Optimal. Leaf size=144 \[ \frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 (a+b \text {ArcCos}(c x))}{2 c^2 d}+\frac {i (a+b \text {ArcCos}(c x))^2}{2 b c^4 d}-\frac {b \text {ArcSin}(c x)}{4 c^4 d}-\frac {(a+b \text {ArcCos}(c x)) \log \left (1-e^{2 i \text {ArcCos}(c x)}\right )}{c^4 d}+\frac {i b \text {PolyLog}\left (2,e^{2 i \text {ArcCos}(c x)}\right )}{2 c^4 d} \]
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Rubi [A]
time = 0.13, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4796, 4766,
3798, 2221, 2317, 2438, 327, 222} \begin {gather*} \frac {i (a+b \text {ArcCos}(c x))^2}{2 b c^4 d}-\frac {\log \left (1-e^{2 i \text {ArcCos}(c x)}\right ) (a+b \text {ArcCos}(c x))}{c^4 d}-\frac {x^2 (a+b \text {ArcCos}(c x))}{2 c^2 d}+\frac {i b \text {Li}_2\left (e^{2 i \text {ArcCos}(c x)}\right )}{2 c^4 d}-\frac {b \text {ArcSin}(c x)}{4 c^4 d}+\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 222
Rule 327
Rule 2221
Rule 2317
Rule 2438
Rule 3798
Rule 4766
Rule 4796
Rubi steps
\begin {align*} \int \frac {x^3 \left (a+b \cos ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {\int \frac {x \left (a+b \cos ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{c^2}-\frac {b \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{2 c d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}-\frac {\text {Subst}\left (\int (a+b x) \cot (x) \, dx,x,\cos ^{-1}(c x)\right )}{c^4 d}-\frac {b \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c^3 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}+\frac {(2 i) \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\cos ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}-\frac {\left (a+b \cos ^{-1}(c x)\right ) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )}{c^4 d}+\frac {b \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}-\frac {\left (a+b \cos ^{-1}(c x)\right ) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )}{c^4 d}-\frac {(i b) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \cos ^{-1}(c x)}\right )}{2 c^4 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}-\frac {\left (a+b \cos ^{-1}(c x)\right ) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )}{c^4 d}+\frac {i b \text {Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )}{2 c^4 d}\\ \end {align*}
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Mathematica [A]
time = 0.15, size = 180, normalized size = 1.25 \begin {gather*} -\frac {2 a c^2 x^2-b c x \sqrt {1-c^2 x^2}+2 b c^2 x^2 \text {ArcCos}(c x)-2 i b \text {ArcCos}(c x)^2+2 b \text {ArcTan}\left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )+4 b \text {ArcCos}(c x) \log \left (1-e^{i \text {ArcCos}(c x)}\right )+4 b \text {ArcCos}(c x) \log \left (1+e^{i \text {ArcCos}(c x)}\right )+2 a \log \left (1-c^2 x^2\right )-4 i b \text {PolyLog}\left (2,-e^{i \text {ArcCos}(c x)}\right )-4 i b \text {PolyLog}\left (2,e^{i \text {ArcCos}(c x)}\right )}{4 c^4 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.89, size = 205, normalized size = 1.42
method | result | size |
derivativedivides | \(\frac {-\frac {a \,c^{2} x^{2}}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {i b \arccos \left (c x \right )^{2}}{2 d}-\frac {b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \polylog \left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \polylog \left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {b \arccos \left (c x \right ) \cos \left (2 \arccos \left (c x \right )\right )}{4 d}+\frac {b \sin \left (2 \arccos \left (c x \right )\right )}{8 d}}{c^{4}}\) | \(205\) |
default | \(\frac {-\frac {a \,c^{2} x^{2}}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {i b \arccos \left (c x \right )^{2}}{2 d}-\frac {b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \polylog \left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \polylog \left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{d}-\frac {b \arccos \left (c x \right ) \cos \left (2 \arccos \left (c x \right )\right )}{4 d}+\frac {b \sin \left (2 \arccos \left (c x \right )\right )}{8 d}}{c^{4}}\) | \(205\) |
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 x^{3}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{3} \operatorname {acos}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \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 \frac {x^3\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}{d-c^2\,d\,x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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