Integrand size = 16, antiderivative size = 89 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {b^4 p x}{5 a^4}+\frac {b^3 p x^2}{10 a^3}-\frac {b^2 p x^3}{15 a^2}+\frac {b p x^4}{20 a}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b^5 p \log (b+a x)}{5 a^5} \]
[Out]
Time = 0.04 (sec) , antiderivative size = 89, 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} \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b^5 p \log (a x+b)}{5 a^5}-\frac {b^4 p x}{5 a^4}+\frac {b^3 p x^2}{10 a^3}-\frac {b^2 p x^3}{15 a^2}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p x^4}{20 a} \]
[In]
[Out]
Rule 45
Rule 269
Rule 2505
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{5} (b p) \int \frac {x^3}{a+\frac {b}{x}} \, dx \\ & = \frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{5} (b p) \int \frac {x^4}{b+a x} \, dx \\ & = \frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{5} (b p) \int \left (-\frac {b^3}{a^4}+\frac {b^2 x}{a^3}-\frac {b x^2}{a^2}+\frac {x^3}{a}+\frac {b^4}{a^4 (b+a x)}\right ) \, dx \\ & = -\frac {b^4 p x}{5 a^4}+\frac {b^3 p x^2}{10 a^3}-\frac {b^2 p x^3}{15 a^2}+\frac {b p x^4}{20 a}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b^5 p \log (b+a x)}{5 a^5} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {a b p x \left (-12 b^3+6 a b^2 x-4 a^2 b x^2+3 a^3 x^3\right )+12 b^5 p \log \left (a+\frac {b}{x}\right )+12 a^5 x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+12 b^5 p \log (x)}{60 a^5} \]
[In]
[Out]
Time = 0.40 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83
method | result | size |
parts | \(\frac {x^{5} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{5}+\frac {p b \left (\frac {\frac {1}{4} a^{3} x^{4}-\frac {1}{3} b \,x^{3} a^{2}+\frac {1}{2} a \,b^{2} x^{2}-x \,b^{3}}{a^{4}}+\frac {b^{4} \ln \left (a x +b \right )}{a^{5}}\right )}{5}\) | \(74\) |
parallelrisch | \(-\frac {-12 x^{5} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{5} p -3 x^{4} a^{4} b \,p^{2}+4 x^{3} a^{3} b^{2} p^{2}-6 x^{2} a^{2} b^{3} p^{2}-12 \ln \left (x \right ) b^{5} p^{2}+12 x a \,b^{4} p^{2}-12 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{5} p -12 b^{5} p^{2}}{60 a^{5} p}\) | \(121\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {12 \, a^{5} p x^{5} \log \left (\frac {a x + b}{x}\right ) + 12 \, a^{5} x^{5} \log \left (c\right ) + 3 \, a^{4} b p x^{4} - 4 \, a^{3} b^{2} p x^{3} + 6 \, a^{2} b^{3} p x^{2} - 12 \, a b^{4} p x + 12 \, b^{5} p \log \left (a x + b\right )}{60 \, a^{5}} \]
[In]
[Out]
Time = 2.51 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.12 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} \frac {x^{5} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{5} + \frac {b p x^{4}}{20 a} - \frac {b^{2} p x^{3}}{15 a^{2}} + \frac {b^{3} p x^{2}}{10 a^{3}} - \frac {b^{4} p x}{5 a^{4}} + \frac {b^{5} p \log {\left (a x + b \right )}}{5 a^{5}} & \text {for}\: a \neq 0 \\\frac {p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) + \frac {1}{60} \, b p {\left (\frac {12 \, b^{4} \log \left (a x + b\right )}{a^{5}} + \frac {3 \, a^{3} x^{4} - 4 \, a^{2} b x^{3} + 6 \, a b^{2} x^{2} - 12 \, b^{3} x}{a^{4}}\right )} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (77) = 154\).
Time = 0.32 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.46 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {\frac {12 \, b^{6} p \log \left (\frac {a x + b}{x}\right )}{a^{5} - \frac {5 \, {\left (a x + b\right )} a^{4}}{x} + \frac {10 \, {\left (a x + b\right )}^{2} a^{3}}{x^{2}} - \frac {10 \, {\left (a x + b\right )}^{3} a^{2}}{x^{3}} + \frac {5 \, {\left (a x + b\right )}^{4} a}{x^{4}} - \frac {{\left (a x + b\right )}^{5}}{x^{5}}} + \frac {12 \, b^{6} p \log \left (-a + \frac {a x + b}{x}\right )}{a^{5}} - \frac {12 \, b^{6} p \log \left (\frac {a x + b}{x}\right )}{a^{5}} - \frac {25 \, a^{4} b^{6} p - 12 \, a^{4} b^{6} \log \left (c\right ) - \frac {77 \, {\left (a x + b\right )} a^{3} b^{6} p}{x} + \frac {94 \, {\left (a x + b\right )}^{2} a^{2} b^{6} p}{x^{2}} - \frac {54 \, {\left (a x + b\right )}^{3} a b^{6} p}{x^{3}} + \frac {12 \, {\left (a x + b\right )}^{4} b^{6} p}{x^{4}}}{a^{9} - \frac {5 \, {\left (a x + b\right )} a^{8}}{x} + \frac {10 \, {\left (a x + b\right )}^{2} a^{7}}{x^{2}} - \frac {10 \, {\left (a x + b\right )}^{3} a^{6}}{x^{3}} + \frac {5 \, {\left (a x + b\right )}^{4} a^{5}}{x^{4}} - \frac {{\left (a x + b\right )}^{5} a^{4}}{x^{5}}}}{60 \, b} \]
[In]
[Out]
Time = 1.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.87 \[ \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {x^5\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{5}-\frac {b^2\,p\,x^3}{15\,a^2}+\frac {b^3\,p\,x^2}{10\,a^3}+\frac {b^5\,p\,\ln \left (b+a\,x\right )}{5\,a^5}+\frac {b\,p\,x^4}{20\,a}-\frac {b^4\,p\,x}{5\,a^4} \]
[In]
[Out]