Optimal. Leaf size=279 \[ \frac {2 (a+b \text {ArcTan}(c+d x))^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {3 i b (a+b \text {ArcTan}(c+d x))^2 \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b (a+b \text {ArcTan}(c+d x))^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}-\frac {3 b^2 (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 b^2 (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (3,-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2}{1+i (c+d x)}\right )}{4 d e}-\frac {3 i b^3 \text {PolyLog}\left (4,-1+\frac {2}{1+i (c+d x)}\right )}{4 d e} \]
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Rubi [A]
time = 0.36, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5151, 12,
4942, 5108, 5004, 5114, 5118, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{i (c+d x)+1}\right ) (a+b \text {ArcTan}(c+d x))}{2 d e}+\frac {3 b^2 \text {Li}_3\left (\frac {2}{i (c+d x)+1}-1\right ) (a+b \text {ArcTan}(c+d x))}{2 d e}-\frac {3 i b \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right ) (a+b \text {ArcTan}(c+d x))^2}{2 d e}+\frac {3 i b \text {Li}_2\left (\frac {2}{i (c+d x)+1}-1\right ) (a+b \text {ArcTan}(c+d x))^2}{2 d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))^3}{d e}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{i (c+d x)+1}\right )}{4 d e}-\frac {3 i b^3 \text {Li}_4\left (\frac {2}{i (c+d x)+1}-1\right )}{4 d e} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 4942
Rule 5004
Rule 5108
Rule 5114
Rule 5118
Rule 5151
Rule 6745
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{c e+d e x} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(6 b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(3 b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2 \log \left (2-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}+\frac {(3 b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2 \log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {\left (3 i b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \text {Li}_2\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}-\frac {\left (3 i b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}-\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{2 d e}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{2 d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}-\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{1+i (c+d x)}\right )}{4 d e}-\frac {3 i b^3 \text {Li}_4\left (-1+\frac {2}{1+i (c+d x)}\right )}{4 d e}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 252, normalized size = 0.90 \begin {gather*} \frac {8 (a+b \text {ArcTan}(c+d x))^3 \tanh ^{-1}\left (\frac {i+c+d x}{-i+c+d x}\right )+6 i b (a+b \text {ArcTan}(c+d x))^2 \text {PolyLog}\left (2,-\frac {i+c+d x}{-i+c+d x}\right )-6 i b (a+b \text {ArcTan}(c+d x))^2 \text {PolyLog}\left (2,\frac {i+c+d x}{-i+c+d x}\right )+6 b^2 (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (3,-\frac {i+c+d x}{-i+c+d x}\right )-6 b^2 (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (3,\frac {i+c+d x}{-i+c+d x}\right )-3 i b^3 \text {PolyLog}\left (4,-\frac {i+c+d x}{-i+c+d x}\right )+3 i b^3 \text {PolyLog}\left (4,\frac {i+c+d x}{-i+c+d x}\right )}{4 d e} \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 = 1.14, size = 2757, normalized size = 9.88
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2757\) |
default | \(\text {Expression too large to display}\) | \(2757\) |
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 [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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{3}}{c + d x}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c + d x}\, dx}{e} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{c\,e+d\,e\,x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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