Optimal. Leaf size=124 \[ -\frac {i \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{8 b^2}-\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {b x^3}{6} \]
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Rubi [A] time = 0.24, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5172, 2184, 2190, 2531, 2282, 6589} \[ -\frac {i \text {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{8 b^2}-\frac {x \text {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {b x^3}{6} \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2282
Rule 2531
Rule 5172
Rule 6589
Rubi steps
\begin {align*} \int x \cot ^{-1}(c-(1-i c) \tan (a+b x)) \, dx &=\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))+\frac {1}{2} (i b) \int \frac {x^2}{i (-1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))+\frac {1}{2} (b c) \int \frac {e^{2 i a+2 i b x} x^2}{i (-1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{2} i \int x \log \left (1+\frac {c e^{2 i a+2 i b x}}{i (-1+i c)+c}\right ) \, dx\\ &=-\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {\int \text {Li}_2\left (-\frac {c e^{2 i a+2 i b x}}{i (-1+i c)+c}\right ) \, dx}{4 b}\\ &=-\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i c x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^2}\\ &=-\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {i \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{8 b^2}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 111, normalized size = 0.90 \[ \frac {1}{2} x^2 \cot ^{-1}(c+i (c+i) \tan (a+b x))-\frac {i \left (2 b^2 x^2 \log \left (1-\frac {i e^{-2 i (a+b x)}}{c}\right )+2 i b x \text {Li}_2\left (\frac {i e^{-2 i (a+b x)}}{c}\right )+\text {Li}_3\left (\frac {i e^{-2 i (a+b x)}}{c}\right )\right )}{8 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.62, size = 270, normalized size = 2.18 \[ -\frac {2 \, b^{3} x^{3} + 3 i \, b^{2} x^{2} \log \left (\frac {{\left (c + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 2 \, a^{3} + 6 \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - {\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) + 6 i \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \]
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 \[ \int x \operatorname {arccot}\left (-{\left (-i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 6.08, size = 1492, normalized size = 12.03 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 221, normalized size = 1.78 \[ -\frac {\frac {{\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \operatorname {arccot}\left ({\left (-i \, c + 1\right )} \tan \left (b x + a\right ) - c\right )}{b} + \frac {2 \, {\left (-4 i \, {\left (b x + a\right )}^{3} + 12 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (6 i \, {\left (b x + a\right )}^{2} - 12 i \, {\left (b x + a\right )} a\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), -c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}(-i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )} {\left (i \, c - 1\right )}}{b {\left (12 \, c + 12 i\right )}}}{2 \, b} \]
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 \[ \int x\,\mathrm {acot}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (-1+c\,1{}\mathrm {i}\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: CoercionFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
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