Optimal. Leaf size=91 \[ \frac {i \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}-\frac {\cot (a+b x)}{2 b^2}-\frac {x \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {x \cot ^2(a+b x)}{2 b}-\frac {x}{2 b}+\frac {i x^2}{2} \]
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Rubi [A] time = 0.11, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3720, 3473, 8, 3717, 2190, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {\cot (a+b x)}{2 b^2}-\frac {x \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {x \cot ^2(a+b x)}{2 b}-\frac {x}{2 b}+\frac {i x^2}{2} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2190
Rule 2279
Rule 2391
Rule 3473
Rule 3717
Rule 3720
Rubi steps
\begin {align*} \int x \cot ^3(a+b x) \, dx &=-\frac {x \cot ^2(a+b x)}{2 b}+\frac {\int \cot ^2(a+b x) \, dx}{2 b}-\int x \cot (a+b x) \, dx\\ &=\frac {i x^2}{2}-\frac {\cot (a+b x)}{2 b^2}-\frac {x \cot ^2(a+b x)}{2 b}+2 i \int \frac {e^{2 i (a+b x)} x}{1-e^{2 i (a+b x)}} \, dx-\frac {\int 1 \, dx}{2 b}\\ &=-\frac {x}{2 b}+\frac {i x^2}{2}-\frac {\cot (a+b x)}{2 b^2}-\frac {x \cot ^2(a+b x)}{2 b}-\frac {x \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {\int \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {x}{2 b}+\frac {i x^2}{2}-\frac {\cot (a+b x)}{2 b^2}-\frac {x \cot ^2(a+b x)}{2 b}-\frac {x \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^2}\\ &=-\frac {x}{2 b}+\frac {i x^2}{2}-\frac {\cot (a+b x)}{2 b^2}-\frac {x \cot ^2(a+b x)}{2 b}-\frac {x \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {i \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}\\ \end {align*}
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Mathematica [A] time = 4.30, size = 179, normalized size = 1.97 \[ \frac {-b^2 x^2 \cot (a)+b^2 x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt {\sec ^2(a)}+i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )-b x \csc ^2(a+b x)-2 b x \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+\csc (a) \sin (b x) \csc (a+b x)+2 \tan ^{-1}(\tan (a)) \left (-\log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+\log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+i b x\right )-i \pi b x-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))}{2 b^2} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.66, size = 291, normalized size = 3.20 \[ \frac {4 \, b x + {\left (i \, \cos \left (2 \, b x + 2 \, a\right ) - i\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + {\left (-i \, \cos \left (2 \, b x + 2 \, a\right ) + i\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \, {\left (a \cos \left (2 \, b x + 2 \, a\right ) - a\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (a \cos \left (2 \, b x + 2 \, a\right ) - a\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b x - {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + a\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left (b x - {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + a\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, \sin \left (2 \, b x + 2 \, a\right )}{4 \, {\left (b^{2} \cos \left (2 \, b x + 2 \, a\right ) - b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cot \left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.80, size = 197, normalized size = 2.16 \[ \frac {i x^{2}}{2}+\frac {2 b x \,{\mathrm e}^{2 i \left (b x +a \right )}-i {\mathrm e}^{2 i \left (b x +a \right )}+i}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {2 i a x}{b}+\frac {i a^{2}}{b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {i \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {i \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.88, size = 591, normalized size = 6.49 \[ \frac {b^{2} x^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, b^{2} x^{2} \sin \left (4 \, b x + 4 \, a\right ) + b^{2} x^{2} - {\left (2 \, b x \cos \left (4 \, b x + 4 \, a\right ) - 4 \, b x \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b x \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, b x \sin \left (2 \, b x + 2 \, a\right ) + 2 \, b x\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + {\left (2 \, b x \cos \left (4 \, b x + 4 \, a\right ) - 4 \, b x \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b x \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, b x \sin \left (2 \, b x + 2 \, a\right ) + 2 \, b x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 2 \, {\left (b^{2} x^{2} + 2 i \, b x + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (2 \, \cos \left (4 \, b x + 4 \, a\right ) - 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + {\left (2 \, \cos \left (4 \, b x + 4 \, a\right ) - 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - {\left (-i \, b x \cos \left (4 \, b x + 4 \, a\right ) + 2 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + b x \sin \left (4 \, b x + 4 \, a\right ) - 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) - i \, b x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (-i \, b x \cos \left (4 \, b x + 4 \, a\right ) + 2 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + b x \sin \left (4 \, b x + 4 \, a\right ) - 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) - i \, b x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (2 i \, b^{2} x^{2} - 4 \, b x + 2 i\right )} \sin \left (2 \, b x + 2 \, a\right ) + 2}{-2 i \, b^{2} \cos \left (4 \, b x + 4 \, a\right ) + 4 i \, b^{2} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, b^{2} \sin \left (4 \, b x + 4 \, a\right ) - 4 \, b^{2} \sin \left (2 \, b x + 2 \, a\right ) - 2 i \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {cot}\left (a+b\,x\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cot ^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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