Optimal. Leaf size=84 \[ -\frac {i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac {i \cos ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {\cos ^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]
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Rubi [A] time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2282, 4626, 3719, 2190, 2279, 2391} \[ -\frac {i \text {PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac {i \cos ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {\cos ^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 3719
Rule 4626
Rubi steps
\begin {align*} \int \cos ^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cos ^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac {i \cos ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac {i \cos ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {\cos ^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{b}-\frac {\operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac {i \cos ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {\cos ^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{b}+\frac {i \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}\\ &=-\frac {i \cos ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac {\cos ^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{b}-\frac {i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}\\ \end {align*}
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Mathematica [F] time = 0.99, size = 0, normalized size = 0.00 \[ \int \cos ^{-1}\left (c e^{a+b x}\right ) \, dx \]
Verification is Not applicable to the result.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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 \[ \int \arccos \left (c e^{\left (b x + a\right )}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.00, size = 107, normalized size = 1.27 \[ -\frac {i \arccos \left (c \,{\mathrm e}^{b x +a}\right )^{2}}{2 b}+\frac {\arccos \left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1+\left (c \,{\mathrm e}^{b x +a}+i \sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )^{2}\right )}{b}-\frac {i \polylog \left (2, -\left (c \,{\mathrm e}^{b x +a}+i \sqrt {1-c^{2} {\mathrm e}^{2 b x +2 a}}\right )^{2}\right )}{2 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {acos}\left (c\,{\mathrm {e}}^{a+b\,x}\right ) \,d x \]
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 \[ \int \operatorname {acos}{\left (c e^{a + b x} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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