Integrand size = 23, antiderivative size = 154 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {2 d f p x}{3 e}-\frac {2 d^2 g p x}{5 e^2}-\frac {2}{9} f p x^3+\frac {2 d g p x^3}{15 e}-\frac {2}{25} g p x^5-\frac {2 d^{3/2} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right ) \]
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Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2526, 2505, 308, 211} \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {2 d^{3/2} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {2 d^2 g p x}{5 e^2}+\frac {2 d f p x}{3 e}+\frac {2 d g p x^3}{15 e}-\frac {2}{9} f p x^3-\frac {2}{25} g p x^5 \]
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Rule 211
Rule 308
Rule 2505
Rule 2526
Rubi steps \begin{align*} \text {integral}& = \int \left (f x^2 \log \left (c \left (d+e x^2\right )^p\right )+g x^4 \log \left (c \left (d+e x^2\right )^p\right )\right ) \, dx \\ & = f \int x^2 \log \left (c \left (d+e x^2\right )^p\right ) \, dx+g \int x^4 \log \left (c \left (d+e x^2\right )^p\right ) \, dx \\ & = \frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{3} (2 e f p) \int \frac {x^4}{d+e x^2} \, dx-\frac {1}{5} (2 e g p) \int \frac {x^6}{d+e x^2} \, dx \\ & = \frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {1}{3} (2 e f p) \int \left (-\frac {d}{e^2}+\frac {x^2}{e}+\frac {d^2}{e^2 \left (d+e x^2\right )}\right ) \, dx-\frac {1}{5} (2 e g p) \int \left (\frac {d^2}{e^3}-\frac {d x^2}{e^2}+\frac {x^4}{e}-\frac {d^3}{e^3 \left (d+e x^2\right )}\right ) \, dx \\ & = \frac {2 d f p x}{3 e}-\frac {2 d^2 g p x}{5 e^2}-\frac {2}{9} f p x^3+\frac {2 d g p x^3}{15 e}-\frac {2}{25} g p x^5+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right )-\frac {\left (2 d^2 f p\right ) \int \frac {1}{d+e x^2} \, dx}{3 e}+\frac {\left (2 d^3 g p\right ) \int \frac {1}{d+e x^2} \, dx}{5 e^2} \\ & = \frac {2 d f p x}{3 e}-\frac {2 d^2 g p x}{5 e^2}-\frac {2}{9} f p x^3+\frac {2 d g p x^3}{15 e}-\frac {2}{25} g p x^5-\frac {2 d^{3/2} f p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{3 e^{3/2}}+\frac {2 d^{5/2} g p \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{5 e^{5/2}}+\frac {1}{3} f x^3 \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{5} g x^5 \log \left (c \left (d+e x^2\right )^p\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {30 d^{3/2} (-5 e f+3 d g) p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )+\sqrt {e} x \left (-2 p \left (45 d^2 g-15 d e \left (5 f+g x^2\right )+e^2 x^2 \left (25 f+9 g x^2\right )\right )+15 e^2 x^2 \left (5 f+3 g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )}{225 e^{5/2}} \]
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Time = 1.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77
method | result | size |
parts | \(\frac {g \,x^{5} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{5}+\frac {f \,x^{3} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{3}-\frac {2 p e \left (\frac {\frac {3}{5} e^{2} g \,x^{5}-d g \,x^{3} e +\frac {5}{3} f \,x^{3} e^{2}+3 d^{2} g x -5 d e f x}{e^{3}}-\frac {d^{2} \left (3 d g -5 e f \right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{e^{3} \sqrt {d e}}\right )}{15}\) | \(118\) |
risch | \(\left (\frac {1}{5} g \,x^{5}+\frac {1}{3} f \,x^{3}\right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )-\frac {i \pi g \,x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}+\frac {i \pi g \,x^{5} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{10}+\frac {i \pi f \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{6}-\frac {i \pi f \,x^{3} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{6}+\frac {i \pi f \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{6}-\frac {i \pi g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{10}+\frac {i \pi g \,x^{5} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi f \,x^{3} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{6}+\frac {\ln \left (c \right ) g \,x^{5}}{5}-\frac {2 g p \,x^{5}}{25}+\frac {\ln \left (c \right ) f \,x^{3}}{3}+\frac {2 d g p \,x^{3}}{15 e}-\frac {2 f p \,x^{3}}{9}+\frac {\sqrt {-d e}\, p \,d^{2} \ln \left (-\sqrt {-d e}\, x +d \right ) g}{5 e^{3}}-\frac {\sqrt {-d e}\, p d \ln \left (-\sqrt {-d e}\, x +d \right ) f}{3 e^{2}}-\frac {\sqrt {-d e}\, p \,d^{2} \ln \left (\sqrt {-d e}\, x +d \right ) g}{5 e^{3}}+\frac {\sqrt {-d e}\, p d \ln \left (\sqrt {-d e}\, x +d \right ) f}{3 e^{2}}-\frac {2 d^{2} g p x}{5 e^{2}}+\frac {2 d f p x}{3 e}\) | \(453\) |
