Optimal. Leaf size=264 \[ -\frac {3 b^2 e \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2}+\frac {3 i b^2 d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}+\frac {3 b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {3 i b^3 e \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c^2}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c} \]
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Rubi [A] time = 0.58, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4864, 4846, 4920, 4854, 2402, 2315, 4984, 4884, 4994, 6610} \[ \frac {3 i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {3 i b^3 e \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^2}+\frac {3 b^3 d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c}-\frac {3 b^2 e \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}+\frac {3 b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c} \]
Antiderivative was successfully verified.
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Rule 2315
Rule 2402
Rule 4846
Rule 4854
Rule 4864
Rule 4884
Rule 4920
Rule 4984
Rule 4994
Rule 6610
Rubi steps
\begin {align*} \int (d+e x) \left (a+b \tan ^{-1}(c x)\right )^3 \, dx &=\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b c) \int \left (\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}+\frac {\left (c^2 d^2-e^2+2 c^2 d e x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 \left (1+c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b) \int \frac {\left (c^2 d^2-e^2+2 c^2 d e x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{2 c e}-\frac {(3 b e) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{2 c}\\ &=-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b) \int \left (\frac {c^2 d^2 \left (1-\frac {e^2}{c^2 d^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}+\frac {2 c^2 d e x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}\right ) \, dx}{2 c e}+\left (3 b^2 e\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-(3 b c d) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx-\frac {\left (3 b^2 e\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c}-\frac {(3 b (c d-e) (c d+e)) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{2 c e}\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+(3 b d) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{i-c x} \, dx+\frac {\left (3 b^3 e\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c}\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c}-\left (6 b^2 d\right ) \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-\frac {\left (3 i b^3 e\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^2}\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^3 e \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^2}+\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c}-\left (3 i b^3 d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^3 e \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^2}+\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c}\\ \end {align*}
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Mathematica [A] time = 0.71, size = 342, normalized size = 1.30 \[ \frac {a^3 c^2 e x^2+3 a^2 b c^2 x \tan ^{-1}(c x) (2 d+e x)-3 a^2 b c d \log \left (c^2 x^2+1\right )+a^2 c x (2 a c d-3 b e)+3 a^2 b e \tan ^{-1}(c x)+3 a b^2 e \left (\log \left (c^2 x^2+1\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x)\right )+6 a b^2 c d \left (\tan ^{-1}(c x) \left ((c x-i) \tan ^{-1}(c x)+2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-i \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )\right )+b^3 e \left (\tan ^{-1}(c x) \left (\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2+(-3 c x+3 i) \tan ^{-1}(c x)-6 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+3 i \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )\right )+b^3 c d \left (-6 i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+3 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+2 \tan ^{-1}(c x)^2 \left ((c x-i) \tan ^{-1}(c x)+3 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )\right )}{2 c^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{3} e x + a^{3} d + {\left (b^{3} e x + b^{3} d\right )} \arctan \left (c x\right )^{3} + 3 \, {\left (a b^{2} e x + a b^{2} d\right )} \arctan \left (c x\right )^{2} + 3 \, {\left (a^{2} b e x + a^{2} b d\right )} \arctan \left (c x\right ), x\right ) \]
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 = 1.42, size = 7462, normalized size = 28.27 \[ \text {output too large to display} \]
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 \[ \frac {7 \, b^{3} d \arctan \left (c x\right )^{4}}{32 \, c} + 56 \, b^{3} c^{2} e \int \frac {x^{3} \arctan \left (c x\right )^{3}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 6 \, b^{3} c^{2} e \int \frac {x^{3} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 192 \, a b^{2} c^{2} e \int \frac {x^{3} \arctan \left (c x\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 56 \, b^{3} c^{2} d \int \frac {x^{2} \arctan \left (c x\right )^{3}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 12 \, b^{3} c^{2} e \int \frac {x^{3} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 6 \, b^{3} c^{2} d \int \frac {x^{2} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 192 \, a b^{2} c^{2} d \int \frac {x^{2} \arctan \left (c x\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 24 \, b^{3} c^{2} d \int \frac {x^{2} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac {1}{2} \, a^{3} e x^{2} + \frac {a b^{2} d \arctan \left (c x\right )^{3}}{c} - 12 \, b^{3} c e \int \frac {x^{2} \arctan \left (c x\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 3 \, b^{3} c e \int \frac {x^{2} \log \left (c^{2} x^{2} + 1\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} - 24 \, b^{3} c d \int \frac {x \arctan \left (c x\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 6 \, b^{3} c d \int \frac {x \log \left (c^{2} x^{2} + 1\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac {3}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} a^{2} b e + a^{3} d x + 56 \, b^{3} e \int \frac {x \arctan \left (c x\right )^{3}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 6 \, b^{3} e \int \frac {x \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 192 \, a b^{2} e \int \frac {x \arctan \left (c x\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 6 \, b^{3} d \int \frac {\arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2}}{64 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac {3 \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} a^{2} b d}{2 \, c} + \frac {1}{16} \, {\left (b^{3} e x^{2} + 2 \, b^{3} d x\right )} \arctan \left (c x\right )^{3} - \frac {3}{64} \, {\left (b^{3} e x^{2} + 2 \, b^{3} d x\right )} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2} \]
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 \[ \int {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3\,\left (d+e\,x\right ) \,d x \]
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 \[ \int \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3} \left (d + e x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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