Optimal. Leaf size=324 \[ \csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\csc ^{-1}(a+b x)^2 \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )-2 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )+2 \text {PolyLog}\left (3,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \text {PolyLog}\left (3,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {1}{2} \text {PolyLog}\left (3,e^{2 i \csc ^{-1}(a+b x)}\right ) \]
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Rubi [A]
time = 0.34, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps
used = 17, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5367, 4648,
4625, 3798, 2221, 2611, 2320, 6724, 4615} \begin {gather*} -2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+2 \text {Li}_3\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )+i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{2 i \csc ^{-1}(a+b x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 i \csc ^{-1}(a+b x)}\right )-\csc ^{-1}(a+b x)^2 \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 3798
Rule 4615
Rule 4625
Rule 4648
Rule 5367
Rule 6724
Rubi steps
\begin {align*} \int \frac {\csc ^{-1}(a+b x)^2}{x} \, dx &=-\text {Subst}\left (\int \frac {x^2 \cot (x) \csc (x)}{-a+\csc (x)} \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=-\text {Subst}\left (\int \frac {x^2 \cot (x)}{1-a \sin (x)} \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=-\left (a \text {Subst}\left (\int \frac {x^2 \cos (x)}{1-a \sin (x)} \, dx,x,\csc ^{-1}(a+b x)\right )\right )-\text {Subst}\left (\int x^2 \cot (x) \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=2 i \text {Subst}\left (\int \frac {e^{2 i x} x^2}{1-e^{2 i x}} \, dx,x,\csc ^{-1}(a+b x)\right )-a \text {Subst}\left (\int \frac {e^{i x} x^2}{1-\sqrt {1-a^2}+i a e^{i x}} \, dx,x,\csc ^{-1}(a+b x)\right )-a \text {Subst}\left (\int \frac {e^{i x} x^2}{1+\sqrt {1-a^2}+i a e^{i x}} \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\csc ^{-1}(a+b x)^2 \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )-2 \text {Subst}\left (\int x \log \left (1+\frac {i a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )-2 \text {Subst}\left (\int x \log \left (1+\frac {i a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )+2 \text {Subst}\left (\int x \log \left (1-e^{2 i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\csc ^{-1}(a+b x)^2 \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )-2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{2 i \csc ^{-1}(a+b x)}\right )-i \text {Subst}\left (\int \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )+2 i \text {Subst}\left (\int \text {Li}_2\left (-\frac {i a e^{i x}}{1-\sqrt {1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )+2 i \text {Subst}\left (\int \text {Li}_2\left (-\frac {i a e^{i x}}{1+\sqrt {1-a^2}}\right ) \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\csc ^{-1}(a+b x)^2 \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )-2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{2 i \csc ^{-1}(a+b x)}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i \csc ^{-1}(a+b x)}\right )+2 \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i a x}{-1+\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )+2 \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {i a x}{1+\sqrt {1-a^2}}\right )}{x} \, dx,x,e^{i \csc ^{-1}(a+b x)}\right )\\ &=\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\csc ^{-1}(a+b x)^2 \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )-2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \text {Li}_2\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \csc ^{-1}(a+b x) \text {Li}_2\left (e^{2 i \csc ^{-1}(a+b x)}\right )+2 \text {Li}_3\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \text {Li}_3\left (-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 i \csc ^{-1}(a+b x)}\right )\\ \end {align*}
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Mathematica [A]
time = 0.24, size = 408, normalized size = 1.26 \begin {gather*} \frac {i \pi ^3}{6}-\frac {1}{3} i \csc ^{-1}(a+b x)^3-\csc ^{-1}(a+b x)^2 \log \left (1-e^{-i \csc ^{-1}(a+b x)}\right )-\csc ^{-1}(a+b x)^2 \log \left (1+e^{i \csc ^{-1}(a+b x)}\right )+\csc ^{-1}(a+b x)^2 \log \left (1-\frac {i a e^{i \csc ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,e^{-i \csc ^{-1}(a+b x)}\right )+2 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,-e^{i \csc ^{-1}(a+b x)}\right )-2 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {i a e^{i \csc ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \text {PolyLog}\left (2,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-2 \text {PolyLog}\left (3,e^{-i \csc ^{-1}(a+b x)}\right )-2 \text {PolyLog}\left (3,-e^{i \csc ^{-1}(a+b x)}\right )+2 \text {PolyLog}\left (3,\frac {i a e^{i \csc ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )+2 \text {PolyLog}\left (3,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 2.38, size = 0, normalized size = 0.00 \[\int \frac {\mathrm {arccsc}\left (b x +a \right )^{2}}{x}\, dx\]
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 {\operatorname {acsc}^{2}{\left (a + b x \right )}}{x}\, dx \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 = -0.00 \begin {gather*} \int \frac {{\mathrm {asin}\left (\frac {1}{a+b\,x}\right )}^2}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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