Optimal. Leaf size=431 \[ \frac {3 b^2 \text {Li}_3\left (1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}-\frac {3 b^2 \text {Li}_3\left (\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}+\frac {3 i b \text {Li}_2\left (1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}-\frac {3 i b \text {Li}_2\left (\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}-\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{c}-\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{4 c}+\frac {3 i b^3 \text {Li}_4\left (\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right )}{4 c} \]
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Rubi [A] time = 0.48, antiderivative size = 431, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6681, 4850, 4988, 4884, 4994, 4998, 6610} \[ \frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}-\frac {3 b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}+\frac {3 i b \text {PolyLog}\left (2,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}-\frac {3 i b \text {PolyLog}\left (2,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}-\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right )}{4 c}+\frac {3 i b^3 \text {PolyLog}\left (4,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right )}{4 c}-\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{c} \]
Antiderivative was successfully verified.
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Rule 4850
Rule 4884
Rule 4988
Rule 4994
Rule 4998
Rule 6610
Rule 6681
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {(6 b) \operatorname {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,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {(3 b) \operatorname {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,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}-\frac {(3 b) \operatorname {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,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {\left (3 i b^2\right ) \operatorname {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,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}+\frac {\left (3 i b^2\right ) \operatorname {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,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b^2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c}+\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b^2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c}+\frac {3 i b^3 \text {Li}_4\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c}\\ \end {align*}
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Mathematica [A] time = 0.20, size = 530, normalized size = 1.23 \[ -\frac {6 b^2 \text {Li}_3\left (-\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )-6 b^2 \text {Li}_3\left (\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )+6 i b \text {Li}_2\left (-\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-6 i b \text {Li}_2\left (\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2+8 \tanh ^{-1}\left (1-\frac {2 i}{-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+i}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-3 i b^3 \text {Li}_4\left (-\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right )+3 i b^3 \text {Li}_4\left (\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right )}{4 c} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{3} \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{3} + 3 \, a b^{2} \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 3 \, a^{2} b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{3}}{c^{2} x^{2} - 1}, x\right ) \]
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 \[ \int -\frac {{\left (b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.99, size = 1631, normalized size = 3.78 \[ \text {result too large to display} \]
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 \[ \frac {1}{2} \, a^{3} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} + \frac {\frac {15}{2} \, {\left (b^{3} \log \left (c x + 1\right ) - b^{3} \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right )^{3} - \frac {45}{8} \, {\left (b^{3} \log \relax (2)^{2} \log \left (c x + 1\right ) - b^{3} \log \relax (2)^{2} \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right ) - \frac {1}{2} \, c \int \frac {784 \, b^{3} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right )^{3} + 3072 \, a b^{2} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right )^{2} - 45 \, {\left (b^{3} \log \relax (2)^{2} \log \left (c x + 1\right ) - b^{3} \log \relax (2)^{2} \log \left (-c x + 1\right ) - 4 \, {\left (b^{3} \log \left (c x + 1\right ) - b^{3} \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right )^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} + 12 \, {\left (15 \, b^{3} \log \relax (2)^{2} + 256 \, a^{2} b\right )} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right )}{8 \, {\left (c^{2} x^{2} - 1\right )}}\,{d x}}{64 \, c} \]
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 \[ \int -\frac {{\left (a+b\,\mathrm {atan}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3}{c^2\,x^2-1} \,d x \]
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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