Integrand size = 30, antiderivative size = 115 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {(f h-e i) (a+b \log (c (e+f x)))^{1+p}}{b d f^2 (1+p)}+\frac {e^{-\frac {a}{b}} i \Gamma \left (1+p,-\frac {a+b \log (c (e+f x))}{b}\right ) (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p}}{c d f^2} \]
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Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2458, 12, 2395, 2336, 2212, 2339, 30} \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {(f h-e i) (a+b \log (c (e+f x)))^{p+1}}{b d f^2 (p+1)}+\frac {i e^{-\frac {a}{b}} (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p} \Gamma \left (p+1,-\frac {a+b \log (c (e+f x))}{b}\right )}{c d f^2} \]
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Rule 12
Rule 30
Rule 2212
Rule 2336
Rule 2339
Rule 2395
Rule 2458
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (\frac {f h-e i}{f}+\frac {i x}{f}\right ) (a+b \log (c x))^p}{d x} \, dx,x,e+f x\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {\left (\frac {f h-e i}{f}+\frac {i x}{f}\right ) (a+b \log (c x))^p}{x} \, dx,x,e+f x\right )}{d f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {i (a+b \log (c x))^p}{f}+\frac {(f h-e i) (a+b \log (c x))^p}{f x}\right ) \, dx,x,e+f x\right )}{d f} \\ & = \frac {i \text {Subst}\left (\int (a+b \log (c x))^p \, dx,x,e+f x\right )}{d f^2}+\frac {(f h-e i) \text {Subst}\left (\int \frac {(a+b \log (c x))^p}{x} \, dx,x,e+f x\right )}{d f^2} \\ & = \frac {i \text {Subst}\left (\int e^x (a+b x)^p \, dx,x,\log (c (e+f x))\right )}{c d f^2}+\frac {(f h-e i) \text {Subst}\left (\int x^p \, dx,x,a+b \log (c (e+f x))\right )}{b d f^2} \\ & = \frac {(f h-e i) (a+b \log (c (e+f x)))^{1+p}}{b d f^2 (1+p)}+\frac {e^{-\frac {a}{b}} i \Gamma \left (1+p,-\frac {a+b \log (c (e+f x))}{b}\right ) (a+b \log (c (e+f x)))^p \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p}}{c d f^2} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=(a+b \log (c (e+f x)))^p \left (\frac {(h+i x) (a+b \log (c (e+f x)))}{b d f+b d f p}+\frac {e^{-\frac {a}{b}} i \Gamma \left (2+p,-\frac {a+b \log (c (e+f x))}{b}\right ) \left (-\frac {a+b \log (c (e+f x))}{b}\right )^{-p}}{c d f^2+c d f^2 p}\right ) \]
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\[\int \frac {\left (i x +h \right ) \left (a +b \ln \left (c \left (f x +e \right )\right )\right )^{p}}{d f x +d e}d x\]
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none
Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.03 \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {{\left (b i p + b i\right )} e^{\left (-\frac {b p \log \left (-\frac {1}{b}\right ) + a}{b}\right )} \Gamma \left (p + 1, -\frac {b \log \left (c f x + c e\right ) + a}{b}\right ) + {\left (a c f h - a c e i + {\left (b c f h - b c e i\right )} \log \left (c f x + c e\right )\right )} {\left (b \log \left (c f x + c e\right ) + a\right )}^{p}}{b c d f^{2} p + b c d f^{2}} \]
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\[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {\int \frac {h \left (a + b \log {\left (c e + c f x \right )}\right )^{p}}{e + f x}\, dx + \int \frac {i x \left (a + b \log {\left (c e + c f x \right )}\right )^{p}}{e + f x}\, dx}{d} \]
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\[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\int { \frac {{\left (i x + h\right )} {\left (b \log \left ({\left (f x + e\right )} c\right ) + a\right )}^{p}}{d f x + d e} \,d x } \]
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\[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\int { \frac {{\left (i x + h\right )} {\left (b \log \left ({\left (f x + e\right )} c\right ) + a\right )}^{p}}{d f x + d e} \,d x } \]
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Timed out. \[ \int \frac {(h+i x) (a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\int \frac {\left (h+i\,x\right )\,{\left (a+b\,\ln \left (c\,\left (e+f\,x\right )\right )\right )}^p}{d\,e+d\,f\,x} \,d x \]
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