Optimal. Leaf size=140 \[ -\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 \text {Li}_3\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {6 \text {Li}_3\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \csc ^{-1}(a+b x)^3}{b}+\frac {6 \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b} \]
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Rubi [A] time = 0.10, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5253, 5217, 3758, 4183, 2531, 2282, 6589} \[ -\frac {6 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 \text {PolyLog}\left (3,-e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {6 \text {PolyLog}\left (3,e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \csc ^{-1}(a+b x)^3}{b}+\frac {6 \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 3758
Rule 4183
Rule 5217
Rule 5253
Rule 6589
Rubi steps
\begin {align*} \int \csc ^{-1}(a+b x)^3 \, dx &=\frac {\operatorname {Subst}\left (\int \csc ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int x^3 \cot (x) \csc (x) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^3}{b}-\frac {3 \operatorname {Subst}\left (\int x^2 \csc (x) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^3}{b}+\frac {6 \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 \operatorname {Subst}\left (\int x \log \left (1-e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}-\frac {6 \operatorname {Subst}\left (\int x \log \left (1+e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^3}{b}+\frac {6 \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}-\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^3}{b}+\frac {6 \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )}{b}\\ &=\frac {(a+b x) \csc ^{-1}(a+b x)^3}{b}+\frac {6 \csc ^{-1}(a+b x)^2 \tanh ^{-1}\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}+\frac {6 \text {Li}_3\left (-e^{i \csc ^{-1}(a+b x)}\right )}{b}-\frac {6 \text {Li}_3\left (e^{i \csc ^{-1}(a+b x)}\right )}{b}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 162, normalized size = 1.16 \[ \frac {-6 i \csc ^{-1}(a+b x) \text {Li}_2\left (-e^{i \csc ^{-1}(a+b x)}\right )+6 i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{i \csc ^{-1}(a+b x)}\right )+6 \text {Li}_3\left (-e^{i \csc ^{-1}(a+b x)}\right )-6 \text {Li}_3\left (e^{i \csc ^{-1}(a+b x)}\right )+a \csc ^{-1}(a+b x)^3+b x \csc ^{-1}(a+b x)^3-3 \csc ^{-1}(a+b x)^2 \log \left (1-e^{i \csc ^{-1}(a+b x)}\right )+3 \csc ^{-1}(a+b x)^2 \log \left (1+e^{i \csc ^{-1}(a+b x)}\right )}{b} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {arccsc}\left (b x + a\right )^{3}, x\right ) \]
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 \[ \int \operatorname {arccsc}\left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.63, size = 247, normalized size = 1.76 \[ x \mathrm {arccsc}\left (b x +a \right )^{3}+\frac {\mathrm {arccsc}\left (b x +a \right )^{3} a}{b}-\frac {3 \mathrm {arccsc}\left (b x +a \right )^{2} \ln \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}+\frac {3 \mathrm {arccsc}\left (b x +a \right )^{2} \ln \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}+\frac {6 i \mathrm {arccsc}\left (b x +a \right ) \polylog \left (2, \frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}-\frac {6 i \mathrm {arccsc}\left (b x +a \right ) \polylog \left (2, -\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}+\frac {6 \polylog \left (3, -\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b}-\frac {6 \polylog \left (3, \frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{b} \]
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 \[ x \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{3} - \frac {3}{4} \, x \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 (4 \, {\left (b^{3} x^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 3 \, a b^{2} x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + a^{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 - a \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )} \log \left (b x + a\right )^{2} - {\left (4 \, b x \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )^{2} - b x \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^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 2 \, a b^{2} x^{2} \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 + {\left (b^{3} x^{3} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + 3 \, a b^{2} x^{2} \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) + a^{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 - a \arctan \left (1, \sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )\right )} \log \left (b x + a\right )\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )\right )}}{4 \, {\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 = -0.01 \[ \int {\mathrm {asin}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]
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 \[ \int \operatorname {acsc}^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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