Optimal. Leaf size=91 \[ \frac {i a}{8 d (a+i a \sinh (c+d x))^2}-\frac {i}{8 d (a-i a \sinh (c+d x))}+\frac {i}{4 d (a+i a \sinh (c+d x))}+\frac {3 \tan ^{-1}(\sinh (c+d x))}{8 a d} \]
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Rubi [A] time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2667, 44, 206} \[ \frac {i a}{8 d (a+i a \sinh (c+d x))^2}-\frac {i}{8 d (a-i a \sinh (c+d x))}+\frac {i}{4 d (a+i a \sinh (c+d x))}+\frac {3 \tan ^{-1}(\sinh (c+d x))}{8 a d} \]
Antiderivative was successfully verified.
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Rule 44
Rule 206
Rule 2667
Rubi steps
\begin {align*} \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^3} \, dx,x,i a \sinh (c+d x)\right )}{d}\\ &=-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{8 a^3 (a-x)^2}+\frac {1}{4 a^2 (a+x)^3}+\frac {1}{4 a^3 (a+x)^2}+\frac {3}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,i a \sinh (c+d x)\right )}{d}\\ &=-\frac {i}{8 d (a-i a \sinh (c+d x))}+\frac {i a}{8 d (a+i a \sinh (c+d x))^2}+\frac {i}{4 d (a+i a \sinh (c+d x))}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,i a \sinh (c+d x)\right )}{8 d}\\ &=\frac {3 \tan ^{-1}(\sinh (c+d x))}{8 a d}-\frac {i}{8 d (a-i a \sinh (c+d x))}+\frac {i a}{8 d (a+i a \sinh (c+d x))^2}+\frac {i}{4 d (a+i a \sinh (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 101, normalized size = 1.11 \[ \frac {\text {sech}^2(c+d x) \left (3 \sinh ^3(c+d x) \tan ^{-1}(\sinh (c+d x))+\sinh ^2(c+d x) \left (3-3 i \tan ^{-1}(\sinh (c+d x))\right )+3 \sinh (c+d x) \left (\tan ^{-1}(\sinh (c+d x))-i\right )-3 i \tan ^{-1}(\sinh (c+d x))+2\right )}{8 a d (\sinh (c+d x)-i)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.58, size = 286, normalized size = 3.14 \[ \frac {{\left (3 i \, e^{\left (6 \, d x + 6 \, c\right )} + 6 \, e^{\left (5 \, d x + 5 \, c\right )} + 3 i \, e^{\left (4 \, d x + 4 \, c\right )} + 12 \, e^{\left (3 \, d x + 3 \, c\right )} - 3 i \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} - 3 i\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + {\left (-3 i \, e^{\left (6 \, d x + 6 \, c\right )} - 6 \, e^{\left (5 \, d x + 5 \, c\right )} - 3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 12 \, e^{\left (3 \, d x + 3 \, c\right )} + 3 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )} + 3 i\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 6 \, e^{\left (5 \, d x + 5 \, c\right )} - 12 i \, e^{\left (4 \, d x + 4 \, c\right )} + 4 \, e^{\left (3 \, d x + 3 \, c\right )} + 12 i \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )}}{8 \, a d e^{\left (6 \, d x + 6 \, c\right )} - 16 i \, a d e^{\left (5 \, d x + 5 \, c\right )} + 8 \, a d e^{\left (4 \, d x + 4 \, c\right )} - 32 i \, a d e^{\left (3 \, d x + 3 \, c\right )} - 8 \, a d e^{\left (2 \, d x + 2 \, c\right )} - 16 i \, a d e^{\left (d x + c\right )} - 8 \, a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.21, size = 177, normalized size = 1.95 \[ -\frac {-\frac {6 i \, \log \left (-i \, e^{\left (d x + c\right )} + i \, e^{\left (-d x - c\right )} + 2\right )}{a} + \frac {6 i \, \log \left (-i \, e^{\left (d x + c\right )} + i \, e^{\left (-d x - c\right )} - 2\right )}{a} - \frac {2 \, {\left (3 \, e^{\left (d x + c\right )} - 3 \, e^{\left (-d x - c\right )} + 10 i\right )}}{a {\left (i \, e^{\left (d x + c\right )} - i \, e^{\left (-d x - c\right )} - 2\right )}} + \frac {-9 i \, {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 52 \, e^{\left (d x + c\right )} + 52 \, e^{\left (-d x - c\right )} + 84 i}{a {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} - 2 i\right )}^{2}}}{32 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.11, size = 180, normalized size = 1.98 \[ \frac {i}{4 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}+\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}{8 d a}-\frac {1}{4 d a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}+\frac {i}{2 d a \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 i \ln \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {3 i}{2 d a \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{d a \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {1}{d a \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.33, size = 180, normalized size = 1.98 \[ -\frac {8 \, {\left (3 \, e^{\left (-d x - c\right )} - 6 i \, e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, e^{\left (-3 \, d x - 3 \, c\right )} + 6 i \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )}\right )}}{{\left (64 i \, a e^{\left (-d x - c\right )} - 32 \, a e^{\left (-2 \, d x - 2 \, c\right )} + 128 i \, a e^{\left (-3 \, d x - 3 \, c\right )} + 32 \, a e^{\left (-4 \, d x - 4 \, c\right )} + 64 i \, a e^{\left (-5 \, d x - 5 \, c\right )} + 32 \, a e^{\left (-6 \, d x - 6 \, c\right )} - 32 \, a\right )} d} - \frac {3 i \, \log \left (e^{\left (-d x - c\right )} + i\right )}{8 \, a d} + \frac {3 i \, \log \left (e^{\left (-d x - c\right )} - i\right )}{8 \, a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.99, size = 137, normalized size = 1.51 \[ \frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2\,d^2}}{a\,d}\right )}{4\,\sqrt {a^2\,d^2}}+\frac {1}{2\,a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}+\frac {1}{4\,a\,d\,\left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}-\frac {1{}\mathrm {i}}{4\,a\,d\,{\left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}^2}-\frac {1{}\mathrm {i}}{a\,d\,{\left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^3}+\frac {1{}\mathrm {i}}{2\,a\,d\,{\left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^4} \]
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 \[ - \frac {i \int \frac {\operatorname {sech}^{3}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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