Integrand size = 21, antiderivative size = 82 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b n}{8 d e \left (d+e x^2\right )}+\frac {b n \log (x)}{4 d^2 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e} \]
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Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2376, 272, 46} \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e}+\frac {b n \log (x)}{4 d^2 e}+\frac {b n}{8 d e \left (d+e x^2\right )} \]
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Rule 46
Rule 272
Rule 2376
Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}+\frac {(b n) \int \frac {1}{x \left (d+e x^2\right )^2} \, dx}{4 e} \\ & = -\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}+\frac {(b n) \text {Subst}\left (\int \frac {1}{x (d+e x)^2} \, dx,x,x^2\right )}{8 e} \\ & = -\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}+\frac {(b n) \text {Subst}\left (\int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx,x,x^2\right )}{8 e} \\ & = \frac {b n}{8 d e \left (d+e x^2\right )}+\frac {b n \log (x)}{4 d^2 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.35 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b n}{8 d e \left (d+e x^2\right )}+\frac {b n \log (x)}{4 d^2 e}-\frac {b n \log (x)}{4 e \left (d+e x^2\right )^2}+\frac {-a-b \left (-n \log (x)+\log \left (c x^n\right )\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e} \]
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Time = 0.56 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(\frac {-\ln \left (e \,x^{2}+d \right ) x^{4} b \,e^{3} n^{2}+2 x^{4} \ln \left (c \,x^{n}\right ) b \,e^{3} n -2 \ln \left (e \,x^{2}+d \right ) x^{2} b d \,e^{2} n^{2}+4 x^{2} \ln \left (c \,x^{n}\right ) b d \,e^{2} n +x^{2} b d \,e^{2} n^{2}-\ln \left (e \,x^{2}+d \right ) b \,d^{2} e \,n^{2}+b \,d^{2} e \,n^{2}-2 a \,d^{2} e n}{8 n \,d^{2} e^{2} \left (e \,x^{2}+d \right )^{2}}\) | \(143\) |
risch | \(-\frac {b \ln \left (x^{n}\right )}{4 e \left (e \,x^{2}+d \right )^{2}}-\frac {\ln \left (e \,x^{2}+d \right ) b \,e^{2} n \,x^{4}-2 \ln \left (x \right ) b \,e^{2} n \,x^{4}-i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 \ln \left (e \,x^{2}+d \right ) b d e n \,x^{2}-4 \ln \left (x \right ) b d e n \,x^{2}-b d e n \,x^{2}+\ln \left (e \,x^{2}+d \right ) b \,d^{2} n -2 \ln \left (x \right ) b \,d^{2} n +2 d^{2} b \ln \left (c \right )-b \,d^{2} n +2 a \,d^{2}}{8 d^{2} e \left (e \,x^{2}+d \right )^{2}}\) | \(243\) |
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Time = 0.29 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.44 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b d e n x^{2} + b d^{2} n - 2 \, b d^{2} \log \left (c\right ) - 2 \, a d^{2} - {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (e x^{2} + d\right ) + 2 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2}\right )} \log \left (x\right )}{8 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (71) = 142\).
Time = 169.10 (sec) , antiderivative size = 619, normalized size of antiderivative = 7.55 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {a}{4 x^{4}} - \frac {b n}{16 x^{4}} - \frac {b \log {\left (c x^{n} \right )}}{4 x^{4}}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {\frac {a x^{2}}{2} - \frac {b n x^{2}}{4} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2}}{d^{3}} & \text {for}\: e = 0 \\\frac {- \frac {a}{4 x^{4}} - \frac {b n}{16 x^{4}} - \frac {b \log {\left (c x^{n} \right )}}{4 x^{4}}}{e^{3}} & \text {for}\: d = 0 \\- \frac {2 a d^{2}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b d^{2} n \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b d^{2} n \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {b d^{2} n}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {2 b d e n x^{2} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {2 b d e n x^{2} \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {b d e n x^{2}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {4 b d e x^{2} \log {\left (c x^{n} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b e^{2} n x^{4} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b e^{2} n x^{4} \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {2 b e^{2} x^{4} \log {\left (c x^{n} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.33 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {1}{8} \, b n {\left (\frac {1}{d e^{2} x^{2} + d^{2} e} - \frac {\log \left (e x^{2} + d\right )}{d^{2} e} + \frac {\log \left (x^{2}\right )}{d^{2} e}\right )} - \frac {b \log \left (c x^{n}\right )}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} - \frac {a}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} \]
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Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.41 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {b d n \log \left (x\right )}{4 \, {\left (d e^{3} x^{4} + 2 \, d^{2} e^{2} x^{2} + d^{3} e\right )}} + \frac {b e n x^{2} + b d n - 2 \, b d \log \left (c\right ) - 2 \, a d}{8 \, {\left (d e^{3} x^{4} + 2 \, d^{2} e^{2} x^{2} + d^{3} e\right )}} - \frac {b n \log \left (e x^{2} + d\right )}{8 \, d^{2} e} + \frac {b n \log \left (x\right )}{4 \, d^{2} e} \]
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Time = 0.41 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.33 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {\frac {b\,n}{2}-a+\frac {b\,e\,n\,x^2}{2\,d}}{4\,d^2\,e+8\,d\,e^2\,x^2+4\,e^3\,x^4}-\frac {b\,\ln \left (c\,x^n\right )}{4\,e\,\left (d^2+2\,d\,e\,x^2+e^2\,x^4\right )}-\frac {b\,n\,\ln \left (e\,x^2+d\right )}{8\,d^2\,e}+\frac {b\,n\,\ln \left (x\right )}{4\,d^2\,e} \]
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