Integrand size = 21, antiderivative size = 154 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac {1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac {x^2 \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3} \]
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Time = 0.18 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5280, 2215, 2221, 2611, 6744, 2320, 6724} \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=-\frac {\operatorname {PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3}+\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {x^2 \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac {b x^4}{12} \]
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Rule 2215
Rule 2221
Rule 2320
Rule 2611
Rule 5280
Rule 6724
Rule 6744
Rubi steps \begin{align*} \text {integral}& = \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 \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {\int x \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \int \operatorname {PolyLog}\left (3,-\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 \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\text {Subst}\left (\int \frac {\operatorname {PolyLog}(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 \operatorname {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.88 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {1}{24} \left (8 x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+4 i x^3 \log \left (1+\frac {i e^{-2 i (a+b x)}}{c}\right )-\frac {6 x^2 \operatorname {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{b}+\frac {6 i x \operatorname {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^2}+\frac {3 \operatorname {PolyLog}\left (4,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^3}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.88 (sec) , antiderivative size = 1448, normalized size of antiderivative = 9.40
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (107) = 214\).
Time = 0.27 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.08 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {b^{4} x^{4} - 2 i \, b^{3} x^{3} \log \left (\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - a^{4} - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) + 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \]
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Exception generated. \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (107) = 214\).
Time = 0.21 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.01 \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\frac {\frac {4 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \operatorname {arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )}{b^{2}} - \frac {{\left (-3 i \, {\left (b x + a\right )}^{4} + 12 i \, {\left (b x + a\right )}^{3} a - 18 i \, {\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (4 i \, {\left (b x + a\right )}^{3} - 9 i \, {\left (b x + a\right )}^{2} a + 9 i \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 i \, a^{2}\right )} {\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 9 \, {\left (b x + a\right )} a^{2}\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )} {\left (-i \, c - 1\right )}}{b^{2} {\left (c - i\right )}}}{12 \, b} \]
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\[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\int { x^{2} \operatorname {arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right ) \,d x } \]
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Timed out. \[ \int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx=\int x^2\,\mathrm {acot}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right ) \,d x \]
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