Optimal. Leaf size=151 \[ -\frac {f \log \left (\frac {f h-e i}{i (e+f x)}+1\right ) (a+b \log (c (e+f x)))}{d (f h-e i)^2}-\frac {i (e+f x) (a+b \log (c (e+f x)))}{d (h+i x) (f h-e i)^2}+\frac {b f \text {Li}_2\left (-\frac {f h-e i}{i (e+f x)}\right )}{d (f h-e i)^2}+\frac {b f \log (h+i x)}{d (f h-e i)^2} \]
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Rubi [A] time = 0.36, antiderivative size = 181, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2411, 12, 2347, 2344, 2301, 2317, 2391, 2314, 31} \[ -\frac {b f \text {PolyLog}\left (2,-\frac {i (e+f x)}{f h-e i}\right )}{d (f h-e i)^2}+\frac {f (a+b \log (c (e+f x)))^2}{2 b d (f h-e i)^2}-\frac {f \log \left (\frac {f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))}{d (f h-e i)^2}-\frac {i (e+f x) (a+b \log (c (e+f x)))}{d (h+i x) (f h-e i)^2}+\frac {b f \log (h+i x)}{d (f h-e i)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 31
Rule 2301
Rule 2314
Rule 2317
Rule 2344
Rule 2347
Rule 2391
Rule 2411
Rubi steps
\begin {align*} \int \frac {a+b \log (c (e+f x))}{(h+181 x)^2 (d e+d f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \log (c x)}{d x \left (\frac {-181 e+f h}{f}+\frac {181 x}{f}\right )^2} \, dx,x,e+f x\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {a+b \log (c x)}{x \left (\frac {-181 e+f h}{f}+\frac {181 x}{f}\right )^2} \, dx,x,e+f x\right )}{d f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {a+b \log (c x)}{x \left (\frac {-181 e+f h}{f}+\frac {181 x}{f}\right )} \, dx,x,e+f x\right )}{d (181 e-f h)}+\frac {181 \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{\left (\frac {-181 e+f h}{f}+\frac {181 x}{f}\right )^2} \, dx,x,e+f x\right )}{d f (181 e-f h)}\\ &=-\frac {181 (e+f x) (a+b \log (c (e+f x)))}{d (181 e-f h)^2 (h+181 x)}-\frac {181 \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{\frac {-181 e+f h}{f}+\frac {181 x}{f}} \, dx,x,e+f x\right )}{d (181 e-f h)^2}+\frac {(181 b) \operatorname {Subst}\left (\int \frac {1}{\frac {-181 e+f h}{f}+\frac {181 x}{f}} \, dx,x,e+f x\right )}{d (181 e-f h)^2}+\frac {f \operatorname {Subst}\left (\int \frac {a+b \log (c x)}{x} \, dx,x,e+f x\right )}{d (181 e-f h)^2}\\ &=\frac {b f \log (h+181 x)}{d (181 e-f h)^2}-\frac {181 (e+f x) (a+b \log (c (e+f x)))}{d (181 e-f h)^2 (h+181 x)}-\frac {f \log \left (-\frac {f (h+181 x)}{181 e-f h}\right ) (a+b \log (c (e+f x)))}{d (181 e-f h)^2}+\frac {f (a+b \log (c (e+f x)))^2}{2 b d (181 e-f h)^2}+\frac {(b f) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {181 x}{-181 e+f h}\right )}{x} \, dx,x,e+f x\right )}{d (181 e-f h)^2}\\ &=\frac {b f \log (h+181 x)}{d (181 e-f h)^2}-\frac {181 (e+f x) (a+b \log (c (e+f x)))}{d (181 e-f h)^2 (h+181 x)}-\frac {f \log \left (-\frac {f (h+181 x)}{181 e-f h}\right ) (a+b \log (c (e+f x)))}{d (181 e-f h)^2}+\frac {f (a+b \log (c (e+f x)))^2}{2 b d (181 e-f h)^2}-\frac {b f \text {Li}_2\left (\frac {181 (e+f x)}{181 e-f h}\right )}{d (181 e-f h)^2}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 141, normalized size = 0.93 \[ \frac {-2 f \log \left (\frac {f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))+\frac {2 (f h-e i) (a+b \log (c (e+f x)))}{h+i x}+\frac {f (a+b \log (c (e+f x)))^2}{b}-2 b f \text {Li}_2\left (\frac {i (e+f x)}{e i-f h}\right )-2 b f (\log (e+f x)-\log (h+i x))}{2 d (f h-e i)^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c f x + c e\right ) + a}{d f i^{2} x^{3} + d e h^{2} + {\left (2 \, d f h i + d e i^{2}\right )} x^{2} + {\left (d f h^{2} + 2 \, d e h i\right )} x}, x\right ) \]
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 \[ \int \frac {b \log \left ({\left (f x + e\right )} c\right ) + a}{{\left (d f x + d e\right )} {\left (i x + h\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 355, normalized size = 2.35 \[ -\frac {b c \,f^{2} i x \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{2} \left (c f i x +c f h \right ) d}-\frac {b c e f i \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{2} \left (c f i x +c f h \right ) d}-\frac {b f \ln \left (\frac {-c e i +c f h +\left (c f x +c e \right ) i}{-c e i +c f h}\right ) \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{2} d}+\frac {b f \ln \left (c f x +c e \right )^{2}}{2 \left (e i -f h \right )^{2} d}-\frac {a c f}{\left (e i -f h \right ) \left (c f i x +c f h \right ) d}-\frac {a f \ln \left (-c e i +c f h +\left (c f x +c e \right ) i \right )}{\left (e i -f h \right )^{2} d}+\frac {a f \ln \left (c f x +c e \right )}{\left (e i -f h \right )^{2} d}-\frac {b f \dilog \left (\frac {-c e i +c f h +\left (c f x +c e \right ) i}{-c e i +c f h}\right )}{\left (e i -f h \right )^{2} d}+\frac {b f \ln \left (-c e i +c f h +\left (c f x +c e \right ) i \right )}{\left (e i -f h \right )^{2} d} \]
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 \[ a {\left (\frac {f \log \left (f x + e\right )}{d f^{2} h^{2} - 2 \, d e f h i + d e^{2} i^{2}} - \frac {f \log \left (i x + h\right )}{d f^{2} h^{2} - 2 \, d e f h i + d e^{2} i^{2}} + \frac {1}{d f h^{2} - d e h i + {\left (d f h i - d e i^{2}\right )} x}\right )} + b \int \frac {\log \left (f x + e\right ) + \log \relax (c)}{d f i^{2} x^{3} + d e h^{2} + {\left (2 \, f h i + e i^{2}\right )} d x^{2} + {\left (f h^{2} + 2 \, e h i\right )} d x}\,{d x} \]
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 \[ \int \frac {a+b\,\ln \left (c\,\left (e+f\,x\right )\right )}{{\left (h+i\,x\right )}^2\,\left (d\,e+d\,f\,x\right )} \,d x \]
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 \[ \frac {\int \frac {a}{e h^{2} + 2 e h i x + e i^{2} x^{2} + f h^{2} x + 2 f h i x^{2} + f i^{2} x^{3}}\, dx + \int \frac {b \log {\left (c e + c f x \right )}}{e h^{2} + 2 e h i x + e i^{2} x^{2} + f h^{2} x + 2 f h i x^{2} + f i^{2} x^{3}}\, dx}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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