Optimal. Leaf size=170 \[ \frac {1}{12} i b x^4+\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \text {PolyLog}\left (2,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {x \text {PolyLog}\left (3,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b^2}-\frac {i \text {PolyLog}\left (4,-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{8 b^3} \]
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Rubi [A]
time = 0.20, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6398, 2215,
2221, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {i \text {Li}_4\left (-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{8 b^3}-\frac {x \text {Li}_3\left (-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b^2}+\frac {i x^2 \text {Li}_2\left (-\left ((1-i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {1}{3} x^3 \tanh ^{-1}(d \tan (a+b x)-i d+1)+\frac {1}{12} i b x^4 \end {gather*}
Antiderivative was successfully verified.
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Rule 2215
Rule 2221
Rule 2320
Rule 2611
Rule 6398
Rule 6724
Rule 6744
Rubi steps
\begin {align*} \int x^2 \tanh ^{-1}(1-i d+d \tan (a+b x)) \, dx &=\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))+\frac {1}{3} (i b) \int \frac {x^3}{1+(1-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{3} (b (i+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+(1-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {1}{2} \int x^2 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right ) \, dx\\ &=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \text {Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {i \int x \text {Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right ) \, dx}{2 b}\\ &=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \text {Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \text {Li}_3\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {\int \text {Li}_3\left ((-1+i d) e^{2 i a+2 i b x}\right ) \, dx}{4 b^2}\\ &=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \text {Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \text {Li}_3\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_3(i (i+d) x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac {1}{12} i b x^4+\frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-i d) e^{2 i a+2 i b x}\right )+\frac {i x^2 \text {Li}_2\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {x \text {Li}_3\left (-(1-i d) e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \text {Li}_4\left (-(1-i d) e^{2 i a+2 i b x}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.45, size = 155, normalized size = 0.91 \begin {gather*} \frac {1}{3} x^3 \tanh ^{-1}(1-i d+d \tan (a+b x))-\frac {4 b^3 x^3 \log \left (1+\frac {i e^{-2 i (a+b x)}}{i+d}\right )+6 i b^2 x^2 \text {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{i+d}\right )+6 b x \text {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{i+d}\right )-3 i \text {PolyLog}\left (4,-\frac {i e^{-2 i (a+b x)}}{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 = 1.80, size = 2346, normalized size = 13.80
method | result | size |
risch | \(\text {Expression too large to display}\) | \(2346\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 343 vs. \(2 (118) = 236\).
time = 0.28, size = 343, normalized size = 2.02 \begin {gather*} \frac {\frac {12 \, {\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 {artanh}\left (d \tan \left (b x + a\right ) - i \, d + 1\right )}{b^{2}} - \frac {-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 (-d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), d \sin \left (2 \, b x + 2 \, a\right ) + \cos \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 ({\left (i \, d - 1\right )} 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 ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left (4 \, b x + a\right )} {\rm Li}_{3}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \, {\rm Li}_{4}({\left (i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{2}}}{36 \, b} \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. 345 vs. \(2 (118) = 236\).
time = 0.40, size = 345, normalized size = 2.03 \begin {gather*} \frac {i \, b^{4} x^{4} + 2 \, b^{3} x^{3} \log \left (-\frac {{\left ({\left (d + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{d}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - i \, a^{4} + 2 \, a^{3} \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) + 2 \, a^{3} \log \left (\frac {2 \, {\left (d + i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, d - 4}}{2 \, {\left (d + i\right )}}\right ) - 12 \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \end {gather*}
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 \begin {gather*} \int x^{2} \operatorname {atanh}{\left (d \tan {\left (a + b x \right )} - i d + 1 \right )}\, dx \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 = -0.01 \begin {gather*} \int x^2\,\mathrm {atanh}\left (d\,\mathrm {tan}\left (a+b\,x\right )+1-d\,1{}\mathrm {i}\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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