Integrand size = 23, antiderivative size = 91 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b d^2 n}{25 x^5}-\frac {2 b d e n}{9 x^3}-\frac {b e^2 n}{x}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x} \]
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Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {276, 2372, 12, 14} \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {b d^2 n}{25 x^5}-\frac {2 b d e n}{9 x^3}-\frac {b e^2 n}{x} \]
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Rule 12
Rule 14
Rule 276
Rule 2372
Rubi steps \begin{align*} \text {integral}& = -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-(b n) \int \frac {-3 d^2-10 d e x^2-15 e^2 x^4}{15 x^6} \, dx \\ & = -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {1}{15} (b n) \int \frac {-3 d^2-10 d e x^2-15 e^2 x^4}{x^6} \, dx \\ & = -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {1}{15} (b n) \int \left (-\frac {3 d^2}{x^6}-\frac {10 d e}{x^4}-\frac {15 e^2}{x^2}\right ) \, dx \\ & = -\frac {b d^2 n}{25 x^5}-\frac {2 b d e n}{9 x^3}-\frac {b e^2 n}{x}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {15 a \left (3 d^2+10 d e x^2+15 e^2 x^4\right )+b n \left (9 d^2+50 d e x^2+225 e^2 x^4\right )+15 b \left (3 d^2+10 d e x^2+15 e^2 x^4\right ) \log \left (c x^n\right )}{225 x^5} \]
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Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\frac {225 x^{4} b \ln \left (c \,x^{n}\right ) e^{2}+225 b \,e^{2} n \,x^{4}+225 x^{4} a \,e^{2}+150 b \ln \left (c \,x^{n}\right ) d e \,x^{2}+50 b d e n \,x^{2}+150 a d e \,x^{2}+45 b \ln \left (c \,x^{n}\right ) d^{2}+9 b \,d^{2} n +45 a \,d^{2}}{225 x^{5}}\) | \(97\) |
risch | \(-\frac {b \left (15 e^{2} x^{4}+10 d e \,x^{2}+3 d^{2}\right ) \ln \left (x^{n}\right )}{15 x^{5}}-\frac {-150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+45 i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+450 \ln \left (c \right ) b \,e^{2} x^{4}+450 b \,e^{2} n \,x^{4}+450 x^{4} a \,e^{2}-45 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-45 i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+300 e \ln \left (c \right ) b d \,x^{2}+100 b d e n \,x^{2}+300 a d e \,x^{2}+150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+45 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+90 d^{2} b \ln \left (c \right )+18 b \,d^{2} n +90 a \,d^{2}}{450 x^{5}}\) | \(419\) |
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Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.22 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {225 \, {\left (b e^{2} n + a e^{2}\right )} x^{4} + 9 \, b d^{2} n + 45 \, a d^{2} + 50 \, {\left (b d e n + 3 \, a d e\right )} x^{2} + 15 \, {\left (15 \, b e^{2} x^{4} + 10 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (15 \, b e^{2} n x^{4} + 10 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )}{225 \, x^{5}} \]
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Time = 0.57 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.23 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=- \frac {a d^{2}}{5 x^{5}} - \frac {2 a d e}{3 x^{3}} - \frac {a e^{2}}{x} - \frac {b d^{2} n}{25 x^{5}} - \frac {b d^{2} \log {\left (c x^{n} \right )}}{5 x^{5}} - \frac {2 b d e n}{9 x^{3}} - \frac {2 b d e \log {\left (c x^{n} \right )}}{3 x^{3}} - \frac {b e^{2} n}{x} - \frac {b e^{2} \log {\left (c x^{n} \right )}}{x} \]
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Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b e^{2} n}{x} - \frac {b e^{2} \log \left (c x^{n}\right )}{x} - \frac {a e^{2}}{x} - \frac {2 \, b d e n}{9 \, x^{3}} - \frac {2 \, b d e \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {2 \, a d e}{3 \, x^{3}} - \frac {b d^{2} n}{25 \, x^{5}} - \frac {b d^{2} \log \left (c x^{n}\right )}{5 \, x^{5}} - \frac {a d^{2}}{5 \, x^{5}} \]
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Time = 0.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {{\left (15 \, b e^{2} n x^{4} + 10 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )}{15 \, x^{5}} - \frac {225 \, b e^{2} n x^{4} + 225 \, b e^{2} x^{4} \log \left (c\right ) + 225 \, a e^{2} x^{4} + 50 \, b d e n x^{2} + 150 \, b d e x^{2} \log \left (c\right ) + 150 \, a d e x^{2} + 9 \, b d^{2} n + 45 \, b d^{2} \log \left (c\right ) + 45 \, a d^{2}}{225 \, x^{5}} \]
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Time = 0.39 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {x^4\,\left (15\,a\,e^2+15\,b\,e^2\,n\right )+x^2\,\left (10\,a\,d\,e+\frac {10\,b\,d\,e\,n}{3}\right )+3\,a\,d^2+\frac {3\,b\,d^2\,n}{5}}{15\,x^5}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{5}+\frac {2\,b\,d\,e\,x^2}{3}+b\,e^2\,x^4\right )}{x^5} \]
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