Integrand size = 20, antiderivative size = 50 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx=a x+\frac {b (f+g x) \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )}{g}+\frac {b e p \log (e+d (f+g x))}{d g} \]
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Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2533, 2498, 269, 31} \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx=a x+\frac {b (f+g x) \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )}{g}+\frac {b e p \log (d (f+g x)+e)}{d g} \]
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Rule 31
Rule 269
Rule 2498
Rule 2533
Rubi steps \begin{align*} \text {integral}& = a x+b \int \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right ) \, dx \\ & = a x+\frac {b \text {Subst}\left (\int \log \left (c \left (d+\frac {e}{x}\right )^p\right ) \, dx,x,f+g x\right )}{g} \\ & = a x+\frac {b (f+g x) \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )}{g}+\frac {(b e p) \text {Subst}\left (\int \frac {1}{\left (d+\frac {e}{x}\right ) x} \, dx,x,f+g x\right )}{g} \\ & = a x+\frac {b (f+g x) \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )}{g}+\frac {(b e p) \text {Subst}\left (\int \frac {1}{e+d x} \, dx,x,f+g x\right )}{g} \\ & = a x+\frac {b (f+g x) \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )}{g}+\frac {b e p \log (e+d (f+g x))}{d g} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.40 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx=a x-b e g p \left (\frac {f \log (f+g x)}{e g^2}-\frac {(e+d f) \log (e+d f+d g x)}{d e g^2}\right )+b x \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right ) \]
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Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.54
method | result | size |
default | \(a x +b \left (\ln \left (c \left (\frac {d g x +d f +e}{g x +f}\right )^{p}\right ) x +e g p \left (-\frac {f \ln \left (g x +f \right )}{g^{2} e}+\frac {\left (d f +e \right ) \ln \left (d g x +d f +e \right )}{e \,g^{2} d}\right )\right )\) | \(77\) |
parts | \(a x +b \left (\ln \left (c \left (\frac {d g x +d f +e}{g x +f}\right )^{p}\right ) x +e g p \left (-\frac {f \ln \left (g x +f \right )}{g^{2} e}+\frac {\left (d f +e \right ) \ln \left (d g x +d f +e \right )}{e \,g^{2} d}\right )\right )\) | \(77\) |
parallelrisch | \(-\frac {b \left (-x \ln \left (c \left (\frac {d g x +d f +e}{g x +f}\right )^{p}\right ) d \,g^{2} p -\ln \left (g x +f \right ) e g \,p^{2}-\ln \left (c \left (\frac {d g x +d f +e}{g x +f}\right )^{p}\right ) d f g p -\ln \left (c \left (\frac {d g x +d f +e}{g x +f}\right )^{p}\right ) e g p \right )}{d \,g^{2} p}+a x\) | \(116\) |
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Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.52 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx=\frac {b d g p x \log \left (\frac {d g x + d f + e}{g x + f}\right ) - b d f p \log \left (g x + f\right ) + b d g x \log \left (c\right ) + a d g x + {\left (b d f + b e\right )} p \log \left (d g x + d f + e\right )}{d g} \]
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Time = 0.76 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.14 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx=a x + b \left (\begin {cases} x \log {\left (c \left (\frac {e}{f}\right )^{p} \right )} & \text {for}\: d = 0 \wedge g = 0 \\\frac {f \log {\left (c \left (\frac {e}{f + g x}\right )^{p} \right )}}{g} + p x + x \log {\left (c \left (\frac {e}{f + g x}\right )^{p} \right )} & \text {for}\: d = 0 \\x \log {\left (c \left (d + \frac {e}{f}\right )^{p} \right )} & \text {for}\: g = 0 \\\frac {f \log {\left (c \left (d + \frac {e}{f + g x}\right )^{p} \right )}}{g} + x \log {\left (c \left (d + \frac {e}{f + g x}\right )^{p} \right )} + \frac {e p \log {\left (d f + d g x + e \right )}}{d g} & \text {otherwise} \end {cases}\right ) \]
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Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.40 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx=-b e g p {\left (\frac {f \log \left (g x + f\right )}{e g^{2}} - \frac {{\left (d f + e\right )} \log \left (d g x + d f + e\right )}{d e g^{2}}\right )} + b x \log \left (c {\left (d + \frac {e}{g x + f}\right )}^{p}\right ) + a x \]
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Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (50) = 100\).
Time = 0.32 (sec) , antiderivative size = 176, normalized size of antiderivative = 3.52 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx={\left (\frac {e^{2} p \log \left (\frac {d g x + d f + e}{g x + f}\right )}{d g^{2} - \frac {{\left (d g x + d f + e\right )} g^{2}}{g x + f}} + \frac {e^{2} \log \left (c\right )}{d g^{2} - \frac {{\left (d g x + d f + e\right )} g^{2}}{g x + f}} + \frac {e^{2} p \log \left (-d + \frac {d g x + d f + e}{g x + f}\right )}{d g^{2}} - \frac {e^{2} p \log \left (\frac {d g x + d f + e}{g x + f}\right )}{d g^{2}}\right )} b {\left (\frac {d f g}{e^{2}} - \frac {{\left (d f + e\right )} g}{e^{2}}\right )} + a x \]
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Time = 1.50 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.22 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )\right ) \, dx=a\,x+b\,x\,\ln \left (c\,{\left (d+\frac {e}{f+g\,x}\right )}^p\right )-\frac {b\,f\,p\,\ln \left (f+g\,x\right )}{g}+\frac {b\,p\,\ln \left (e+d\,f+d\,g\,x\right )\,\left (e+d\,f\right )}{d\,g} \]
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