Optimal. Leaf size=277 \[ \frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {2 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}+\frac {i a b x}{c^2 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}+\frac {i b^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c^3 d}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c^3 d}+\frac {i b^2 x \tan ^{-1}(c x)}{c^2 d}-\frac {i b^2 \log \left (c^2 x^2+1\right )}{2 c^3 d} \]
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Rubi [A] time = 0.51, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {4866, 4852, 4916, 4846, 260, 4884, 4920, 4854, 2402, 2315, 4994, 6610} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}+\frac {i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {i a b x}{c^2 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {2 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}-\frac {i b^2 \log \left (c^2 x^2+1\right )}{2 c^3 d}+\frac {i b^2 x \tan ^{-1}(c x)}{c^2 d} \]
Antiderivative was successfully verified.
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Rule 260
Rule 2315
Rule 2402
Rule 4846
Rule 4852
Rule 4854
Rule 4866
Rule 4884
Rule 4916
Rule 4920
Rule 4994
Rule 6610
Rubi steps
\begin {align*} \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d+i c d x} \, dx &=\frac {i \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{d+i c d x} \, dx}{c}-\frac {i \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}-\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d+i c d x} \, dx}{c^2}+\frac {(i b) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{d}+\frac {\int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c^2 d}\\ &=\frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {(i b) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^2 d}-\frac {(i b) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^2 d}+\frac {(2 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}-\frac {(2 b) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c d}\\ &=\frac {i a b x}{c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}+\frac {(2 b) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c^2 d}+\frac {\left (i b^2\right ) \int \tan ^{-1}(c x) \, dx}{c^2 d}-\frac {b^2 \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}\\ &=\frac {i a b x}{c^2 d}+\frac {i b^2 x \tan ^{-1}(c x)}{c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}+\frac {2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}-\frac {\left (i b^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{c d}\\ &=\frac {i a b x}{c^2 d}+\frac {i b^2 x \tan ^{-1}(c x)}{c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}+\frac {2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \log \left (1+c^2 x^2\right )}{2 c^3 d}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^3 d}\\ &=\frac {i a b x}{c^2 d}+\frac {i b^2 x \tan ^{-1}(c x)}{c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}+\frac {2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \log \left (1+c^2 x^2\right )}{2 c^3 d}+\frac {i b^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}\\ \end {align*}
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Mathematica [A] time = 0.60, size = 330, normalized size = 1.19 \[ -\frac {i \left (3 a^2 c^2 x^2-3 a^2 \log \left (c^2 x^2+1\right )+6 i a^2 c x-6 i a^2 \tan ^{-1}(c x)-6 i a b \log \left (c^2 x^2+1\right )+6 a b c^2 x^2 \tan ^{-1}(c x)+6 b \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right ) \left (-i a-i b \tan ^{-1}(c x)+b\right )-6 a b c x-12 i a b \tan ^{-1}(c x)^2+6 a b \tan ^{-1}(c x)+12 i a b c x \tan ^{-1}(c x)+12 a b \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+3 b^2 \log \left (c^2 x^2+1\right )+3 b^2 c^2 x^2 \tan ^{-1}(c x)^2+3 b^2 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )-4 i b^2 \tan ^{-1}(c x)^3+9 b^2 \tan ^{-1}(c x)^2+6 i b^2 c x \tan ^{-1}(c x)^2-6 b^2 c x \tan ^{-1}(c x)+6 b^2 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+12 i b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{6 c^3 d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {i \, b^{2} x^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 4 \, a b x^{2} \log \left (-\frac {c x + i}{c x - i}\right ) - 4 i \, a^{2} x^{2}}{4 \, c d x - 4 i \, d}, x\right ) \]
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 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 4.06, size = 1212, normalized size = 4.38 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int \frac {x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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