Optimal. Leaf size=499 \[ \frac {i c (a+b \text {ArcTan}(c x))^3}{c^2 d^2+e^2}+\frac {c^2 d (a+b \text {ArcTan}(c x))^3}{e \left (c^2 d^2+e^2\right )}-\frac {(a+b \text {ArcTan}(c x))^3}{e (d+e x)}-\frac {3 b c (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \text {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )} \]
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Rubi [A]
time = 0.36, antiderivative size = 499, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4974, 4968,
5104, 5004, 5040, 4964, 5114, 6745} \begin {gather*} \frac {3 i b^2 c \text {Li}_2\left (1-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))}{c^2 d^2+e^2}+\frac {3 i b^2 c \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))}{c^2 d^2+e^2}-\frac {3 i b^2 c (a+b \text {ArcTan}(c x)) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {i c (a+b \text {ArcTan}(c x))^3}{c^2 d^2+e^2}+\frac {c^2 d (a+b \text {ArcTan}(c x))^3}{e \left (c^2 d^2+e^2\right )}-\frac {3 b c \log \left (\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))^2}{c^2 d^2+e^2}+\frac {3 b c \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2}{c^2 d^2+e^2}+\frac {3 b c (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{c^2 d^2+e^2}-\frac {(a+b \text {ArcTan}(c x))^3}{e (d+e x)}-\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 4964
Rule 4968
Rule 4974
Rule 5004
Rule 5040
Rule 5104
Rule 5114
Rule 6745
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{(d+e x)^2} \, dx &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}+\frac {(3 b c) \int \left (\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac {c^2 (d-e x) \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right ) \left (1+c^2 x^2\right )}\right ) \, dx}{e}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}+\frac {\left (3 b c^3\right ) \int \frac {(d-e x) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{e \left (c^2 d^2+e^2\right )}+\frac {(3 b c e) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d+e x} \, dx}{c^2 d^2+e^2}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {\left (3 b c^3\right ) \int \left (\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}-\frac {e x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}\right ) \, dx}{e \left (c^2 d^2+e^2\right )}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )}-\frac {\left (3 b c^3\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{c^2 d^2+e^2}+\frac {\left (3 b c^3 d\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{e \left (c^2 d^2+e^2\right )}\\ &=\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d^2+e^2}+\frac {c^2 d \left (a+b \tan ^{-1}(c x)\right )^3}{e \left (c^2 d^2+e^2\right )}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {\left (3 b c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{i-c x} \, dx}{c^2 d^2+e^2}\\ &=\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d^2+e^2}+\frac {c^2 d \left (a+b \tan ^{-1}(c x)\right )^3}{e \left (c^2 d^2+e^2\right )}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )}-\frac {\left (6 b^2 c^2\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}{c^2 d^2+e^2}\\ &=\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d^2+e^2}+\frac {c^2 d \left (a+b \tan ^{-1}(c x)\right )^3}{e \left (c^2 d^2+e^2\right )}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )}-\frac {\left (3 i b^3 c^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d^2+e^2}\\ &=\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d^2+e^2}+\frac {c^2 d \left (a+b \tan ^{-1}(c x)\right )^3}{e \left (c^2 d^2+e^2\right )}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{e (d+e x)}-\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^2 d^2+e^2}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{c^2 d^2+e^2}-\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 \left (c^2 d^2+e^2\right )}+\frac {3 b^3 c \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 \left (c^2 d^2+e^2\right )}\\ \end {align*}
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Mathematica [F]
time = 85.55, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b \text {ArcTan}(c x))^3}{(d+e x)^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 2.88, size = 3050, normalized size = 6.11
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(3050\) |
default | \(\text {Expression too large to display}\) | \(3050\) |
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*} \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}}{\left (d + e x\right )^{2}}\, dx \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\,x\right )\right )}^3}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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