Optimal. Leaf size=629 \[ \frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{PolyLog}(3,c (a+b x))}{2 a^2}-\frac{b^2 \text{PolyLog}(2,c (a+b x))}{2 a^2}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{2 a^2}-\frac{b^2 \text{PolyLog}\left (3,-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \text{PolyLog}\left (3,-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac{b^2 \text{PolyLog}(3,1-c (a+b x))}{2 a^2}-\frac{b^2 \text{PolyLog}\left (2,-\frac{b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right )}{2 a^2}-\frac{b^2 \log (x) \text{PolyLog}(2,c (a+b x))}{2 a^2}-\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{PolyLog}\left (2,-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{PolyLog}\left (2,-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac{b^2 \left (\log \left (-\frac{a (1-c (a+b x))}{b x}\right )+\log (x)\right ) \text{PolyLog}(2,1-c (a+b x))}{2 a^2}+\frac{b^2 \text{PolyLog}\left (3,-\frac{b x}{a}\right )}{2 a^2}-\frac{b \text{PolyLog}(2,c (a+b x))}{2 a x}-\frac{b^2 \left (\log \left (\frac{1-a c}{1-c (a+b x)}\right )-\log \left (\frac{(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac{b x}{a}+1\right )\right ) \log ^2\left (-\frac{a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac{b^2 \left (\log (c (a+b x))-\log \left (\frac{b x}{a}+1\right )\right ) \left (\log \left (-\frac{a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2}{4 a^2}-\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}-\frac{b^2 \log (x) \log \left (\frac{b x}{a}+1\right ) \log (1-c (a+b x))}{2 a^2} \]
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Rubi [A] time = 0.663908, antiderivative size = 629, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 13, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6599, 6598, 36, 29, 31, 2416, 2394, 2315, 2393, 2391, 6597, 2440, 2435} \[ \frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{PolyLog}(3,c (a+b x))}{2 a^2}-\frac{b^2 \text{PolyLog}(2,c (a+b x))}{2 a^2}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{2 a^2}-\frac{b^2 \text{PolyLog}\left (3,-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \text{PolyLog}\left (3,-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac{b^2 \text{PolyLog}(3,1-c (a+b x))}{2 a^2}-\frac{b^2 \text{PolyLog}\left (2,-\frac{b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right )}{2 a^2}-\frac{b^2 \log (x) \text{PolyLog}(2,c (a+b x))}{2 a^2}-\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{PolyLog}\left (2,-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{PolyLog}\left (2,-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac{b^2 \left (\log \left (-\frac{a (1-c (a+b x))}{b x}\right )+\log (x)\right ) \text{PolyLog}(2,1-c (a+b x))}{2 a^2}+\frac{b^2 \text{PolyLog}\left (3,-\frac{b x}{a}\right )}{2 a^2}-\frac{b \text{PolyLog}(2,c (a+b x))}{2 a x}-\frac{b^2 \left (\log \left (\frac{1-a c}{1-c (a+b x)}\right )-\log \left (\frac{(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac{b x}{a}+1\right )\right ) \log ^2\left (-\frac{a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac{b^2 \left (\log (c (a+b x))-\log \left (\frac{b x}{a}+1\right )\right ) \left (\log \left (-\frac{a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2}{4 a^2}-\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}-\frac{b^2 \log (x) \log \left (\frac{b x}{a}+1\right ) \log (1-c (a+b x))}{2 a^2} \]
Antiderivative was successfully verified.
