Integrand size = 16, antiderivative size = 75 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b^3 p x}{4 a^3}-\frac {b^2 p x^2}{8 a^2}+\frac {b p x^3}{12 a}+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )-\frac {b^4 p \log (b+a x)}{4 a^4} \]
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Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2505, 269, 45} \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {b^4 p \log (a x+b)}{4 a^4}+\frac {b^3 p x}{4 a^3}-\frac {b^2 p x^2}{8 a^2}+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p x^3}{12 a} \]
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Rule 45
Rule 269
Rule 2505
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{4} (b p) \int \frac {x^2}{a+\frac {b}{x}} \, dx \\ & = \frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{4} (b p) \int \frac {x^3}{b+a x} \, dx \\ & = \frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{4} (b p) \int \left (\frac {b^2}{a^3}-\frac {b x}{a^2}+\frac {x^2}{a}-\frac {b^3}{a^3 (b+a x)}\right ) \, dx \\ & = \frac {b^3 p x}{4 a^3}-\frac {b^2 p x^2}{8 a^2}+\frac {b p x^3}{12 a}+\frac {1}{4} x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )-\frac {b^4 p \log (b+a x)}{4 a^4} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a b p x \left (6 b^2-3 a b x+2 a^2 x^2\right )-6 b^4 p \log \left (a+\frac {b}{x}\right )+6 a^4 x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )-6 b^4 p \log (x)}{24 a^4} \]
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Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84
method | result | size |
parts | \(\frac {x^{4} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{4}+\frac {p b \left (\frac {\frac {1}{3} x^{3} a^{2}-\frac {1}{2} a b \,x^{2}+b^{2} x}{a^{3}}-\frac {b^{3} \ln \left (a x +b \right )}{a^{4}}\right )}{4}\) | \(63\) |
parallelrisch | \(-\frac {-6 x^{4} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{4} p -2 x^{3} a^{3} b \,p^{2}+3 x^{2} a^{2} b^{2} p^{2}+6 \ln \left (x \right ) b^{4} p^{2}-6 x a \,b^{3} p^{2}+6 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{4} p +6 b^{4} p^{2}}{24 a^{4} p}\) | \(107\) |
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Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {6 \, a^{4} p x^{4} \log \left (\frac {a x + b}{x}\right ) + 6 \, a^{4} x^{4} \log \left (c\right ) + 2 \, a^{3} b p x^{3} - 3 \, a^{2} b^{2} p x^{2} + 6 \, a b^{3} p x - 6 \, b^{4} p \log \left (a x + b\right )}{24 \, a^{4}} \]
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Time = 1.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.16 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} \frac {x^{4} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{4} + \frac {b p x^{3}}{12 a} - \frac {b^{2} p x^{2}}{8 a^{2}} + \frac {b^{3} p x}{4 a^{3}} - \frac {b^{4} p \log {\left (a x + b \right )}}{4 a^{4}} & \text {for}\: a \neq 0 \\\frac {p x^{4}}{16} + \frac {x^{4} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{4} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.85 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{4} \, x^{4} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) - \frac {1}{24} \, b p {\left (\frac {6 \, b^{3} \log \left (a x + b\right )}{a^{4}} - \frac {2 \, a^{2} x^{3} - 3 \, a b x^{2} + 6 \, b^{2} x}{a^{3}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (65) = 130\).
Time = 0.33 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.43 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {\frac {6 \, b^{5} p \log \left (\frac {a x + b}{x}\right )}{a^{4} - \frac {4 \, {\left (a x + b\right )} a^{3}}{x} + \frac {6 \, {\left (a x + b\right )}^{2} a^{2}}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} a}{x^{3}} + \frac {{\left (a x + b\right )}^{4}}{x^{4}}} + \frac {6 \, b^{5} p \log \left (-a + \frac {a x + b}{x}\right )}{a^{4}} - \frac {6 \, b^{5} p \log \left (\frac {a x + b}{x}\right )}{a^{4}} - \frac {11 \, a^{3} b^{5} p - 6 \, a^{3} b^{5} \log \left (c\right ) - \frac {26 \, {\left (a x + b\right )} a^{2} b^{5} p}{x} + \frac {21 \, {\left (a x + b\right )}^{2} a b^{5} p}{x^{2}} - \frac {6 \, {\left (a x + b\right )}^{3} b^{5} p}{x^{3}}}{a^{7} - \frac {4 \, {\left (a x + b\right )} a^{6}}{x} + \frac {6 \, {\left (a x + b\right )}^{2} a^{5}}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} a^{4}}{x^{3}} + \frac {{\left (a x + b\right )}^{4} a^{3}}{x^{4}}}}{24 \, b} \]
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Time = 1.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {x^4\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{4}-\frac {b^2\,p\,x^2}{8\,a^2}-\frac {b^4\,p\,\ln \left (b+a\,x\right )}{4\,a^4}+\frac {b\,p\,x^3}{12\,a}+\frac {b^3\,p\,x}{4\,a^3} \]
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