Optimal. Leaf size=119 \[ \frac {i f \sinh (c+d x)}{a d^2}+\frac {2 f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{a d^2}-\frac {i (e+f x) \cosh (c+d x)}{a d}-\frac {(e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}+\frac {e x}{a}+\frac {f x^2}{2 a} \]
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Rubi [A] time = 0.18, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {5557, 3296, 2637, 3318, 4184, 3475} \[ \frac {i f \sinh (c+d x)}{a d^2}+\frac {2 f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{a d^2}-\frac {i (e+f x) \cosh (c+d x)}{a d}-\frac {(e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}+\frac {e x}{a}+\frac {f x^2}{2 a} \]
Antiderivative was successfully verified.
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Rule 2637
Rule 3296
Rule 3318
Rule 3475
Rule 4184
Rule 5557
Rubi steps
\begin {align*} \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac {i \int (e+f x) \sinh (c+d x) \, dx}{a}\\ &=-\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {\int (e+f x) \, dx}{a}+\frac {(i f) \int \cosh (c+d x) \, dx}{a d}-\int \frac {e+f x}{a+i a \sinh (c+d x)} \, dx\\ &=\frac {e x}{a}+\frac {f x^2}{2 a}-\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {i f \sinh (c+d x)}{a d^2}-\frac {\int (e+f x) \csc ^2\left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {i d x}{2}\right ) \, dx}{2 a}\\ &=\frac {e x}{a}+\frac {f x^2}{2 a}-\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {i f \sinh (c+d x)}{a d^2}-\frac {(e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {f \int \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}\\ &=\frac {e x}{a}+\frac {f x^2}{2 a}-\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {2 f \log \left (\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right )}{a d^2}+\frac {i f \sinh (c+d x)}{a d^2}-\frac {(e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}\\ \end {align*}
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Mathematica [A] time = 1.02, size = 238, normalized size = 2.00 \[ \frac {\left (\sinh \left (\frac {1}{2} (c+d x)\right )-i \cosh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right ) \left (c^2 (-f)-2 i d (e+f x) \cosh (c+d x)+2 c d e+2 i f \sinh (c+d x)+4 i f \tan ^{-1}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+2 f \log (\cosh (c+d x))-2 i c f+2 d^2 e x+d^2 f x^2-2 i d f x\right )+\sinh \left (\frac {1}{2} (c+d x)\right ) \left (i (c+d x+2 i) (-c f+2 d e+d f x)+2 d (e+f x) \cosh (c+d x)-2 f \sinh (c+d x)-4 f \tan ^{-1}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+2 i f \log (\cosh (c+d x))\right )\right )}{2 a d^2 (\sinh (c+d x)-i)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 174, normalized size = 1.46 \[ -\frac {d f x + d e - {\left (-i \, d f x - i \, d e + i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} - {\left (d^{2} f x^{2} - d e + {\left (2 \, d^{2} e - 5 \, d f\right )} x + f\right )} e^{\left (2 \, d x + 2 \, c\right )} - {\left (-i \, d^{2} f x^{2} - 5 i \, d e + {\left (-2 i \, d^{2} e - i \, d f\right )} x - i \, f\right )} e^{\left (d x + c\right )} - {\left (4 \, f e^{\left (2 \, d x + 2 \, c\right )} - 4 i \, f e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + f}{2 \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.33, size = 272, normalized size = 2.29 \[ \frac {d^{2} f x^{2} e^{\left (2 \, d x + 3 \, c\right )} - i \, d^{2} f x^{2} e^{\left (d x + 2 \, c\right )} - i \, d f x e^{\left (3 \, d x + 4 \, c\right )} + 2 \, d^{2} x e^{\left (2 \, d x + 3 \, c + 1\right )} - 5 \, d