Optimal. Leaf size=264 \[ -\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \text {Li}_2\left (e^{2 i \csc ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \text {Li}_3\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {Li}_3\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {3 i \csc ^{-1}(a+b x)^2}{2 b^2}-\frac {3 \csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 a \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3 \]
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Rubi [A] time = 0.26, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {5259, 4427, 4190, 4183, 2531, 2282, 6589, 4184, 3717, 2190, 2279, 2391} \[ \frac {6 i a \csc ^{-1}(a+b x) \text {PolyLog}\left (2,-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \csc ^{-1}(a+b x) \text {PolyLog}\left (2,e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \text {PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \text {PolyLog}\left (3,-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {PolyLog}\left (3,e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {3 i \csc ^{-1}(a+b x)^2}{2 b^2}-\frac {3 \csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 a \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 3717
Rule 4183
Rule 4184
Rule 4190
Rule 4427
Rule 5259
Rule 6589
Rubi steps
\begin {align*} \int x \csc ^{-1}(a+b x)^3 \, dx &=-\frac {\operatorname {Subst}\left (\int x^3 \cot (x) \csc (x) (-a+\csc (x)) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int x^2 (-a+\csc (x))^2 \, dx,x,\csc ^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int \left (a^2 x^2-2 a x^2 \csc (x)+x^2 \csc ^2(x)\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int x^2 \csc ^2(x) \, dx,x,\csc ^{-1}(a+b x)\right )}{2 b^2}+\frac {(3 a) \operatorname {Subst}\left (\int x^2 \csc (x) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {6 a \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \operatorname {Subst}\left (\int x \cot (x) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}-\frac {(6 a) \operatorname {Subst}\left (\int x \log \left (1-e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}+\frac {(6 a) \operatorname {Subst}\left (\int x \log \left (1+e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {3 i \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {6 a \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}-\frac {(6 i a) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}+\frac {(6 i a) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {3 i \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {6 a \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {3 \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b^2}-\frac {(6 a) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {(6 a) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{b^2}\\ &=\frac {3 i \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {6 a \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 a \text {Li}_3\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {Li}_3\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \csc ^{-1}(a+b x)}\right )}{2 b^2}\\ &=\frac {3 i \csc ^{-1}(a+b x)^2}{2 b^2}+\frac {3 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \csc ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \csc ^{-1}(a+b x)^3-\frac {6 a \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {3 \csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {3 i \text {Li}_2\left (e^{2 i \csc ^{-1}(a+b x)}\right )}{2 b^2}-\frac {6 a \text {Li}_3\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b^2}+\frac {6 a \text {Li}_3\left (e^{i \csc ^{-1}(a+b x)}\right )}{b^2}\\ \end {align*}
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Mathematica [A] time = 0.80, size = 314, normalized size = 1.19 \[ \frac {3 a \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2+3 b x \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \csc ^{-1}(a+b x)^2-a^2 \csc ^{-1}(a+b x)^3+b^2 x^2 \csc ^{-1}(a+b x)^3+12 i a \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )-12 i a \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )+3 i \text {Li}_2\left (e^{2 i \csc ^{-1}(a+b x)}\right )-12 a \text {Li}_3\left (-e^{i \csc ^{-1}(a+b x)}\right )+12 a \text {Li}_3\left (e^{i \csc ^{-1}(a+b x)}\right )+3 i \csc ^{-1}(a+b x)^2+6 a \csc ^{-1}(a+b x)^2 \log \left (1-e^{i \csc ^{-1}(a+b x)}\right )-6 a \csc ^{-1}(a+b x)^2 \log \left (1+e^{i \csc ^{-1}(a+b x)}\right )-6 \csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{2 b^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {arccsc}\left (b x + a\right )^{3}, x\right ) \]
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 \[ \int x \operatorname {arccsc}\left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.62, size = 481, normalized size = 1.82 \[ \frac {x^{2} \mathrm {arccsc}\left (b x +a \right )^{3}}{2}+\frac {3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arccsc}\left (b x +a \right )^{2} x}{2 b}-\frac {a^{2} \mathrm {arccsc}\left (b x +a \right )^{3}}{2 b^{2}}+\frac {3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \mathrm {arccsc}\left (b x +a \right )^{2} a}{2 b^{2}}+\frac {3 i \mathrm {arccsc}\left (b x +a \right )^{2}}{2 b^{2}}-\frac {3 a \mathrm {arccsc}\left (b x +a \right )^{2} \ln \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}+\frac {3 a \mathrm {arccsc}\left (b x +a \right )^{2} \ln \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}-\frac {6 i a \,\mathrm {arccsc}\left (b x +a \right ) \polylog \left (2, \frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}+\frac {6 a \polylog \left (3, \frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}-\frac {3 \,\mathrm {arccsc}\left (b x +a \right ) \ln \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}+\frac {6 i a \,\mathrm {arccsc}\left (b x +a \right ) \polylog \left (2, -\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}-\frac {6 a \polylog \left (3, -\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}-\frac {3 \,\mathrm {arccsc}\left (b x +a \right ) \ln \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}+\frac {3 i \polylog \left (2, \frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}}+\frac {3 i \polylog \left (2, -\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b^{2}} \]
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} \, x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{3} - \frac {3}{8} \, x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2} - \int \frac {3 \, {\left (8 \, {\left (b^{3} x^{4} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 3 \, a b^{2} x^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + {\left (3 \, a^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )} b x^{2} + {\left (a^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - a \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )} x\right )} \log \left (b x + a\right )^{2} - {\left (4 \, b x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - b x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )^{2}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} - 4 \, {\left (b^{3} x^{4} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 2 \, a b^{2} x^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + {\left (a^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )} b x^{2} + 2 \, {\left (b^{3} x^{4} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 3 \, a b^{2} x^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + {\left (3 \, a^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )} b x^{2} + {\left (a^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) - a \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )} x\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )}}{8 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} - 1\right )} b x - a\right )}}\,{d x} \]
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 \[ \int x\,{\mathrm {asin}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]
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 \[ \int x \operatorname {acsc}^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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