Optimal. Leaf size=85 \[ \frac {i \csc ^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac {\csc ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \csc ^{-1}\left (c e^{a+b x}\right )}\right )}{b}+\frac {i \text {PolyLog}\left (2,e^{2 i \csc ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b} \]
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Rubi [A]
time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {2320, 5327,
4721, 3798, 2221, 2317, 2438} \begin {gather*} \frac {i \text {Li}_2\left (e^{2 i \csc ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac {i \csc ^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac {\csc ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 i \csc ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2317
Rule 2320
Rule 2438
Rule 3798
Rule 4721
Rule 5327
Rubi steps
\begin {align*} \int \csc ^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac {\text {Subst}\left (\int \frac {\csc ^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \frac {\sin ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,e^{-a-b x}\right )}{b}\\ &=-\frac {\text {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac {i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}+\frac {(2 i) \text {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac {i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}-\frac {\sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{b}+\frac {\text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac {i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}-\frac {\sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{b}-\frac {i \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{2 b}\\ &=\frac {i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}-\frac {\sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{b}+\frac {i \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{2 b}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(85)=170\).
time = 0.46, size = 280, normalized size = 3.29 \begin {gather*} x \csc ^{-1}\left (c e^{a+b x}\right )+\frac {e^{-a-b x} \left (4 \sqrt {-1+c^2 e^{2 (a+b x)}} \text {ArcTan}\left (\sqrt {-1+c^2 e^{2 (a+b x)}}\right ) \left (2 b x-\log \left (c^2 e^{2 (a+b x)}\right )\right )+\sqrt {1-c^2 e^{2 (a+b x)}} \left (\log ^2\left (c^2 e^{2 (a+b x)}\right )-4 \log \left (c^2 e^{2 (a+b x)}\right ) \log \left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )+2 \log ^2\left (\frac {1}{2} \left (1+\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )\right )-4 \sqrt {1-c^2 e^{2 (a+b x)}} \text {PolyLog}\left (2,\frac {1}{2} \left (1-\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )\right )}{8 b c \sqrt {1-\frac {e^{-2 (a+b x)}}{c^2}}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.21, size = 188, normalized size = 2.21
method | result | size |
derivativedivides | \(\frac {\frac {i \mathrm {arccsc}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\mathrm {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1-\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )-\mathrm {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \polylog \left (2, \frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \polylog \left (2, -\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )}{b}\) | \(188\) |
default | \(\frac {\frac {i \mathrm {arccsc}\left ({\mathrm e}^{b x +a} c \right )^{2}}{2}-\mathrm {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1-\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )-\mathrm {arccsc}\left ({\mathrm e}^{b x +a} c \right ) \ln \left (1+\frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \polylog \left (2, \frac {i {\mathrm e}^{-b x -a}}{c}+\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )+i \polylog \left (2, -\frac {i {\mathrm e}^{-b x -a}}{c}-\sqrt {1-\frac {{\mathrm e}^{-2 b x -2 a}}{c^{2}}}\right )}{b}\) | \(188\) |
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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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 \operatorname {acsc}{\left (c e^{a + b x} \right )}\, 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 [B]
time = 1.25, size = 91, normalized size = 1.07 \begin {gather*} \frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2\,b}+\frac {{\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )}^2\,1{}\mathrm {i}}{2\,b}-\frac {\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right )}{b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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