Optimal. Leaf size=130 \[ -\frac {4 i f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}+\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f} \]
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Rubi [A] time = 0.27, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {5557, 32, 3318, 4184, 3716, 2190, 2279, 2391} \[ -\frac {4 i f^2 \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}+\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f} \]
Antiderivative was successfully verified.
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Rule 32
Rule 2190
Rule 2279
Rule 2391
Rule 3318
Rule 3716
Rule 4184
Rule 5557
Rubi steps
\begin {align*} \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac {(e+f x)^2}{a+i a \sinh (c+d x)} \, dx-\frac {i \int (e+f x)^2 \, dx}{a}\\ &=-\frac {i (e+f x)^3}{3 a f}+\frac {i \int (e+f x)^2 \csc ^2\left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {i d x}{2}\right ) \, dx}{2 a}\\ &=-\frac {i (e+f x)^3}{3 a f}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {(2 i f) \int (e+f x) \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}\\ &=\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(4 f) \int \frac {e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )} (e+f x)}{1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}} \, dx}{a d}\\ &=\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {\left (4 i f^2\right ) \int \log \left (1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {\left (4 i f^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{a d^3}\\ &=\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {4 i f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {i (e+f x)^2 \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 = 2.40, size = 188, normalized size = 1.45 \[ \frac {\frac {3 \left (4 \left (1+i e^c\right ) f^2 \text {Li}_2\left (i e^{-c-d x}\right )-2 d (e+f x) \left (d (e+f x)+2 \left (1+i e^c\right ) f \log \left (1-i e^{-c-d x}\right )\right )\right )}{\left (e^c-i\right ) d^3}+\frac {6 i \sinh \left (\frac {d x}{2}\right ) (e+f x)^2}{d \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}-i x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 a} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.53, size = 262, normalized size = 2.02 \[ -\frac {d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 6 \, d^{2} e^{2} - 12 \, c d e f + 6 \, c^{2} f^{2} + 12 \, {\left (i \, f^{2} e^{\left (d x + c\right )} + f^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - {\left (-i \, d^{3} f^{2} x^{3} + 12 i \, c d e f - 6 i \, c^{2} f^{2} + {\left (-3 i \, d^{3} e f + 6 i \, d^{2} f^{2}\right )} x^{2} + {\left (-3 i \, d^{3} e^{2} + 12 i \, d^{2} e f\right )} x\right )} e^{\left (d x + c\right )} + {\left (12 \, d e f - 12 \, c f^{2} - {\left (-12 i \, d e f + 12 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (12 \, d f^{2} x + 12 \, c f^{2} - {\left (-12 i \, d f^{2} x - 12 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{3 \, {\left (a d^{3} e^{\left (d x + c\right )} - i \, a d^{3}\right )}} \]
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 \[ \int \frac {{\left (f x + e\right )}^{2} \sinh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 269, normalized size = 2.07 \[ -\frac {i x^{3} f^{2}}{3 a}-\frac {i e f \,x^{2}}{a}-\frac {i e^{2} x}{a}-\frac {2 \left (x^{2} f^{2}+2 e f x +e^{2}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {4 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e f}{a \,d^{2}}+\frac {4 i \ln \left ({\mathrm e}^{d x +c}\right ) e f}{a \,d^{2}}+\frac {2 i f^{2} x^{2}}{a d}+\frac {4 i f^{2} c x}{a \,d^{2}}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {4 i f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}-\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}} \]
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 \[ -\frac {1}{6} \, f^{2} {\left (\frac {2 i \, d x^{3} e^{\left (d x + c\right )} + 2 \, d x^{3} + 12 \, x^{2}}{a d e^{\left (d x + c\right )} - i \, a d} - 24 \, \int \frac {x}{a d e^{\left (d x + c\right )} - i \, a d}\,{d x}\right )} + e f {\left (\frac {-i \, d x^{2} + {\left (d x^{2} e^{c} - 4 \, x e^{c}\right )} e^{\left (d x\right )}}{i \, a d e^{\left (d x + c\right )} + a d} - \frac {4 i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {1}{2} \, e^{2} {\left (-\frac {2 i \, {\left (d x + c\right )}}{a d} - \frac {4}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} \]
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 \[ \int \frac {\mathrm {sinh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]
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 \[ \frac {2 e^{2} e^{c} + 4 e f x e^{c} + 2 f^{2} x^{2} e^{c}}{- i a d e^{c} - a d e^{- d x}} - \frac {i \left (\int \left (- \frac {i d e^{2}}{e^{c} e^{d x} - i}\right )\, dx + \int \left (- \frac {i d f^{2} x^{2}}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {d e^{2} e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \frac {4 e f e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \frac {4 f^{2} x e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \left (- \frac {2 i d e f x}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {d f^{2} x^{2} e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \frac {2 d e f x e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx\right )}{a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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