Integrand size = 15, antiderivative size = 137 \[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {1}{4} i e^{-\frac {2 a}{b n}} x^2 \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {(2-i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )+\frac {1}{4} i e^{-\frac {2 a}{b n}} x^2 \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {(2+i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )+\frac {1}{2} x^2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \]
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Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6661, 12, 4585, 2347, 2209} \[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {1}{4} i x^2 e^{-\frac {2 a}{b n}} \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {(2-i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )+\frac {1}{4} i x^2 e^{-\frac {2 a}{b n}} \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {(i b d n+2) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )+\frac {1}{2} x^2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \]
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Rule 12
Rule 2209
Rule 2347
Rule 4585
Rule 6661
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )-\frac {1}{2} (b d n) \int \frac {x \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{d \left (a+b \log \left (c x^n\right )\right )} \, dx \\ & = \frac {1}{2} x^2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )-\frac {1}{2} (b n) \int \frac {x \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{a+b \log \left (c x^n\right )} \, dx \\ & = \frac {1}{2} x^2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )-\frac {1}{4} \left (i b e^{-i a d} n x^{i b d n} \left (c x^n\right )^{-i b d}\right ) \int \frac {x^{1-i b d n}}{a+b \log \left (c x^n\right )} \, dx+\frac {1}{4} \left (i b e^{i a d} n x^{-i b d n} \left (c x^n\right )^{i b d}\right ) \int \frac {x^{1+i b d n}}{a+b \log \left (c x^n\right )} \, dx \\ & = \frac {1}{2} x^2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )-\frac {1}{4} \left (i b e^{-i a d} x^2 \left (c x^n\right )^{-i b d-\frac {2-i b d n}{n}}\right ) \text {Subst}\left (\int \frac {e^{\frac {(2-i b d n) x}{n}}}{a+b x} \, dx,x,\log \left (c x^n\right )\right )+\frac {1}{4} \left (i b e^{i a d} x^2 \left (c x^n\right )^{i b d-\frac {2+i b d n}{n}}\right ) \text {Subst}\left (\int \frac {e^{\frac {(2+i b d n) x}{n}}}{a+b x} \, dx,x,\log \left (c x^n\right )\right ) \\ & = -\frac {1}{4} i e^{-\frac {2 a}{b n}} x^2 \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {(2-i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )+\frac {1}{4} i e^{-\frac {2 a}{b n}} x^2 \left (c x^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {(2+i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )+\frac {1}{2} x^2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \\ \end{align*}
Time = 1.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77 \[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {1}{4} x^2 \left (-i e^{-\frac {2 a}{b n}} \left (c x^n\right )^{-2/n} \left (\operatorname {ExpIntegralEi}\left (\frac {(2-i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )-\operatorname {ExpIntegralEi}\left (\frac {(2+i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )\right )+2 \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )\right ) \]
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\[\int x \,\operatorname {Si}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )d x\]
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none
Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.02 \[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {1}{2} \, x^{2} \operatorname {Si}\left (b d \log \left (c x^{n}\right ) + a d\right ) + \frac {1}{4} \, {\left (i \, {\rm Ei}\left (\frac {i \, a b d n + {\left (i \, b^{2} d n + 2 \, b\right )} \log \left (c\right ) + {\left (i \, b^{2} d n^{2} + 2 \, b n\right )} \log \left (x\right ) + 2 \, a}{b n}\right ) - i \, {\rm Ei}\left (\frac {-i \, a b d n + {\left (-i \, b^{2} d n + 2 \, b\right )} \log \left (c\right ) + {\left (-i \, b^{2} d n^{2} + 2 \, b n\right )} \log \left (x\right ) + 2 \, a}{b n}\right )\right )} e^{\left (-\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} \]
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\[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x \operatorname {Si}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]
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\[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { x \operatorname {Si}\left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \]
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Timed out. \[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]
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Timed out. \[ \int x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int x\,\mathrm {sinint}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right ) \,d x \]
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