Optimal. Leaf size=629 \[ -\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac {b^2 \log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac {b^2 \left (\log \left (1+\frac {b x}{a}\right )+\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac {b^2 \left (\log (c (a+b x))-\log \left (1+\frac {b x}{a}\right )\right ) \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )^2}{4 a^2}-\frac {b^2 \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \text {PolyLog}\left (2,-\frac {b x}{a}\right )}{2 a^2}-\frac {b^2 \text {PolyLog}(2,c (a+b x))}{2 a^2}-\frac {b \text {PolyLog}(2,c (a+b x))}{2 a x}-\frac {b^2 \log (x) \text {PolyLog}(2,c (a+b x))}{2 a^2}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {b c x}{1-a c}\right )}{2 a^2}-\frac {b^2 \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac {b^2 \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \text {PolyLog}(2,1-c (a+b x))}{2 a^2}+\frac {b^2 \text {PolyLog}\left (3,-\frac {b x}{a}\right )}{2 a^2}+\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {PolyLog}(3,c (a+b x))}{2 a^2}-\frac {b^2 \text {PolyLog}\left (3,-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \text {PolyLog}\left (3,-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac {b^2 \text {PolyLog}(3,1-c (a+b x))}{2 a^2} \]
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Rubi [A]
time = 0.47, antiderivative size = 629, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 13, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6734, 6733,
36, 29, 31, 2463, 2441, 2352, 2440, 2438, 6732, 2490, 2485} \begin {gather*} \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}_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 \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 \text {Li}_2\left (-\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 {Li}_2(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_2\left (-\frac {b x}{a (1-c (a+b x))}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )}{2 a^2}+\frac {b^2 \text {Li}_2\left (-\frac {b c x}{1-c (a+b x)}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )}{2 a^2}-\frac {b^2 \text {Li}_2(1-c (a+b x)) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )}{2 a^2}-\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}+\frac {b^2 \text {Li}_3\left (-\frac {b x}{a}\right )}{2 a^2}-\frac {b \text {Li}_2(c (a+b x))}{2 a x} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 2352
Rule 2438
Rule 2440
Rule 2441
Rule 2463
Rule 2485
Rule 2490
Rule 6732
Rule 6733
Rule 6734
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 \text {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 \text {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*}
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Mathematica [A]
time = 1.33, size = 573, normalized size = 0.91 \begin {gather*} \frac {-\text {PolyLog}(3,c (a+b x))+\frac {b x \left (-((a+b x \log (x)-b x \log (a+b x)) \text {PolyLog}(2,c (a+b x)))+b x \left (\log (c (a+b x)) \log (1-a c-b c x)-\log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-a c-b c x)+\frac {1}{2} \left (\log (c (a+b x))-\log \left (1+\frac {b x}{a}\right )\right ) \log (1-a c-b c x) (-2 \log (x)+\log (1-a c-b c x))-\left (\log (c (a+b x))-\log \left (1+\frac {b x}{a}\right )\right ) \log (1-a c-b c x) \log \left (\frac {a (-1+a c+b c x)}{b x}\right )-\frac {1}{2} \left (\log \left (\frac {1-a c}{b c x}\right )-\log \left (\frac {(1-a c) (a+b x)}{b x}\right )+\log \left (1+\frac {b x}{a}\right )\right ) \log ^2\left (\frac {a (-1+a c+b c x)}{b x}\right )-\log (x) \left (\log (1-a c-b c x)-\log \left (1+\frac {b c x}{-1+a c}\right )\right )-\left (\log (1-a c-b c x)-\log \left (\frac {a (-1+a c+b c x)}{b x}\right )\right ) \text {PolyLog}\left (2,-\frac {b x}{a}\right )+\text {PolyLog}\left (2,\frac {b c x}{1-a c}\right )-\log (a+b x) \text {PolyLog}(2,c (a+b x))+\text {PolyLog}(2,1-a c-b c x)-\left (\log (x)+\log \left (\frac {a (-1+a c+b c x)}{b x}\right )\right ) \text {PolyLog}(2,1-a c-b c x)+\log \left (\frac {a (-1+a c+b c x)}{b x}\right ) \left (\text {PolyLog}\left (2,\frac {a (-1+a c+b c x)}{b x}\right )-\text {PolyLog}\left (2,\frac {-1+a c+b c x}{b c x}\right )\right )+\text {PolyLog}\left (3,-\frac {b x}{a}\right )+\text {PolyLog}(3,c (a+b x))+\text {PolyLog}(3,1-a c-b c x)-\text {PolyLog}\left (3,\frac {a (-1+a c+b c x)}{b x}\right )+\text {PolyLog}\left (3,\frac {-1+a c+b c x}{b c x}\right )\right )\right )}{a^2}}{2 x^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\polylog \left (3, c \left (b x +a \right )\right )}{x^{3}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {Li}_{3}\left (a c + b c x\right )}{x^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right )}{x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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