Optimal. Leaf size=287 \[ \frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}-\frac {e^2 \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 p \text {Li}_2\left (\frac {b}{a x}+1\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (\frac {a (d+e x)}{a d-b e}\right )}{d^3}+\frac {e^2 p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{d^3}-\frac {a p}{2 b d x}-\frac {e^2 p \text {Li}_2\left (\frac {e x}{d}+1\right )}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}-\frac {e p}{d^2 x}+\frac {p}{4 d x^2} \]
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Rubi [A] time = 0.33, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {2466, 2454, 2395, 43, 2389, 2295, 2394, 2315, 2462, 260, 2416, 2393, 2391} \[ -\frac {e^2 p \text {PolyLog}\left (2,\frac {b}{a x}+1\right )}{d^3}+\frac {e^2 p \text {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^3}-\frac {e^2 p \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d^3}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}-\frac {e^2 \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}+\frac {e^2 p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{d^3}-\frac {a p}{2 b d x}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}-\frac {e p}{d^2 x}+\frac {p}{4 d x^2} \]
Antiderivative was successfully verified.
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Rule 43
Rule 260
Rule 2295
Rule 2315
Rule 2389
Rule 2391
Rule 2393
Rule 2394
Rule 2395
Rule 2416
Rule 2454
Rule 2462
Rule 2466
Rubi steps
\begin {align*} \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx &=\int \left (\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d x^3}-\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2 x^2}+\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3 x}-\frac {e^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3} \, dx}{d}-\frac {e \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x} \, dx}{d^3}-\frac {e^3 \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx}{d^3}\\ &=-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {\operatorname {Subst}\left (\int x \log \left (c (a+b x)^p\right ) \, dx,x,\frac {1}{x}\right )}{d}+\frac {e \operatorname {Subst}\left (\int \log \left (c (a+b x)^p\right ) \, dx,x,\frac {1}{x}\right )}{d^2}-\frac {e^2 \operatorname {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x} \, dx,x,\frac {1}{x}\right )}{d^3}-\frac {\left (b e^2 p\right ) \int \frac {\log (d+e x)}{\left (a+\frac {b}{x}\right ) x^2} \, dx}{d^3}\\ &=-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}+\frac {e \operatorname {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,a+\frac {b}{x}\right )}{b d^2}+\frac {(b p) \operatorname {Subst}\left (\int \frac {x^2}{a+b x} \, dx,x,\frac {1}{x}\right )}{2 d}-\frac {\left (b e^2 p\right ) \int \left (\frac {\log (d+e x)}{b x}-\frac {a \log (d+e x)}{b (b+a x)}\right ) \, dx}{d^3}+\frac {\left (b e^2 p\right ) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right )}{a+b x} \, dx,x,\frac {1}{x}\right )}{d^3}\\ &=-\frac {e p}{d^2 x}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}+\frac {(b p) \operatorname {Subst}\left (\int \left (-\frac {a}{b^2}+\frac {x}{b}+\frac {a^2}{b^2 (a+b x)}\right ) \, dx,x,\frac {1}{x}\right )}{2 d}-\frac {\left (e^2 p\right ) \int \frac {\log (d+e x)}{x} \, dx}{d^3}+\frac {\left (a e^2 p\right ) \int \frac {\log (d+e x)}{b+a x} \, dx}{d^3}\\ &=\frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}+\frac {\left (e^3 p\right ) \int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx}{d^3}-\frac {\left (e^3 p\right ) \int \frac {\log \left (\frac {e (b+a x)}{-a d+b e}\right )}{d+e x} \, dx}{d^3}\\ &=\frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {e x}{d}\right )}{d^3}-\frac {\left (e^2 p\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {a x}{-a d+b e}\right )}{x} \, dx,x,d+e x\right )}{d^3}\\ &=\frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (\frac {a (d+e x)}{a d-b e}\right )}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {e x}{d}\right )}{d^3}\\ \end {align*}
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Mathematica [A] time = 0.24, size = 241, normalized size = 0.84 \[ -\frac {-\frac {d^2 p \left (2 a^2 x^2 \log \left (a+\frac {b}{x}\right )+b (b-2 a x)\right )}{b^2 x^2}+\frac {2 d^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2}+4 e^2 \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )-\frac {4 d e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b}+4 e^2 \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )+4 e^2 p \left (-\text {Li}_2\left (\frac {a (d+e x)}{a d-b e}\right )+\log (d+e x) \left (\log \left (-\frac {e x}{d}\right )-\log \left (\frac {e (a x+b)}{b e-a d}\right )\right )+\text {Li}_2\left (\frac {e x}{d}+1\right )\right )+4 e^2 p \text {Li}_2\left (\frac {b}{a x}+1\right )+\frac {4 d e p}{x}}{4 d^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (c \left (\frac {a x + b}{x}\right )^{p}\right )}{e x^{4} + d x^{3}}, x\right ) \]
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 \[ \int \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{\left (e x +d \right ) x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.01, size = 307, normalized size = 1.07 \[ \frac {1}{4} \, {\left (4 \, e {\left (\frac {a \log \left (a x + b\right )}{b^{2} d^{2}} - \frac {a \log \relax (x)}{b^{2} d^{2}} - \frac {1}{b d^{2} x}\right )} - \frac {4 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} e^{2}}{b d^{3}} + \frac {4 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} e^{2}}{b d^{3}} + \frac {4 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {a e x + a d}{a d - b e} + 1\right ) + {\rm Li}_2\left (\frac {a e x + a d}{a d - b e}\right )\right )} e^{2}}{b d^{3}} + \frac {2 \, a^{2} \log \left (a x + b\right )}{b^{3} d} - \frac {2 \, a^{2} \log \relax (x)}{b^{3} d} - \frac {2 \, {\left (2 \, e^{2} \log \left (e x + d\right ) \log \relax (x) - e^{2} \log \relax (x)^{2}\right )}}{b d^{3}} - \frac {2 \, a x - b}{b^{2} d x^{2}}\right )} b p - \frac {1}{2} \, {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \relax (x)}{d^{3}} - \frac {2 \, e x - d}{d^{2} x^{2}}\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \]
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 \[ \int \frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{x^3\,\left (d+e\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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