Optimal. Leaf size=399 \[ \frac {1}{3} x^3 \text {ArcTan}(c+d \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {x^2 \text {PolyLog}\left (2,\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac {x^2 \text {PolyLog}\left (2,\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac {i x \text {PolyLog}\left (3,\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}-\frac {i x \text {PolyLog}\left (3,\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}-\frac {\text {PolyLog}\left (4,\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{8 b^3}+\frac {\text {PolyLog}\left (4,\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^3} \]
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Rubi [A]
time = 0.39, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5285, 2221,
2611, 6744, 2320, 6724} \begin {gather*} \frac {1}{3} x^3 \text {ArcTan}(d \cot (a+b x)+c)-\frac {\text {Li}_4\left (\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{8 b^3}+\frac {\text {Li}_4\left (\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^3}+\frac {i x \text {Li}_3\left (\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b^2}-\frac {i x \text {Li}_3\left (\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}+\frac {x^2 \text {Li}_2\left (\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac {1}{6} i x^3 \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 5285
Rule 6724
Rule 6744
Rubi steps
\begin {align*} \int x^2 \tan ^{-1}(c+d \cot (a+b x)) \, dx &=\frac {1}{3} x^3 \tan ^{-1}(c+d \cot (a+b x))+\frac {1}{3} (b (1+i c-d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+i c+d+(-1-i c+d) e^{2 i a+2 i b x}} \, dx-\frac {1}{3} (b (1-i c+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1-i c-d+(-1+i c-d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{3} x^3 \tan ^{-1}(c+d \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {1}{2} i \int x^2 \log \left (1+\frac {(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx-\frac {1}{2} i \int x^2 \log \left (1+\frac {(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx\\ &=\frac {1}{3} x^3 \tan ^{-1}(c+d \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac {\int x \text {Li}_2\left (-\frac {(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx}{2 b}-\frac {\int x \text {Li}_2\left (-\frac {(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \tan ^{-1}(c+d \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac {i x \text {Li}_3\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}-\frac {i x \text {Li}_3\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}+\frac {i \int \text {Li}_3\left (-\frac {(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx}{4 b^2}-\frac {i \int \text {Li}_3\left (-\frac {(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \tan ^{-1}(c+d \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac {i x \text {Li}_3\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}-\frac {i x \text {Li}_3\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}-\frac {\text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {(-1-i c+d) x}{1+i c+d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}+\frac {\text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(c+i (1+d)) x}{c-i (-1+d)}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac {1}{3} x^3 \tan ^{-1}(c+d \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac {i x \text {Li}_3\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}-\frac {i x \text {Li}_3\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}-\frac {\text {Li}_4\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{8 b^3}+\frac {\text {Li}_4\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.71, size = 359, normalized size = 0.90 \begin {gather*} \frac {1}{3} x^3 \text {ArcTan}(c+d \cot (a+b x))+\frac {4 i b^3 x^3 \log \left (1-\frac {(c+i (-1+d)) e^{2 i (a+b x)}}{c-i (1+d)}\right )-4 i b^3 x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i (a+b x)}}{i+c-i d}\right )+6 b^2 x^2 \text {PolyLog}\left (2,\frac {(c+i (-1+d)) e^{2 i (a+b x)}}{c-i (1+d)}\right )-6 b^2 x^2 \text {PolyLog}\left (2,\frac {(c+i (1+d)) e^{2 i (a+b x)}}{i+c-i d}\right )+6 i b x \text {PolyLog}\left (3,\frac {(c+i (-1+d)) e^{2 i (a+b x)}}{c-i (1+d)}\right )-6 i b x \text {PolyLog}\left (3,\frac {(c+i (1+d)) e^{2 i (a+b x)}}{i+c-i d}\right )-3 \text {PolyLog}\left (4,\frac {(c+i (-1+d)) e^{2 i (a+b x)}}{c-i (1+d)}\right )+3 \text {PolyLog}\left (4,\frac {(c+i (1+d)) e^{2 i (a+b x)}}{i+c-i d}\right )}{24 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 20.12, size = 7924, normalized size = 19.86
method | result | size |
risch | \(\text {Expression too large to display}\) | \(7924\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1589 vs. \(2 (283) = 566\).
time = 1.08, size = 1589, normalized size = 3.98 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\mathrm {atan}\left (c+d\,\mathrm {cot}\left (a+b\,x\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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