Optimal. Leaf size=159 \[ -\frac {i f \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac {i f \text {Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac {i (e+f x) \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x) \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac {(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f} \]
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Rubi [A] time = 0.10, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5183, 4180, 2531, 2282, 6589} \[ -\frac {i f \text {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac {i f \text {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}+\frac {i (e+f x) \text {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x) \text {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac {(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac {(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4180
Rule 5183
Rule 6589
Rubi steps
\begin {align*} \int (e+f x) \tan ^{-1}(\tanh (a+b x)) \, dx &=\frac {(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f}-\frac {b \int (e+f x)^2 \text {sech}(2 a+2 b x) \, dx}{2 f}\\ &=-\frac {(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac {(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f}+\frac {1}{2} i \int (e+f x) \log \left (1-i e^{2 a+2 b x}\right ) \, dx-\frac {1}{2} i \int (e+f x) \log \left (1+i e^{2 a+2 b x}\right ) \, dx\\ &=-\frac {(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac {(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f}+\frac {i (e+f x) \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x) \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {(i f) \int \text {Li}_2\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b}+\frac {(i f) \int \text {Li}_2\left (i e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=-\frac {(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac {(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f}+\frac {i (e+f x) \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x) \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {(i f) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}+\frac {(i f) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=-\frac {(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac {(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f}+\frac {i (e+f x) \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x) \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac {i f \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac {i f \text {Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}\\ \end {align*}
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Mathematica [A] time = 1.94, size = 278, normalized size = 1.75 \[ -\frac {i f \left (2 b^2 x^2 \log \left (1-i e^{2 (a+b x)}\right )-2 b^2 x^2 \log \left (1+i e^{2 (a+b x)}\right )-2 b x \text {Li}_2\left (-i e^{2 (a+b x)}\right )+2 b x \text {Li}_2\left (i e^{2 (a+b x)}\right )+\text {Li}_3\left (-i e^{2 (a+b x)}\right )-\text {Li}_3\left (i e^{2 (a+b x)}\right )\right )}{8 b^2}-\frac {e \left (-2 i \left (\text {Li}_2\left (-i e^{2 (a+b x)}\right )-\text {Li}_2\left (i e^{2 (a+b x)}\right )\right )-\left ((-4 i a-4 i b x+\pi ) \left (\log \left (1-i e^{2 (a+b x)}\right )-\log \left (1+i e^{2 (a+b x)}\right )\right )\right )+(\pi -4 i a) \log \left (\cot \left (\frac {1}{4} (4 i a+4 i b x+\pi )\right )\right )\right )}{8 b}+e x \tan ^{-1}(\tanh (a+b x))+\frac {1}{2} f x^2 \tan ^{-1}(\tanh (a+b x)) \]
Antiderivative was successfully verified.
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fricas [C] time = 0.64, size = 596, normalized size = 3.75 \[ \frac {2 \, {\left (b^{2} f x^{2} + 2 \, b^{2} e x\right )} \arctan \left (\frac {\sinh \left (b x + a\right )}{\cosh \left (b x + a\right )}\right ) + {\left (-2 i \, b f x - 2 i \, b e\right )} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + {\left (-2 i \, b f x - 2 i \, b e\right )} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + {\left (2 i \, b f x + 2 i \, b e\right )} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + {\left (2 i \, b f x + 2 i \, b e\right )} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + {\left (-i \, b^{2} f x^{2} - 2 i \, b^{2} e x - 2 i \, a b e + i \, a^{2} f\right )} \log \left (\frac {1}{2} \, \sqrt {4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (-i \, b^{2} f x^{2} - 2 i \, b^{2} e x - 2 i \, a b e + i \, a^{2} f\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (i \, b^{2} f x^{2} + 2 i \, b^{2} e x + 2 i \, a b e - i \, a^{2} f\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (i \, b^{2} f x^{2} + 2 i \, b^{2} e x + 2 i \, a b e - i \, a^{2} f\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (2 i \, a b e - i \, a^{2} f\right )} \log \left (i \, \sqrt {4 i} + 2 \, \cosh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )\right ) + {\left (2 i \, a b e - i \, a^{2} f\right )} \log \left (-i \, \sqrt {4 i} + 2 \, \cosh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )\right ) + {\left (-2 i \, a b e + i \, a^{2} f\right )} \log \left (i \, \sqrt {-4 i} + 2 \, \cosh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )\right ) + {\left (-2 i \, a b e + i \, a^{2} f\right )} \log \left (-i \, \sqrt {-4 i} + 2 \, \cosh \left (b x + a\right ) + 2 \, \sinh \left (b x + a\right )\right ) + 2 i \, f {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 2 i \, f {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 i \, f {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 i \, f {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{4 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [C] time = 4.75, size = 2414, normalized size = 15.18 \[ \text {Expression too large to display} \]
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 \[ \frac {1}{2} \, {\left (f x^{2} + 2 \, e x\right )} \arctan \left (\frac {e^{\left (2 \, b x + 2 \, a\right )} - 1}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \int \frac {{\left (b f x^{2} e^{\left (2 \, a\right )} + 2 \, b e x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{e^{\left (4 \, b x + 4 \, a\right )} + 1}\,{d x} \]
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 \[ \int \mathrm {atan}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\,\left (e+f\,x\right ) \,d x \]
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 \[ \int \left (e + f x\right ) \operatorname {atan}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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