f x e^{\left (2 \, d x + 3 \, c\right )} - 2 i \, d^{2} x e^{\left (d x + 2 \, c + 1\right )} - i \, d f x e^{\left (d x + 2 \, c\right )} - d f x e^{c} + 4 \, f e^{\left (2 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 4 i \, f e^{\left (d x + 2 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - i \, d e^{\left (3 \, d x + 4 \, c + 1\right )} + i \, f e^{\left (3 \, d x + 4 \, c\right )} - d e^{\left (2 \, d x + 3 \, c + 1\right )} + f e^{\left (2 \, d x + 3 \, c\right )} - 5 i \, d e^{\left (d x + 2 \, c + 1\right )} - i \, f e^{\left (d x + 2 \, c\right )} - d e^{\left (c + 1\right )} - f e^{c}}{2 \, a d^{2} e^{\left (2 \, d x + 3 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + 2 \, c\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.19, size = 134, normalized size = 1.13 \[ \frac {f \,x^{2}}{2 a}+\frac {e x}{a}-\frac {i \left (d f x +d e -f \right ) {\mathrm e}^{d x +c}}{2 a \,d^{2}}-\frac {i \left (d f x +d e +f \right ) {\mathrm e}^{-d x -c}}{2 a \,d^{2}}-\frac {2 f x}{a d}-\frac {2 f c}{a \,d^{2}}-\frac {2 i \left (f x +e \right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {2 f \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.47, size = 240, normalized size = 2.02 \[ -\frac {1}{4} \, f {\left (\frac {4 \, x e^{\left (d x + c\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac {-2 i \, d^{2} x^{2} e^{c} - 2 i \, d x e^{c} - {\left (2 i \, d x e^{\left (3 \, c\right )} - 2 i \, e^{\left (3 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \, {\left (d^{2} x^{2} e^{\left (2 \, c\right )} - 3 \, d x e^{\left (2 \, c\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} - 2 \, {\left (d x + 1\right )} e^{\left (-d x\right )} - 2 i \, e^{c}}{a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}} - \frac {8 \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {1}{2} \, e {\left (\frac {2 \, {\left (d x + c\right )}}{a d} + \frac {-5 i \, e^{\left (-d x - c\right )} + 1}{{\left (i \, a e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac {i \, e^{\left (-d x - c\right )}}{a d}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.58, size = 143, normalized size = 1.20 \[ \frac {f\,x^2}{2\,a}+{\mathrm {e}}^{c+d\,x}\,\left (\frac {\left (f-d\,e\right )\,1{}\mathrm {i}}{2\,a\,d^2}-\frac {f\,x\,1{}\mathrm {i}}{2\,a\,d}\right )-{\mathrm {e}}^{-c-d\,x}\,\left (\frac {\left (f+d\,e\right )\,1{}\mathrm {i}}{2\,a\,d^2}+\frac {f\,x\,1{}\mathrm {i}}{2\,a\,d}\right )-\frac {\left (e+f\,x\right )\,2{}\mathrm {i}}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}-\frac {x\,\left (2\,f-d\,e\right )}{a\,d}+\frac {2\,f\,\ln \left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-\mathrm {i}\right )}{a\,d^2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.62, size = 235, normalized size = 1.97 \[ \frac {2 i e e^{c} + 2 i f x e^{c}}{- i a d e^{c} - a d e^{- d x}} + \begin {cases} \frac {\left (\left (- 2 i a d^{3} e - 2 i a d^{3} f x - 2 i a d^{2} f\right ) e^{- d x} + \left (- 2 i a d^{3} e e^{2 c} - 2 i a d^{3} f x e^{2 c} + 2 i a d^{2} f e^{2 c}\right ) e^{d x}\right ) e^{- c}}{4 a^{2} d^{4}} & \text {for}\: 4 a^{2} d^{4} e^{c} \neq 0 \\\frac {x^{2} \left (- i f e^{2 c} + i f\right ) e^{- c}}{4 a} + \frac {x \left (- i e e^{2 c} + i e\right ) e^{- c}}{2 a} & \text {otherwise} \end {cases} + \frac {f x^{2}}{2 a} + \frac {x \left (d e + 2 f\right )}{a d} + \frac {2 f \log {\left (i e^{c} + e^{- d x} \right )}}{a d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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