Optimal. Leaf size=169 \[ \frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{32 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3} \]
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Rubi [A] time = 0.169124, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {43, 2350, 12, 897, 1261, 208} \[ \frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{32 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3} \]
Antiderivative was successfully verified.
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Rule 43
Rule 2350
Rule 12
Rule 897
Rule 1261
Rule 208
Rubi steps
\begin{align*} \int \frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt{d+e x}} \, dx &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-(b n) \int \frac{2 \sqrt{d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{15 e^3 x} \, dx\\ &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(2 b n) \int \frac{\sqrt{d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{x} \, dx}{15 e^3}\\ &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(4 b n) \operatorname{Subst}\left (\int \frac{x^2 \left (15 d^2-10 d x^2+3 x^4\right )}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{15 e^4}\\ &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(4 b n) \operatorname{Subst}\left (\int \left (8 d^2 e-7 d e x^2+3 e x^4+\frac{8 d^3}{-\frac{d}{e}+\frac{x^2}{e}}\right ) \, dx,x,\sqrt{d+e x}\right )}{15 e^4}\\ &=-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3}+\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{\left (32 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{15 e^4}\\ &=-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3}+\frac{32 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^3}+\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\\ \end{align*}
Mathematica [A] time = 0.175731, size = 118, normalized size = 0.7 \[ \frac{2 \sqrt{d+e x} \left (15 a \left (8 d^2-4 d e x+3 e^2 x^2\right )+15 b \left (8 d^2-4 d e x+3 e^2 x^2\right ) \log \left (c x^n\right )-2 b n \left (94 d^2-17 d e x+9 e^2 x^2\right )\right )+480 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{225 e^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.598, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\frac{1}{\sqrt{ex+d}}}}\, 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 = 1.45459, size = 748, normalized size = 4.43 \begin{align*} \left [\frac{2 \,{\left (120 \, b d^{\frac{5}{2}} n \log \left (\frac{e x + 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) -{\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \,{\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \,{\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{225 \, e^{3}}, -\frac{2 \,{\left (240 \, b \sqrt{-d} d^{2} n \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \,{\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \,{\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{225 \, e^{3}}\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{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{\sqrt{e x + d}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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