Integrand size = 16, antiderivative size = 80 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {2 a^2 p x}{5 b^2}+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25}+\frac {2 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right ) \]
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Time = 0.03 (sec) , antiderivative size = 80, 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, 308, 211} \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {2 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}-\frac {2 a^2 p x}{5 b^2}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25} \]
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Rule 211
Rule 308
Rule 2505
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{5} (2 b p) \int \frac {x^6}{a+b x^2} \, dx \\ & = \frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{5} (2 b p) \int \left (\frac {a^2}{b^3}-\frac {a x^2}{b^2}+\frac {x^4}{b}-\frac {a^3}{b^3 \left (a+b x^2\right )}\right ) \, dx \\ & = -\frac {2 a^2 p x}{5 b^2}+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {\left (2 a^3 p\right ) \int \frac {1}{a+b x^2} \, dx}{5 b^2} \\ & = -\frac {2 a^2 p x}{5 b^2}+\frac {2 a p x^3}{15 b}-\frac {2 p x^5}{25}+\frac {2 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^2\right )^p\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{75} \left (-\frac {30 a^2 p x}{b^2}+\frac {10 a p x^3}{b}-6 p x^5+\frac {30 a^{5/2} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}+15 x^5 \log \left (c \left (a+b x^2\right )^p\right )\right ) \]
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Time = 0.65 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89
method | result | size |
parts | \(\frac {x^{5} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{5}-\frac {2 p b \left (\frac {\frac {1}{5} x^{5} b^{2}-\frac {1}{3} a b \,x^{3}+a^{2} x}{b^{3}}-\frac {a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\right )}{5}\) | \(71\) |
risch | \(\frac {x^{5} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{5}-\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{10}+\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}+\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{10}-\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}+\frac {\ln \left (c \right ) x^{5}}{5}-\frac {2 p \,x^{5}}{25}+\frac {2 a p \,x^{3}}{15 b}+\frac {\sqrt {-a b}\, a^{2} p \ln \left (-\sqrt {-a b}\, x +a \right )}{5 b^{3}}-\frac {\sqrt {-a b}\, a^{2} p \ln \left (\sqrt {-a b}\, x +a \right )}{5 b^{3}}-\frac {2 a^{2} p x}{5 b^{2}}\) | \(229\) |
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Time = 0.32 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.35 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\left [\frac {15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 15 \, a^{2} p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}, \frac {15 \, b^{2} p x^{5} \log \left (b x^{2} + a\right ) - 6 \, b^{2} p x^{5} + 15 \, b^{2} x^{5} \log \left (c\right ) + 10 \, a b p x^{3} + 30 \, a^{2} p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 30 \, a^{2} p x}{75 \, b^{2}}\right ] \]
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Time = 31.80 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.95 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \frac {x^{5} \log {\left (0^{p} c \right )}}{5} & \text {for}\: a = 0 \wedge b = 0 \\\frac {x^{5} \log {\left (a^{p} c \right )}}{5} & \text {for}\: b = 0 \\- \frac {2 p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{5} & \text {for}\: a = 0 \\\frac {2 a^{3} p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{5 b^{3} \sqrt {- \frac {a}{b}}} - \frac {a^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{5 b^{3} \sqrt {- \frac {a}{b}}} - \frac {2 a^{2} p x}{5 b^{2}} + \frac {2 a p x^{3}}{15 b} - \frac {2 p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \]
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Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) + \frac {2}{75} \, b p {\left (\frac {15 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {3 \, b^{2} x^{5} - 5 \, a b x^{3} + 15 \, a^{2} x}{b^{3}}\right )} \]
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Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{5} \, p x^{5} \log \left (b x^{2} + a\right ) - \frac {1}{25} \, {\left (2 \, p - 5 \, \log \left (c\right )\right )} x^{5} + \frac {2 \, a p x^{3}}{15 \, b} + \frac {2 \, a^{3} p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{5 \, \sqrt {a b} b^{2}} - \frac {2 \, a^{2} p x}{5 \, b^{2}} \]
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Time = 1.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {x^5\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{5}-\frac {2\,p\,x^5}{25}+\frac {2\,a^{5/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{5\,b^{5/2}}+\frac {2\,a\,p\,x^3}{15\,b}-\frac {2\,a^2\,p\,x}{5\,b^2} \]
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