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Rule 6599
Rule 6598
Rule 36
Rule 29
Rule 31
Rule 2416
Rule 2394
Rule 2315
Rule 2393
Rule 2391
Rule 6597
Rule 2440
Rule 2435
Rubi steps
\begin{align*} \int \frac{\text{Li}_3(c (a+b x))}{x^3} \, dx &=\frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{Li}_3(c (a+b x))}{2 a^2}-\frac{1}{2} b^3 \int \left (-\frac{\text{Li}_2(c (a+b x))}{a b^2 x^2}+\frac{\text{Li}_2(c (a+b x))}{a^2 b x}\right ) \, dx\\ &=\frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{Li}_3(c (a+b x))}{2 a^2}+\frac{b \int \frac{\text{Li}_2(c (a+b x))}{x^2} \, dx}{2 a}-\frac{b^2 \int \frac{\text{Li}_2(c (a+b x))}{x} \, dx}{2 a^2}\\ &=-\frac{b \text{Li}_2(c (a+b x))}{2 a x}-\frac{b^2 \log (x) \text{Li}_2(c (a+b x))}{2 a^2}+\frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{Li}_3(c (a+b x))}{2 a^2}-\frac{b^2 \int \frac{\log (1-a c-b c x)}{x (a+b x)} \, dx}{2 a}-\frac{b^3 \int \frac{\log (x) \log (1-a c-b c x)}{a+b x} \, dx}{2 a^2}\\ &=-\frac{b \text{Li}_2(c (a+b x))}{2 a x}-\frac{b^2 \log (x) \text{Li}_2(c (a+b x))}{2 a^2}+\frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{Li}_3(c (a+b x))}{2 a^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (-\frac{a}{b}+\frac{x}{b}\right ) \log \left (-\frac{-a b c-b (1-a c)}{b}-c x\right )}{x} \, dx,x,a+b x\right )}{2 a^2}-\frac{b^2 \int \left (\frac{\log (1-a c-b c x)}{a x}-\frac{b \log (1-a c-b c x)}{a (a+b x)}\right ) \, dx}{2 a}\\ &=-\frac{b^2 \log (x) \log \left (1+\frac{b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac{b^2 \left (\log \left (1+\frac{b x}{a}\right )+\log \left (\frac{1-a c}{1-c (a+b x)}\right )-\log \left (\frac{(1-a c) (a+b x)}{a (1-c (a+b x))}\right )\right ) \log ^2\left (-\frac{a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac{b^2 \left (\log (c (a+b x))-\log \left (1+\frac{b x}{a}\right )\right ) \left (\log (x)+\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right )^2}{4 a^2}-\frac{b^2 \left (\log (1-c (a+b x))-\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right ) \text{Li}_2\left (-\frac{b x}{a}\right )}{2 a^2}-\frac{b \text{Li}_2(c (a+b x))}{2 a x}-\frac{b^2 \log (x) \text{Li}_2(c (a+b x))}{2 a^2}-\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{Li}_2\left (-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{Li}_2\left (-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac{b^2 \left (\log (x)+\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right ) \text{Li}_2(1-c (a+b x))}{2 a^2}+\frac{b^2 \text{Li}_3\left (-\frac{b x}{a}\right )}{2 a^2}+\frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{Li}_3(c (a+b x))}{2 a^2}-\frac{b^2 \text{Li}_3\left (-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \text{Li}_3\left (-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac{b^2 \text{Li}_3(1-c (a+b x))}{2 a^2}-\frac{b^2 \int \frac{\log (1-a c-b c x)}{x} \, dx}{2 a^2}+\frac{b^3 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{2 a^2}\\ &=-\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac{b^2 \log (x) \log \left (1+\frac{b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac{b^2 \left (\log \left (1+\frac{b x}{a}\right )+\log \left (\frac{1-a c}{1-c (a+b x)}\right )-\log \left (\frac{(1-a c) (a+b x)}{a (1-c (a+b x))}\right )\right ) \log ^2\left (-\frac{a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac{b^2 \left (\log (c (a+b x))-\log \left (1+\frac{b x}{a}\right )\right ) \left (\log (x)+\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right )^2}{4 a^2}-\frac{b^2 \left (\log (1-c (a+b x))-\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right ) \text{Li}_2\left (-\frac{b x}{a}\right )}{2 a^2}-\frac{b \text{Li}_2(c (a+b x))}{2 a x}-\frac{b^2 \log (x) \text{Li}_2(c (a+b x))}{2 a^2}-\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{Li}_2\left (-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{Li}_2\left (-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac{b^2 \left (\log (x)+\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right ) \text{Li}_2(1-c (a+b x))}{2 a^2}+\frac{b^2 \text{Li}_3\left (-\frac{b x}{a}\right )}{2 a^2}+\frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{Li}_3(c (a+b x))}{2 a^2}-\frac{b^2 \text{Li}_3\left (-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \text{Li}_3\left (-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac{b^2 \text{Li}_3(1-c (a+b x))}{2 a^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 a^2}-\frac{\left (b^3 c\right ) \int \frac{\log \left (-\frac{b c x}{-1+a c}\right )}{1-a c-b c x} \, dx}{2 a^2}\\ &=-\frac{b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac{b^2 \log (x) \log \left (1+\frac{b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac{b^2 \left (\log \left (1+\frac{b x}{a}\right )+\log \left (\frac{1-a c}{1-c (a+b x)}\right )-\log \left (\frac{(1-a c) (a+b x)}{a (1-c (a+b x))}\right )\right ) \log ^2\left (-\frac{a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac{b^2 \left (\log (c (a+b x))-\log \left (1+\frac{b x}{a}\right )\right ) \left (\log (x)+\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right )^2}{4 a^2}-\frac{b^2 \left (\log (1-c (a+b x))-\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right ) \text{Li}_2\left (-\frac{b x}{a}\right )}{2 a^2}-\frac{b^2 \text{Li}_2(c (a+b x))}{2 a^2}-\frac{b \text{Li}_2(c (a+b x))}{2 a x}-\frac{b^2 \log (x) \text{Li}_2(c (a+b x))}{2 a^2}-\frac{b^2 \text{Li}_2\left (1-\frac{b c x}{1-a c}\right )}{2 a^2}-\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{Li}_2\left (-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \log \left (-\frac{a (1-c (a+b x))}{b x}\right ) \text{Li}_2\left (-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac{b^2 \left (\log (x)+\log \left (-\frac{a (1-c (a+b x))}{b x}\right )\right ) \text{Li}_2(1-c (a+b x))}{2 a^2}+\frac{b^2 \text{Li}_3\left (-\frac{b x}{a}\right )}{2 a^2}+\frac{\left (b^2-\frac{a^2}{x^2}\right ) \text{Li}_3(c (a+b x))}{2 a^2}-\frac{b^2 \text{Li}_3\left (-\frac{b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac{b^2 \text{Li}_3\left (-\frac{b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac{b^2 \text{Li}_3(1-c (a+b x))}{2 a^2}\\ \end{align*}
Mathematica [A] time = 1.93103, size = 573, normalized size = 0.91 \[ \frac{\frac{b x \left (b x \left (\text{PolyLog}\left (2,\frac{b c x}{1-a c}\right )+\text{PolyLog}(2,-a c-b c x+1)+\text{PolyLog}(3,c (a+b x))+\text{PolyLog}(3,-a c-b c x+1)-\text{PolyLog}\left (3,\frac{a (a c+b c x-1)}{b x}\right )+\text{PolyLog}\left (3,\frac{a c+b c x-1}{b c x}\right )+\log \left (\frac{a (a c+b c x-1)}{b x}\right ) \left (\text{PolyLog}\left (2,\frac{a (a c+b c x-1)}{b x}\right )-\text{PolyLog}\left (2,\frac{a c+b c x-1}{b c x}\right )\right )-\text{PolyLog}\left (2,-\frac{b x}{a}\right ) \left (\log (-a c-b c x+1)-\log \left (\frac{a (a c+b c x-1)}{b x}\right )\right )-\log (a+b x) \text{PolyLog}(2,c (a+b x))-\left (\log \left (\frac{a (a c+b c x-1)}{b x}\right )+\log (x)\right ) \text{PolyLog}(2,-a c-b c x+1)+\text{PolyLog}\left (3,-\frac{b x}{a}\right )-\frac{1}{2} \left (\log \left (\frac{1-a c}{b c x}\right )-\log \left (-\frac{(a c-1) (a+b x)}{b x}\right )+\log \left (\frac{b x}{a}+1\right )\right ) \log ^2\left (\frac{a (a c+b c x-1)}{b x}\right )-\left (\log (c (a+b x))-\log \left (\frac{b x}{a}+1\right )\right ) \log (-a c-b c x+1) \log \left (\frac{a (a c+b c x-1)}{b x}\right )+\log (c (a+b x)) \log (-a c-b c x+1)-\log (x) \log \left (\frac{b x}{a}+1\right ) \log (-a c-b c x+1)+\frac{1}{2} \left (\log (c (a+b x))-\log \left (\frac{b x}{a}+1\right )\right ) \log (-a c-b c x+1) (\log (-a c-b c x+1)-2 \log (x))-\log (x) \left (\log (-a c-b c x+1)-\log \left (\frac{b c x}{a c-1}+1\right )\right )\right )-(-b x \log (a+b x)+a+b x \log (x)) \text{PolyLog}(2,c (a+b x))\right )}{a^2}-\text{PolyLog}(3,c (a+b x))}{2 x^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.004, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\it polylog} \left ( 3,c \left ( bx+a \right ) \right ) }{{x}^{3}}}\, 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*} b \int \frac{{\rm Li}_2\left (b c x + a c\right )}{2 \,{\left (b x^{3} + a x^{2}\right )}}\,{d x} - \frac{{\rm Li}_{3}(b c x + a c)}{2 \, x^{2}} \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{{\rm polylog}\left (3, b c x + a c\right )}{x^{3}}, 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{\operatorname{Li}_{3}\left (a c + b c x\right )}{x^{3}}\, 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{{\rm Li}_{3}({\left (b x + a\right )} c)}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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