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Time = 0.34 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.95 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {18 \, e^{2} g p x^{5} + 10 \, {\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 15 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) - 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \, {\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}, -\frac {18 \, e^{2} g p x^{5} + 10 \, {\left (5 \, e^{2} f - 3 \, d e g\right )} p x^{3} + 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) - 30 \, {\left (5 \, d e f - 3 \, d^{2} g\right )} p x - 15 \, {\left (3 \, e^{2} g p x^{5} + 5 \, e^{2} f p x^{3}\right )} \log \left (e x^{2} + d\right ) - 15 \, {\left (3 \, e^{2} g x^{5} + 5 \, e^{2} f x^{3}\right )} \log \left (c\right )}{225 \, e^{2}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (158) = 316\).
Time = 31.34 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.08 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \left (\frac {f x^{3}}{3} + \frac {g x^{5}}{5}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (\frac {f x^{3}}{3} + \frac {g x^{5}}{5}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- \frac {2 f p x^{3}}{9} + \frac {f x^{3} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{3} - \frac {2 g p x^{5}}{25} + \frac {g x^{5} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{5} & \text {for}\: d = 0 \\\frac {2 d^{3} g p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {d^{3} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5 e^{3} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} f p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} + \frac {d^{2} f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3 e^{2} \sqrt {- \frac {d}{e}}} - \frac {2 d^{2} g p x}{5 e^{2}} + \frac {2 d f p x}{3 e} + \frac {2 d g p x^{3}}{15 e} - \frac {2 f p x^{3}}{9} + \frac {f x^{3} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{3} - \frac {2 g p x^{5}}{25} + \frac {g x^{5} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{5} & \text {otherwise} \end {cases} \]
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Exception generated. \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \]
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Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{25} \, {\left (2 \, g p - 5 \, g \log \left (c\right )\right )} x^{5} - \frac {{\left (10 \, e f p - 6 \, d g p - 15 \, e f \log \left (c\right )\right )} x^{3}}{45 \, e} + \frac {1}{15} \, {\left (3 \, g p x^{5} + 5 \, f p x^{3}\right )} \log \left (e x^{2} + d\right ) + \frac {2 \, {\left (5 \, d e f p - 3 \, d^{2} g p\right )} x}{15 \, e^{2}} - \frac {2 \, {\left (5 \, d^{2} e f p - 3 \, d^{3} g p\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{15 \, \sqrt {d e} e^{2}} \]
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Time = 1.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int x^2 \left (f+g x^2\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )\,\left (\frac {g\,x^5}{5}+\frac {f\,x^3}{3}\right )-x^3\,\left (\frac {2\,f\,p}{9}-\frac {2\,d\,g\,p}{15\,e}\right )-\frac {2\,g\,p\,x^5}{25}+\frac {d\,x\,\left (\frac {2\,f\,p}{3}-\frac {2\,d\,g\,p}{5\,e}\right )}{e}+\frac {2\,d^{3/2}\,p\,\mathrm {atan}\left (\frac {d^{3/2}\,\sqrt {e}\,p\,x\,\left (3\,d\,g-5\,e\,f\right )}{3\,d^3\,g\,p-5\,d^2\,e\,f\,p}\right )\,\left (3\,d\,g-5\,e\,f\right )}{15\,e^{5/2}} \]
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