Integrand size = 20, antiderivative size = 81 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {p \log (x)}{d e}+\frac {a p \log (b+a x)}{e (a d-b e)}-\frac {b p \log (d+e x)}{d (a d-b e)} \]
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Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2513, 528, 84} \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}+\frac {a p \log (a x+b)}{e (a d-b e)}-\frac {b p \log (d+e x)}{d (a d-b e)}-\frac {p \log (x)}{d e} \]
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Rule 84
Rule 528
Rule 2513
Rubi steps \begin{align*} \text {integral}& = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {(b p) \int \frac {1}{\left (a+\frac {b}{x}\right ) x^2 (d+e x)} \, dx}{e} \\ & = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {(b p) \int \frac {1}{x (b+a x) (d+e x)} \, dx}{e} \\ & = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {(b p) \int \left (\frac {1}{b d x}+\frac {a^2}{b (-a d+b e) (b+a x)}+\frac {e^2}{d (a d-b e) (d+e x)}\right ) \, dx}{e} \\ & = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {p \log (x)}{d e}+\frac {a p \log (b+a x)}{e (a d-b e)}-\frac {b p \log (d+e x)}{d (a d-b e)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e (d+e x)}-\frac {p \log (x)}{d e}+\frac {a p \log (b+a x)}{e (a d-b e)}-\frac {b p \log (d+e x)}{d (a d-b e)} \]
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Time = 0.89 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06
method | result | size |
parts | \(-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{e \left (e x +d \right )}-\frac {p b \left (\frac {\ln \left (x \right )}{b d}+\frac {e \ln \left (e x +d \right )}{d \left (a d -b e \right )}-\frac {a \ln \left (a x +b \right )}{b \left (a d -b e \right )}\right )}{e}\) | \(86\) |
parallelrisch | \(-\frac {-\ln \left (x \right ) x \,b^{2} e \,p^{2}+\ln \left (e x +d \right ) x \,b^{2} e \,p^{2}-\ln \left (x \right ) b^{2} d \,p^{2}+\ln \left (e x +d \right ) b^{2} d \,p^{2}-x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a b d p -\ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{2} d p}{\left (e x +d \right ) p b d \left (a d -b e \right )}\) | \(124\) |
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Time = 0.38 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.83 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {{\left (a d^{2} - b d e\right )} p \log \left (\frac {a x + b}{x}\right ) - {\left (a d e p x + a d^{2} p\right )} \log \left (a x + b\right ) + {\left (b e^{2} p x + b d e p\right )} \log \left (e x + d\right ) + {\left (a d^{2} - b d e\right )} \log \left (c\right ) + {\left ({\left (a d e - b e^{2}\right )} p x + {\left (a d^{2} - b d e\right )} p\right )} \log \left (x\right )}{a d^{3} e - b d^{2} e^{2} + {\left (a d^{2} e^{2} - b d e^{3}\right )} x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (61) = 122\).
Time = 3.72 (sec) , antiderivative size = 425, normalized size of antiderivative = 5.25 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=\begin {cases} \frac {d p \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac {e p x \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac {e x \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{d^{2} e + d e^{2} x} & \text {for}\: a = 0 \\\frac {x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )} + \frac {b p \log {\left (a x + b \right )}}{a}}{d^{2}} & \text {for}\: e = 0 \\- \frac {d p}{d^{2} e + d e^{2} x} + \frac {e x \log {\left (c \left (\frac {b}{x} + \frac {b e}{d}\right )^{p} \right )}}{d^{2} e + d e^{2} x} & \text {for}\: a = \frac {b e}{d} \\\frac {- \frac {a \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{b} + \frac {p}{x} - \frac {\log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{x}}{e^{2}} & \text {for}\: d = 0 \\\frac {a d x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} + \frac {b d p \log {\left (x + \frac {b}{a} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} - \frac {b d p \log {\left (\frac {d}{e} + x \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} + \frac {b e p x \log {\left (x + \frac {b}{a} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} - \frac {b e p x \log {\left (\frac {d}{e} + x \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} - \frac {b e x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{a d^{3} + a d^{2} e x - b d^{2} e - b d e^{2} x} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=\frac {b p {\left (\frac {a \log \left (a x + b\right )}{a b d - b^{2} e} - \frac {e \log \left (e x + d\right )}{a d^{2} - b d e} - \frac {\log \left (x\right )}{b d}\right )}}{e} - \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} e} \]
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Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.79 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\frac {b^{2} p \log \left (-a d + b e + \frac {{\left (a x + b\right )} d}{x}\right )}{a d^{2} - b d e} + \frac {b^{2} p \log \left (\frac {a x + b}{x}\right )}{a d^{2} - b d e - \frac {{\left (a x + b\right )} d^{2}}{x}} - \frac {b^{2} p \log \left (\frac {a x + b}{x}\right )}{a d^{2} - b d e} + \frac {b^{2} \log \left (c\right )}{a d^{2} - b d e - \frac {{\left (a x + b\right )} d^{2}}{x}}}{b} \]
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Time = 1.52 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{(d+e x)^2} \, dx=-\frac {\ln \left (c\,{\left (\frac {b+a\,x}{x}\right )}^p\right )}{x\,e^2+d\,e}-\frac {p\,\ln \left (x\right )}{d\,e}-\frac {a\,p\,\ln \left (b+a\,x\right )}{b\,e^2-a\,d\,e}-\frac {b\,p\,\ln \left (d+e\,x\right )}{a\,d^2-b\,d\,e} \]
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