Optimal. Leaf size=91 \[ -\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac {d p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{e^2}-\frac {p x}{e} \]
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Rubi [A] time = 0.11, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {43, 2416, 2389, 2295, 2394, 2393, 2391} \[ -\frac {d p \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e^2}-\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac {p x}{e} \]
Antiderivative was successfully verified.
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Rule 43
Rule 2295
Rule 2389
Rule 2391
Rule 2393
Rule 2394
Rule 2416
Rubi steps
\begin {align*} \int \frac {x \log \left (c (a+b x)^p\right )}{d+e x} \, dx &=\int \left (\frac {\log \left (c (a+b x)^p\right )}{e}-\frac {d \log \left (c (a+b x)^p\right )}{e (d+e x)}\right ) \, dx\\ &=\frac {\int \log \left (c (a+b x)^p\right ) \, dx}{e}-\frac {d \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx}{e}\\ &=-\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}+\frac {\operatorname {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,a+b x\right )}{b e}+\frac {(b d p) \int \frac {\log \left (\frac {b (d+e x)}{b d-a e}\right )}{a+b x} \, dx}{e^2}\\ &=-\frac {p x}{e}+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}+\frac {(d p) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {e x}{b d-a e}\right )}{x} \, dx,x,a+b x\right )}{e^2}\\ &=-\frac {p x}{e}+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}-\frac {d p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{e^2}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 79, normalized size = 0.87 \[ \frac {\log \left (c (a+b x)^p\right ) \left (-b d \log \left (\frac {b (d+e x)}{b d-a e}\right )+a e+b e x\right )-b d p \text {Li}_2\left (\frac {e (a+b x)}{a e-b d}\right )-b e p x}{b e^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}, x\right ) \]
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 \[ \int \frac {x \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.32, size = 427, normalized size = 4.69 \[ \frac {i \pi d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{2 e^{2}}-\frac {i \pi d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 e^{2}}-\frac {i \pi d \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \ln \left (e x +d \right )}{2 e^{2}}+\frac {i \pi d \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3} \ln \left (e x +d \right )}{2 e^{2}}-\frac {i \pi x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{2 e}+\frac {i \pi x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2 e}+\frac {i \pi x \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2 e}-\frac {i \pi x \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2 e}+\frac {d p \ln \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right ) \ln \left (e x +d \right )}{e^{2}}+\frac {a p \ln \left (a e -b d +\left (e x +d \right ) b \right )}{b e}+\frac {d p \dilog \left (\frac {a e -b d +\left (e x +d \right ) b}{a e -b d}\right )}{e^{2}}-\frac {d \ln \relax (c ) \ln \left (e x +d \right )}{e^{2}}-\frac {d \ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{e^{2}}-\frac {p x}{e}+\frac {x \ln \relax (c )}{e}+\frac {x \ln \left (\left (b x +a \right )^{p}\right )}{e}-\frac {d p}{e^{2}} \]
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 \[ \int \frac {x \log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{d+e\,x} \,d x \]
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 \[ \int \frac {x \log {\left (c \left (a + b x\right )^{p} \right )}}{d + e x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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