Integrand size = 16, antiderivative size = 159 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {\sqrt {3} a^{5/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 b^{5/3}}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right ) \]
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Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2505, 308, 298, 31, 648, 631, 210, 642} \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {\sqrt {3} a^{5/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )+\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25} \]
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Rule 31
Rule 210
Rule 298
Rule 308
Rule 631
Rule 642
Rule 648
Rule 2505
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{5} (3 b p) \int \frac {x^7}{a+b x^3} \, dx \\ & = \frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{5} (3 b p) \int \left (-\frac {a x}{b^2}+\frac {x^4}{b}+\frac {a^2 x}{b^2 \left (a+b x^3\right )}\right ) \, dx \\ & = \frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (3 a^2 p\right ) \int \frac {x}{a+b x^3} \, dx}{5 b} \\ & = \frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )+\frac {\left (a^{5/3} p\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{5 b^{4/3}}-\frac {\left (a^{5/3} p\right ) \int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{5 b^{4/3}} \\ & = \frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (a^{5/3} p\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{10 b^{5/3}}-\frac {\left (3 a^2 p\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{10 b^{4/3}} \\ & = \frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {\left (3 a^{5/3} p\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{5 b^{5/3}} \\ & = \frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {\sqrt {3} a^{5/3} p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 b^{5/3}}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right ) \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.00 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.43 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}-\frac {3 a p x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )}{10 b}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right ) \]
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Time = 0.87 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87
method | result | size |
parts | \(\frac {x^{5} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{5}-\frac {3 p b \left (-\frac {-\frac {1}{5} b \,x^{5}+\frac {1}{2} x^{2} a}{b^{2}}+\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) a^{2}}{b^{2}}\right )}{5}\) | \(139\) |
risch | \(\frac {x^{5} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{5}+\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{10}-\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{10}+\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}+\frac {\ln \left (c \right ) x^{5}}{5}-\frac {3 p \,x^{5}}{25}+\frac {3 a p \,x^{2}}{10 b}-\frac {a^{2} p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{5 b^{2}}\) | \(196\) |
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Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.01 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {10 \, b p x^{5} \log \left (b x^{3} + a\right ) - 6 \, b p x^{5} + 10 \, b x^{5} \log \left (c\right ) + 15 \, a p x^{2} - 10 \, \sqrt {3} a p \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - 5 \, a p \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + 10 \, a p \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{50 \, b} \]
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Timed out. \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\text {Timed out} \]
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Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) - \frac {1}{50} \, b p {\left (\frac {10 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, a^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {10 \, a^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {3 \, {\left (2 \, b x^{5} - 5 \, a x^{2}\right )}}{b^{2}}\right )} \]
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Time = 0.33 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{10} \, a^{2} b^{4} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{5}} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{7}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{7}}\right )} + \frac {1}{5} \, p x^{5} \log \left (b x^{3} + a\right ) - \frac {1}{25} \, {\left (3 \, p - 5 \, \log \left (c\right )\right )} x^{5} + \frac {3 \, a p x^{2}}{10 \, b} \]
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Time = 3.71 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {x^5\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{5}-\frac {3\,p\,x^5}{25}+\frac {a^{5/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{5\,b^{5/3}}+\frac {3\,a\,p\,x^2}{10\,b}+\frac {a^{5/3}\,p\,\ln \left (\frac {9\,a^4\,p^2\,x}{25\,b}+\frac {9\,a^{13/3}\,p^2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{25\,b^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,b^{5/3}}-\frac {a^{5/3}\,p\,\ln \left (\frac {9\,a^4\,p^2\,x}{25\,b}+\frac {9\,a^{13/3}\,p^2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{25\,b^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,b^{5/3}} \]
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