Optimal. Leaf size=248 \[ -\frac{1}{10} b c^5 e \text{PolyLog}\left (2,\frac{1}{c^2 x^2+1}\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{5 x^5}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{1}{10} b c^5 \log \left (1-\frac{1}{c^2 x^2+1}\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac{b c^3 \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 x^2}-\frac{b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 x^4}-\frac{7 b c^3 e}{60 x^2}+\frac{19}{60} b c^5 e \log \left (c^2 x^2+1\right )-\frac{5}{6} b c^5 e \log (x) \]
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Rubi [A] time = 0.625143, antiderivative size = 245, normalized size of antiderivative = 0.99, number of steps used = 26, number of rules used = 18, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {5017, 2475, 2411, 2347, 2344, 2301, 2316, 2315, 2314, 31, 2319, 44, 4918, 4852, 266, 36, 29, 4884} \[ -\frac{1}{10} b c^5 e \text{PolyLog}\left (2,-c^2 x^2\right )-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{5 x^5}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}-\frac{b c^5 \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{20 e}+\frac{b c^3 \left (c^2 x^2+1\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{10 x^2}-\frac{b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{20 x^4}+\frac{1}{5} b c^5 d \log (x)-\frac{7 b c^3 e}{60 x^2}+\frac{19}{60} b c^5 e \log \left (c^2 x^2+1\right )-\frac{5}{6} b c^5 e \log (x) \]
Antiderivative was successfully verified.
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Rule 5017
Rule 2475
Rule 2411
Rule 2347
Rule 2344
Rule 2301
Rule 2316
Rule 2315
Rule 2314
Rule 31
Rule 2319
Rule 44
Rule 4918
Rule 4852
Rule 266
Rule 36
Rule 29
Rule 4884
Rubi steps
\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^6} \, dx &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{1}{5} (b c) \int \frac{d+e \log \left (1+c^2 x^2\right )}{x^5 \left (1+c^2 x^2\right )} \, dx+\frac{1}{5} \left (2 c^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{d+e \log \left (1+c^2 x\right )}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{5} \left (2 c^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^4} \, dx-\frac{1}{5} \left (2 c^4 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^3} \, dx,x,1+c^2 x^2\right )}{10 c}+\frac{1}{15} \left (2 b c^3 e\right ) \int \frac{1}{x^3 \left (1+c^2 x^2\right )} \, dx-\frac{1}{5} \left (2 c^4 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx+\frac{1}{5} \left (2 c^6 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{\left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^3} \, dx,x,1+c^2 x^2\right )}{10 c}-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )+\frac{1}{15} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{5} \left (2 b c^5 e\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{\left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )+\frac{1}{10} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )} \, dx,x,1+c^2 x^2\right )+\frac{1}{20} (b c e) \operatorname{Subst}\left (\int \frac{1}{x \left (-\frac{1}{c^2}+\frac{x}{c^2}\right )^2} \, dx,x,1+c^2 x^2\right )+\frac{1}{15} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}-\frac{c^2}{x}+\frac{c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )-\frac{1}{5} \left (b c^5 e\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{b c^3 e}{15 x^2}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}-\frac{2}{15} b c^5 e \log (x)+\frac{1}{15} b c^5 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}+\frac{b c^3 \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}+\frac{1}{10} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )-\frac{1}{10} \left (b c^5\right ) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x} \, dx,x,1+c^2 x^2\right )+\frac{1}{20} (b c e) \operatorname{Subst}\left (\int \left (\frac{c^4}{(-1+x)^2}-\frac{c^4}{-1+x}+\frac{c^4}{x}\right ) \, dx,x,1+c^2 x^2\right )-\frac{1}{10} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )-\frac{1}{5} \left (b c^5 e\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{5} \left (b c^7 e\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{7 b c^3 e}{60 x^2}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{1}{5} b c^5 d \log (x)-\frac{5}{6} b c^5 e \log (x)+\frac{19}{60} b c^5 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}+\frac{b c^3 \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}-\frac{b c^5 \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{20 e}+\frac{1}{10} \left (b c^3 e\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{-\frac{1}{c^2}+\frac{x}{c^2}} \, dx,x,1+c^2 x^2\right )\\ &=-\frac{7 b c^3 e}{60 x^2}-\frac{2 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{15 x^3}+\frac{2 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{5 b}+\frac{1}{5} b c^5 d \log (x)-\frac{5}{6} b c^5 e \log (x)+\frac{19}{60} b c^5 e \log \left (1+c^2 x^2\right )-\frac{b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{20 x^4}+\frac{b c^3 \left (1+c^2 x^2\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{10 x^2}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \left (d+e \log \left (1+c^2 x^2\right )\right )}{5 x^5}-\frac{b c^5 \left (d+e \log \left (1+c^2 x^2\right )\right )^2}{20 e}-\frac{1}{10} b c^5 e \text{Li}_2\left (-c^2 x^2\right )\\ \end{align*}
