Optimal. Leaf size=155 \[ \frac {\text {Li}_4\left (-i c e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {b x^4}{12} \]
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Rubi [A] time = 0.26, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5172, 2184, 2190, 2531, 6609, 2282, 6589} \[ -\frac {i x \text {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {\text {PolyLog}\left (4,-i c e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {x^2 \text {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {b x^4}{12} \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2282
Rule 2531
Rule 5172
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^2 \cot ^{-1}(c-(1-i c) \tan (a+b x)) \, dx &=\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))+\frac {1}{3} (i b) \int \frac {x^3}{i (-1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))+\frac {1}{3} (b c) \int \frac {e^{2 i a+2 i b x} x^3}{i (-1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{2} i \int x^2 \log \left (1+\frac {c e^{2 i a+2 i b x}}{i (-1+i c)+c}\right ) \, dx\\ &=-\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {\int x \text {Li}_2\left (-\frac {c e^{2 i a+2 i b x}}{i (-1+i c)+c}\right ) \, dx}{2 b}\\ &=-\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {i \int \text {Li}_3\left (-\frac {c e^{2 i a+2 i b x}}{i (-1+i c)+c}\right ) \, dx}{4 b^2}\\ &=-\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3(-i c x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=-\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {\text {Li}_4\left (-i c e^{2 i a+2 i b x}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 141, normalized size = 0.91 \[ \frac {1}{3} x^3 \cot ^{-1}(c+i (c+i) \tan (a+b x))-\frac {4 i b^3 x^3 \log \left (1-\frac {i e^{-2 i (a+b x)}}{c}\right )-6 b^2 x^2 \text {Li}_2\left (\frac {i e^{-2 i (a+b x)}}{c}\right )+6 i b x \text {Li}_3\left (\frac {i e^{-2 i (a+b x)}}{c}\right )+3 \text {Li}_4\left (\frac {i e^{-2 i (a+b x)}}{c}\right )}{24 b^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.72, size = 321, normalized size = 2.07 \[ -\frac {b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (\frac {{\left (c + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - a^{4} - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - {\left (-2 i \, b^{3} x^{3} - 2 i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (-2 i \, b^{3} x^{3} - 2 i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \]
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 \[ \int x^{2} \operatorname {arccot}\left (-{\left (-i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 6.51, size = 1527, normalized size = 9.85 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 312, normalized size = 2.01 \[ -\frac {\frac {{\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {arccot}\left ({\left (-i \, c + 1\right )} \tan \left (b x + a\right ) - c\right )}{b^{2}} + \frac {3 \, {\left (-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} + {\left (8 i \, {\left (b x + a\right )}^{3} - 18 i \, {\left (b x + a\right )}^{2} a + 18 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), -c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + {\left (-12 i \, {\left (b x + a\right )}^{2} + 18 i \, {\left (b x + a\right )} a - 9 i \, a^{2}\right )} {\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}(-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}(-i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )} {\left (i \, c - 1\right )}}{b^{2} {\left (12 \, c + 12 i\right )}}}{3 \, b} \]
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 \[ \int x^2\,\mathrm {acot}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (-1+c\,1{}\mathrm {i}\right )\right ) \,d x \]
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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