Optimal. Leaf size=267 \[ -\frac{i b f^{5/2} m n \text{PolyLog}\left (2,-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{PolyLog}\left (2,\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}+\frac{12 b f^2 m n}{25 e^2 x}+\frac{2 b f^{5/2} m n \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{25 e^{5/2}}-\frac{16 b f m n}{225 e x^3} \]
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Rubi [A] time = 0.189025, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2455, 325, 205, 2376, 4848, 2391} \[ -\frac{i b f^{5/2} m n \text{PolyLog}\left (2,-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{PolyLog}\left (2,\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}+\frac{12 b f^2 m n}{25 e^2 x}+\frac{2 b f^{5/2} m n \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{25 e^{5/2}}-\frac{16 b f m n}{225 e x^3} \]
Antiderivative was successfully verified.
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Rule 2455
Rule 325
Rule 205
Rule 2376
Rule 4848
Rule 2391
Rubi steps
\begin{align*} \int \frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx &=-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-(b n) \int \left (-\frac{2 f m}{15 e x^4}+\frac{2 f^2 m}{5 e^2 x^2}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2} x}-\frac{\log \left (d \left (e+f x^2\right )^m\right )}{5 x^6}\right ) \, dx\\ &=-\frac{2 b f m n}{45 e x^3}+\frac{2 b f^2 m n}{5 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{1}{5} (b n) \int \frac{\log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx-\frac{\left (2 b f^{5/2} m n\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{x} \, dx}{5 e^{5/2}}\\ &=-\frac{2 b f m n}{45 e x^3}+\frac{2 b f^2 m n}{5 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{1}{25} (2 b f m n) \int \frac{1}{x^4 \left (e+f x^2\right )} \, dx-\frac{\left (i b f^{5/2} m n\right ) \int \frac{\log \left (1-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{x} \, dx}{5 e^{5/2}}+\frac{\left (i b f^{5/2} m n\right ) \int \frac{\log \left (1+\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{x} \, dx}{5 e^{5/2}}\\ &=-\frac{16 b f m n}{225 e x^3}+\frac{2 b f^2 m n}{5 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac{i b f^{5/2} m n \text{Li}_2\left (-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{Li}_2\left (\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}-\frac{\left (2 b f^2 m n\right ) \int \frac{1}{x^2 \left (e+f x^2\right )} \, dx}{25 e}\\ &=-\frac{16 b f m n}{225 e x^3}+\frac{12 b f^2 m n}{25 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac{i b f^{5/2} m n \text{Li}_2\left (-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{Li}_2\left (\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{\left (2 b f^3 m n\right ) \int \frac{1}{e+f x^2} \, dx}{25 e^2}\\ &=-\frac{16 b f m n}{225 e x^3}+\frac{12 b f^2 m n}{25 e^2 x}+\frac{2 b f^{5/2} m n \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{25 e^{5/2}}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac{i b f^{5/2} m n \text{Li}_2\left (-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{Li}_2\left (\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.164178, size = 399, normalized size = 1.49 \[ -\frac{45 i b f^{5/2} m n x^5 \text{PolyLog}\left (2,-\frac{i \sqrt{f} x}{\sqrt{e}}\right )-45 i b f^{5/2} m n x^5 \text{PolyLog}\left (2,\frac{i \sqrt{f} x}{\sqrt{e}}\right )+45 a e^{5/2} \log \left (d \left (e+f x^2\right )^m\right )+30 a e^{3/2} f m x^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{f x^2}{e}\right )+45 b e^{5/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+30 b e^{3/2} f m x^2 \log \left (c x^n\right )-90 b \sqrt{e} f^2 m x^4 \log \left (c x^n\right )-90 b f^{5/2} m x^5 \log \left (c x^n\right ) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )+9 b e^{5/2} n \log \left (d \left (e+f x^2\right )^m\right )+16 b e^{3/2} f m n x^2-108 b \sqrt{e} f^2 m n x^4-45 i b f^{5/2} m n x^5 \log (x) \log \left (1-\frac{i \sqrt{f} x}{\sqrt{e}}\right )+45 i b f^{5/2} m n x^5 \log (x) \log \left (1+\frac{i \sqrt{f} x}{\sqrt{e}}\right )-18 b f^{5/2} m n x^5 \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )+90 b f^{5/2} m n x^5 \log (x) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{225 e^{5/2} x^5} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.194, size = 2385, normalized size = 8.9 \begin{align*} \text{result too large to display} \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{{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}}, 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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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