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 23 \, b d^{2} n - 15 \, a d^{2} +{\left (11 \, b d e n - 30 \, a d e\right )} x^{2} - 15 \,{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{75 \, e}\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{\left (e x^{2} + d\right )}^{\frac{3}{2}}{\left (b \log \left (c x^{n}\right ) + a\right )} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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