Optimal. Leaf size=312 \[ -\frac{8 b^2 \sqrt{e} n^2 \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}+\frac{8 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{g \sqrt{e f-d g}}-\frac{16 b^2 \sqrt{e} n^2 \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{g \sqrt{e f-d g}} \]
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Rubi [A] time = 0.765924, antiderivative size = 312, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 12, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2398, 2411, 63, 208, 2348, 12, 1587, 6741, 5984, 5918, 2402, 2315} \[ -\frac{8 b^2 \sqrt{e} n^2 \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}+\frac{8 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{g \sqrt{e f-d g}}-\frac{16 b^2 \sqrt{e} n^2 \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{g \sqrt{e f-d g}} \]
Antiderivative was successfully verified.
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Rule 2398
Rule 2411
Rule 63
Rule 208
Rule 2348
Rule 12
Rule 1587
Rule 6741
Rule 5984
Rule 5918
Rule 2402
Rule 2315
Rubi steps
\begin{align*} \int \frac{\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{3/2}} \, dx &=-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}+\frac{(4 b e n) \int \frac{a+b \log \left (c (d+e x)^n\right )}{(d+e x) \sqrt{f+g x}} \, dx}{g}\\ &=-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}+\frac{(4 b n) \operatorname{Subst}\left (\int \frac{a+b \log \left (c x^n\right )}{x \sqrt{\frac{e f-d g}{e}+\frac{g x}{e}}} \, dx,x,d+e x\right )}{g}\\ &=-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}-\frac{\left (4 b^2 n^2\right ) \operatorname{Subst}\left (\int -\frac{2 \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f-\frac{d g}{e}+\frac{g x}{e}}}{\sqrt{e f-d g}}\right )}{\sqrt{e f-d g} x} \, dx,x,d+e x\right )}{g}\\ &=-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}+\frac{\left (8 b^2 \sqrt{e} n^2\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f-\frac{d g}{e}+\frac{g x}{e}}}{\sqrt{e f-d g}}\right )}{x} \, dx,x,d+e x\right )}{g \sqrt{e f-d g}}\\ &=-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}+\frac{\left (16 b^2 e^{3/2} n^2\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{e f-d g}}\right )}{d g+e \left (-f+x^2\right )} \, dx,x,\sqrt{f+g x}\right )}{g \sqrt{e f-d g}}\\ &=-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}+\frac{\left (16 b^2 e^{3/2} n^2\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{e f-d g}}\right )}{-e f+d g+e x^2} \, dx,x,\sqrt{f+g x}\right )}{g \sqrt{e f-d g}}\\ &=\frac{8 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{g \sqrt{e f-d g}}-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}-\frac{\left (16 b^2 e n^2\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{e f-d g}}\right )}{1-\frac{\sqrt{e} x}{\sqrt{e f-d g}}} \, dx,x,\sqrt{f+g x}\right )}{g (e f-d g)}\\ &=\frac{8 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{g \sqrt{e f-d g}}-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}-\frac{16 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{g \sqrt{e f-d g}}+\frac{\left (16 b^2 e n^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-\frac{\sqrt{e} x}{\sqrt{e f-d g}}}\right )}{1-\frac{e x^2}{e f-d g}} \, dx,x,\sqrt{f+g x}\right )}{g (e f-d g)}\\ &=\frac{8 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{g \sqrt{e f-d g}}-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}-\frac{16 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{g \sqrt{e f-d g}}-\frac{\left (16 b^2 \sqrt{e} n^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{g \sqrt{e f-d g}}\\ &=\frac{8 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )^2}{g \sqrt{e f-d g}}-\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt{e f-d g}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g \sqrt{f+g x}}-\frac{16 b^2 \sqrt{e} n^2 \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) \log \left (\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{g \sqrt{e f-d g}}-\frac{8 b^2 \sqrt{e} n^2 \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}}\right )}{g \sqrt{e f-d g}}\\ \end{align*}
Mathematica [A] time = 0.624339, size = 424, normalized size = 1.36 \[ \frac{2 \left (\frac{b \sqrt{e} n \left (-b n \left (2 \text{PolyLog}\left (2,\frac{1}{2}-\frac{\sqrt{e} \sqrt{f+g x}}{2 \sqrt{e f-d g}}\right )+\log \left (\sqrt{e f-d g}-\sqrt{e} \sqrt{f+g x}\right ) \left (\log \left (\sqrt{e f-d g}-\sqrt{e} \sqrt{f+g x}\right )+2 \log \left (\frac{1}{2} \left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}+1\right )\right )\right )\right )+b n \left (2 \text{PolyLog}\left (2,\frac{1}{2} \left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}+1\right )\right )+\log \left (\sqrt{e f-d g}+\sqrt{e} \sqrt{f+g x}\right ) \left (\log \left (\sqrt{e f-d g}+\sqrt{e} \sqrt{f+g x}\right )+2 \log \left (\frac{1}{2}-\frac{\sqrt{e} \sqrt{f+g x}}{2 \sqrt{e f-d g}}\right )\right )\right )+2 \log \left (\sqrt{e f-d g}-\sqrt{e} \sqrt{f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-2 \log \left (\sqrt{e f-d g}+\sqrt{e} \sqrt{f+g x}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{\sqrt{e f-d g}}-\frac{\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{\sqrt{f+g x}}\right )}{g} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.905, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) \right ) ^{2} \left ( gx+f \right ) ^{-{\frac{3}{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 [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{g x + f} b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 2 \, \sqrt{g x + f} a b \log \left ({\left (e x + d\right )}^{n} c\right ) + \sqrt{g x + f} a^{2}}{g^{2} x^{2} + 2 \, f g x + f^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \log{\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}{\left (f + g x\right )^{\frac{3}{2}}}\, dx \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 ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{{\left (g x + f\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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