Integrand size = 18, antiderivative size = 99 \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-2 d p x-\frac {1}{2} e p x^2+\frac {2 \sqrt {a} d p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {\left (b d^2-a e^2\right ) p \log \left (a+b x^2\right )}{2 b e}+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e} \]
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Time = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2513, 815, 649, 211, 266} \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {2 \sqrt {a} d p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}-\frac {p \left (b d^2-a e^2\right ) \log \left (a+b x^2\right )}{2 b e}-2 d p x-\frac {1}{2} e p x^2 \]
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Rule 211
Rule 266
Rule 649
Rule 815
Rule 2513
Rubi steps \begin{align*} \text {integral}& = \frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}-\frac {(b p) \int \frac {x (d+e x)^2}{a+b x^2} \, dx}{e} \\ & = \frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}-\frac {(b p) \int \left (\frac {2 d e}{b}+\frac {e^2 x}{b}-\frac {2 a d e-\left (b d^2-a e^2\right ) x}{b \left (a+b x^2\right )}\right ) \, dx}{e} \\ & = -2 d p x-\frac {1}{2} e p x^2+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}+\frac {p \int \frac {2 a d e-\left (b d^2-a e^2\right ) x}{a+b x^2} \, dx}{e} \\ & = -2 d p x-\frac {1}{2} e p x^2+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e}+(2 a d p) \int \frac {1}{a+b x^2} \, dx+\frac {\left (\left (-b d^2+a e^2\right ) p\right ) \int \frac {x}{a+b x^2} \, dx}{e} \\ & = -2 d p x-\frac {1}{2} e p x^2+\frac {2 \sqrt {a} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {\left (b d^2-a e^2\right ) p \log \left (a+b x^2\right )}{2 b e}+\frac {(d+e x)^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 e} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=-2 d p x+\frac {2 \sqrt {a} d p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}+d x \log \left (c \left (a+b x^2\right )^p\right )+\frac {1}{2} e \left (-p x^2+\frac {\left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b}\right ) \]
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Time = 0.48 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.79
method | result | size |
default | \(d \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) x -2 d p x +\frac {2 d p a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}+\frac {e \left (\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) \left (b \,x^{2}+a \right )-\left (b \,x^{2}+a \right ) p \right )}{2 b}\) | \(78\) |
parts | \(\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) e \,x^{2}}{2}+d \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) x -p b \left (\frac {\frac {1}{2} e \,x^{2}+2 d x}{b}-\frac {a \left (\frac {e \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {2 d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{b}\right )\) | \(93\) |
risch | \(\left (\frac {1}{2} e \,x^{2}+d x \right ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right )-\frac {i x \pi d \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}+\frac {i \operatorname {csgn}\left (i c \right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} x^{2} e \pi }{4}-\frac {i x \pi d {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{2}-\frac {i \pi e \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{4}+\frac {i x \pi d {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) x^{2} e \pi }{4}+\frac {i x \pi d \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi e \,x^{2} \operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{4}+\frac {\ln \left (c \right ) e \,x^{2}}{2}-\frac {e p \,x^{2}}{2}+x \ln \left (c \right ) d +\frac {p \ln \left (-\sqrt {-a b}\, x +a \right ) d \sqrt {-a b}}{b}-\frac {p \ln \left (\sqrt {-a b}\, x +a \right ) d \sqrt {-a b}}{b}+\frac {p \ln \left (-\sqrt {-a b}\, x +a \right ) a e}{2 b}+\frac {p \ln \left (\sqrt {-a b}\, x +a \right ) a e}{2 b}-2 d p x\) | \(387\) |
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Time = 0.34 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.00 \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\left [-\frac {b e p x^{2} - 2 \, b d p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 4 \, b d p x - {\left (b e p x^{2} + 2 \, b d p x + a e p\right )} \log \left (b x^{2} + a\right ) - {\left (b e x^{2} + 2 \, b d x\right )} \log \left (c\right )}{2 \, b}, -\frac {b e p x^{2} - 4 \, b d p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 4 \, b d p x - {\left (b e p x^{2} + 2 \, b d p x + a e p\right )} \log \left (b x^{2} + a\right ) - {\left (b e x^{2} + 2 \, b d x\right )} \log \left (c\right )}{2 \, b}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (92) = 184\).
Time = 4.32 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.01 \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} \left (d x + \frac {e x^{2}}{2}\right ) \log {\left (0^{p} c \right )} & \text {for}\: a = 0 \wedge b = 0 \\\left (d x + \frac {e x^{2}}{2}\right ) \log {\left (a^{p} c \right )} & \text {for}\: b = 0 \\- 2 d p x + d x \log {\left (c \left (b x^{2}\right )^{p} \right )} - \frac {e p x^{2}}{2} + \frac {e x^{2} \log {\left (c \left (b x^{2}\right )^{p} \right )}}{2} & \text {for}\: a = 0 \\\frac {2 a d p \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{b \sqrt {- \frac {a}{b}}} - \frac {a d \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{b \sqrt {- \frac {a}{b}}} + \frac {a e \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2 b} - 2 d p x + d x \log {\left (c \left (a + b x^{2}\right )^{p} \right )} - \frac {e p x^{2}}{2} + \frac {e x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{2} & \text {otherwise} \end {cases} \]
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Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.81 \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{2} \, {\left (\frac {4 \, a d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {a e \log \left (b x^{2} + a\right )}{b^{2}} - \frac {e x^{2} + 4 \, d x}{b}\right )} b p + \frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \]
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Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.86 \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {2 \, a d p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}} - \frac {1}{2} \, {\left (e p - e \log \left (c\right )\right )} x^{2} + \frac {a e p \log \left (b x^{2} + a\right )}{2 \, b} - {\left (2 \, d p - d \log \left (c\right )\right )} x + \frac {1}{2} \, {\left (e p x^{2} + 2 \, d p x\right )} \log \left (b x^{2} + a\right ) \]
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Time = 2.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.82 \[ \int (d+e x) \log \left (c \left (a+b x^2\right )^p\right ) \, dx=d\,x\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )-\frac {e\,p\,x^2}{2}-2\,d\,p\,x+\frac {e\,x^2\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{2}+\frac {a\,e\,p\,\ln \left (b\,x^2+a\right )}{2\,b}+\frac {2\,\sqrt {a}\,d\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{\sqrt {b}} \]
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