Mathematica [A] time = 0.246785, size = 259, normalized size = 1.04 \[ \frac{1}{60} \left (6 b c^5 e \text{PolyLog}\left (2,c^2 x^2+1\right )-\frac{12 \left (a+b \tan ^{-1}(c x)\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^5}-\frac{8 c^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac{12 c^5 e \left (a+b \tan ^{-1}(c x)\right )^2}{b}+\frac{24 c^4 e \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{3 b c^5 \left (e \log \left (c^2 x^2+1\right )+d\right )^2}{e}+6 b c^5 \log \left (-c^2 x^2\right ) \left (e \log \left (c^2 x^2+1\right )+d\right )+\frac{6 b c^3 \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^2}-\frac{3 b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{x^4}-18 b c^5 e \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )+7 b c^3 e \left (c^2 \log \left (c^2 x^2+1\right )-2 c^2 \log (x)-\frac{1}{x^2}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 14.653, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( cx \right ) \right ) \left ( d+e\ln \left ({c}^{2}{x}^{2}+1 \right ) \right ) }{{x}^{6}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{20} \,{\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac{2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac{4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d + \frac{1}{15} \,{\left (2 \,{\left (3 \, c^{3} \arctan \left (c x\right ) + \frac{3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c^{2} - \frac{3 \, \log \left (c^{2} x^{2} + 1\right )}{x^{5}}\right )} a e + b e \int \frac{\arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{x^{6}}\,{d x} - \frac{a d}{5 \, x^{5}} \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{b d \arctan \left (c x\right ) + a d +{\left (b e \arctan \left (c x\right ) + a e\right )} \log \left (c^{2} x^{2} + 1\right )}{x^{6}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 87.3639, size = 474, normalized size = 1.91 \begin{align*} \frac{2 a c^{4} e \operatorname{atan}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{5 \sqrt{\frac{1}{c^{2}}}} + \frac{2 a c^{4} e}{5 x} - \frac{2 a c^{2} e}{15 x^{3}} - \frac{a d}{5 x^{5}} - \frac{a e \log{\left (c^{2} x^{2} + 1 \right )}}{5 x^{5}} + 4 b c^{9} e \left (\begin{cases} \frac{x^{2}}{40 c^{2}} - \frac{\log{\left (c^{2} x^{2} + 1 \right )}}{40 c^{4}} & \text{for}\: c = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}^{2}}{80 c^{4}} & \text{otherwise} \end{cases}\right ) - \frac{b c^{7} d \left (\begin{cases} x^{2} & \text{for}\: c^{2} = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}}{c^{2}} & \text{otherwise} \end{cases}\right )}{10} - \frac{b c^{7} e \left (\begin{cases} x^{2} & \text{for}\: c^{2} = 0 \\\frac{\log{\left (c^{2} x^{2} + 1 \right )}}{c^{2}} & \text{otherwise} \end{cases}\right ) \log{\left (c^{2} x^{2} + 1 \right )}}{10} + \frac{b c^{5} d \log{\left (x^{2} \right )}}{10} - \frac{5 b c^{5} e \log{\left (x \right )}}{6} + \frac{5 b c^{5} e \log{\left (c^{2} x^{2} + 1 \right )}}{12} - \frac{b c^{5} e \operatorname{atan}^{2}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{5} - \frac{b c^{5} e \operatorname{Li}_{2}\left (c^{2} x^{2} e^{i \pi }\right )}{10} + \frac{2 b c^{4} e \operatorname{atan}{\left (c x \right )} \operatorname{atan}{\left (\frac{x}{\sqrt{\frac{1}{c^{2}}}} \right )}}{5 \sqrt{\frac{1}{c^{2}}}} + \frac{2 b c^{4} e \operatorname{atan}{\left (c x \right )}}{5 x} + \frac{b c^{3} d}{10 x^{2}} + \frac{b c^{3} e \log{\left (c^{2} x^{2} + 1 \right )}}{10 x^{2}} - \frac{7 b c^{3} e}{60 x^{2}} - \frac{2 b c^{2} e \operatorname{atan}{\left (c x \right )}}{15 x^{3}} - \frac{b c d}{20 x^{4}} - \frac{b c e \log{\left (c^{2} x^{2} + 1 \right )}}{20 x^{4}} - \frac{b d \operatorname{atan}{\left (c x \right )}}{5 x^{5}} - \frac{b e \log{\left (c^{2} x^{2} + 1 \right )} \operatorname{atan}{\left (c x \right )}}{5 x^{5}} \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 \arctan \left (c x\right ) + a\right )}{\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{6}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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