Optimal. Leaf size=50 \[ \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac {b n \log \left (d+e x^2\right )}{4 d e} \]
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Rubi [A] time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2335, 260} \[ \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac {b n \log \left (d+e x^2\right )}{4 d e} \]
Antiderivative was successfully verified.
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Rule 260
Rule 2335
Rubi steps
\begin {align*} \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx &=\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac {(b n) \int \frac {x}{d+e x^2} \, dx}{2 d}\\ &=\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac {b n \log \left (d+e x^2\right )}{4 d e}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 74, normalized size = 1.48 \[ -\frac {2 a d+2 b d \log \left (c x^n\right )+b e n x^2 \log \left (d+e x^2\right )-2 b n \log (x) \left (d+e x^2\right )+b d n \log \left (d+e x^2\right )}{4 d e \left (d+e x^2\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.74, size = 61, normalized size = 1.22 \[ \frac {2 \, b e n x^{2} \log \relax (x) - 2 \, b d \log \relax (c) - 2 \, a d - {\left (b e n x^{2} + b d n\right )} \log \left (e x^{2} + d\right )}{4 \, {\left (d e^{2} x^{2} + d^{2} e\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 70, normalized size = 1.40 \[ -\frac {b n x^{2} e \log \left (x^{2} e + d\right ) - 2 \, b n x^{2} e \log \relax (x) + b d n \log \left (x^{2} e + d\right ) + 2 \, b d \log \relax (c) + 2 \, a d}{4 \, {\left (d x^{2} e^{2} + d^{2} e\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.20, size = 179, normalized size = 3.58 \[ -\frac {b \ln \left (x^{n}\right )}{2 \left (e \,x^{2}+d \right ) e}-\frac {-2 b e n \,x^{2} \ln \relax (x )+b e n \,x^{2} \ln \left (e \,x^{2}+d \right )-i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b d \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b d \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-2 b d n \ln \relax (x )+b d n \ln \left (e \,x^{2}+d \right )+2 b d \ln \relax (c )+2 a d}{4 \left (e \,x^{2}+d \right ) d e} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 71, normalized size = 1.42 \[ -\frac {1}{4} \, b n {\left (\frac {\log \left (e x^{2} + d\right )}{d e} - \frac {\log \left (x^{2}\right )}{d e}\right )} - \frac {b \log \left (c x^{n}\right )}{2 \, {\left (e^{2} x^{2} + d e\right )}} - \frac {a}{2 \, {\left (e^{2} x^{2} + d e\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.50, size = 73, normalized size = 1.46 \[ \frac {b\,n\,\ln \relax (x)}{2\,d\,e}-\frac {b\,\ln \left (c\,x^n\right )}{2\,\left (e^2\,x^2+d\,e\right )}-\frac {b\,n\,\ln \left (e\,x^2+d\right )}{4\,d\,e}-\frac {a}{2\,e^2\,x^2+2\,d\,e} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 59.95, size = 366, normalized size = 7.32 \[ \begin {cases} \tilde {\infty } \left (- \frac {a}{2 x^{2}} - \frac {b n \log {\relax (x )}}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\relax (c )}}{2 x^{2}}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {- \frac {a}{2 x^{2}} - \frac {b n \log {\relax (x )}}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\relax (c )}}{2 x^{2}}}{e^{2}} & \text {for}\: d = 0 \\\frac {\frac {a x^{2}}{2} + \frac {b n x^{2} \log {\relax (x )}}{2} - \frac {b n x^{2}}{4} + \frac {b x^{2} \log {\relax (c )}}{2}}{d^{2}} & \text {for}\: e = 0 \\- \frac {2 a d}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac {b d n \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac {b d n \log {\left (i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} + \frac {2 b e n x^{2} \log {\relax (x )}}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac {b e n x^{2} \log {\left (- i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac {b e n x^{2} \log {\left (i \sqrt {d} \sqrt {\frac {1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} + \frac {2 b e x^{2} \log {\relax (c )}}{4 d^{2} e + 4 d e^{2} x^{2}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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