Optimal. Leaf size=180 \[ -\frac {\log \left (e^{2 c (a+b x)}-\sqrt {2} e^{a c+b c x}+1\right )}{2 \sqrt {2} b c}+\frac {\log \left (e^{2 c (a+b x)}+\sqrt {2} e^{a c+b c x}+1\right )}{2 \sqrt {2} b c}+\frac {\tan ^{-1}\left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}-\frac {\tan ^{-1}\left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2} b c}+\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c} \]
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Rubi [A] time = 0.18, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2194, 5207, 12, 2249, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\log \left (e^{2 c (a+b x)}-\sqrt {2} e^{a c+b c x}+1\right )}{2 \sqrt {2} b c}+\frac {\log \left (e^{2 c (a+b x)}+\sqrt {2} e^{a c+b c x}+1\right )}{2 \sqrt {2} b c}+\frac {\tan ^{-1}\left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}-\frac {\tan ^{-1}\left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2} b c}+\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 297
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 2194
Rule 2249
Rule 5207
Rubi steps
\begin {align*} \int e^{c (a+b x)} \tan ^{-1}(\tanh (a c+b c x)) \, dx &=\frac {\operatorname {Subst}\left (\int e^x \tan ^{-1}(\tanh (x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {\operatorname {Subst}\left (\int \frac {2 e^{3 x}}{1+e^{4 x}} \, dx,x,a c+b c x\right )}{b c}\\ &=\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {2 \operatorname {Subst}\left (\int \frac {e^{3 x}}{1+e^{4 x}} \, dx,x,a c+b c x\right )}{b c}\\ &=\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c}+\frac {\operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,e^{a c+b c x}\right )}{b c}-\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,e^{a c+b c x}\right )}{2 b c}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,e^{a c+b c x}\right )}{2 b c}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,e^{a c+b c x}\right )}{2 \sqrt {2} b c}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,e^{a c+b c x}\right )}{2 \sqrt {2} b c}\\ &=\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {\log \left (1-\sqrt {2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt {2} b c}+\frac {\log \left (1+\sqrt {2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt {2} b c}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}\\ &=\frac {\tan ^{-1}\left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}-\frac {\tan ^{-1}\left (1+\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}+\frac {e^{a c+b c x} \tan ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {\log \left (1-\sqrt {2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt {2} b c}+\frac {\log \left (1+\sqrt {2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt {2} b c}\\ \end {align*}
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Mathematica [C] time = 0.13, size = 89, normalized size = 0.49 \[ \frac {\text {RootSum}\left [\text {$\#$1}^4+1\& ,\frac {-\log \left (e^{c (a+b x)}-\text {$\#$1}\right )+a c+b c x}{\text {$\#$1}}\& \right ]+2 e^{c (a+b x)} \tan ^{-1}\left (\frac {e^{2 c (a+b x)}-1}{e^{2 c (a+b x)}+1}\right )}{2 b c} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.51, size = 431, normalized size = 2.39 \[ \frac {4 \, \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} e^{\left (b c x + a c\right )} + \sqrt {2} \sqrt {\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} - 1\right ) + 4 \, \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} e^{\left (b c x + a c\right )} + \sqrt {2} \sqrt {-\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} + 1\right ) + \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}\right ) + 4 \, \arctan \left (\frac {e^{\left (2 \, b c x + 2 \, a c\right )} - 1}{e^{\left (2 \, b c x + 2 \, a c\right )} + 1}\right ) e^{\left (b c x + a c\right )}}{4 \, b c} \]
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 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.62, size = 1355, normalized size = 7.53 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 167, normalized size = 0.93 \[ \frac {\arctan \left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} - \frac {\sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} + \frac {\sqrt {2} \log \left (\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} - \frac {\sqrt {2} \log \left (-\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.42, size = 164, normalized size = 0.91 \[ \frac {4\,{\mathrm {e}}^{a\,c+b\,c\,x}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}-1}{{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}+1}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4-4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4+4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1+1{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4-4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4+4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1+1{}\mathrm {i}\right )}{4\,b\,c} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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