3.169 \(\int \frac{\text{sech}^4(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=37 \[ \frac{4}{15} i \tanh ^3(x)-\frac{4}{5} i \tanh (x)-\frac{i \text{sech}^3(x)}{5 (\sinh (x)+i)} \]

[Out]

((-I/5)*Sech[x]^3)/(I + Sinh[x]) - ((4*I)/5)*Tanh[x] + ((4*I)/15)*Tanh[x]^3

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Rubi [A]  time = 0.0437954, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2672, 3767} \[ \frac{4}{15} i \tanh ^3(x)-\frac{4}{5} i \tanh (x)-\frac{i \text{sech}^3(x)}{5 (\sinh (x)+i)} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(I + Sinh[x]),x]

[Out]

((-I/5)*Sech[x]^3)/(I + Sinh[x]) - ((4*I)/5)*Tanh[x] + ((4*I)/15)*Tanh[x]^3

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{\text{sech}^4(x)}{i+\sinh (x)} \, dx &=-\frac{i \text{sech}^3(x)}{5 (i+\sinh (x))}-\frac{4}{5} i \int \text{sech}^4(x) \, dx\\ &=-\frac{i \text{sech}^3(x)}{5 (i+\sinh (x))}+\frac{4}{5} \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-i \tanh (x)\right )\\ &=-\frac{i \text{sech}^3(x)}{5 (i+\sinh (x))}-\frac{4}{5} i \tanh (x)+\frac{4}{15} i \tanh ^3(x)\\ \end{align*}

Mathematica [A]  time = 0.0642261, size = 35, normalized size = 0.95 \[ -\frac{1}{15} i \left (8 \tanh ^3(x)+\frac{3 \text{sech}^3(x)}{\sinh (x)+i}+12 \tanh (x) \text{sech}^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^4/(I + Sinh[x]),x]

[Out]

(-I/15)*((3*Sech[x]^3)/(I + Sinh[x]) + 12*Sech[x]^2*Tanh[x] + 8*Tanh[x]^3)

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Maple [B]  time = 0.04, size = 93, normalized size = 2.5 \begin{align*}{{\frac{i}{6}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}}-{{\frac{5\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}-{{\frac{2\,i}{5}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}+{{\frac{5\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{{\frac{11\,i}{8}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}-{\frac{3}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^4/(I+sinh(x)),x)

[Out]

1/6*I/(tanh(1/2*x)-I)^3-5/8*I/(tanh(1/2*x)-I)+1/4/(tanh(1/2*x)-I)^2-2/5*I/(tanh(1/2*x)+I)^5+5/3*I/(tanh(1/2*x)
+I)^3-11/8*I/(tanh(1/2*x)+I)+1/(tanh(1/2*x)+I)^4-3/2/(tanh(1/2*x)+I)^2

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Maxima [B]  time = 1.21073, size = 277, normalized size = 7.49 \begin{align*} -\frac{32 \, e^{\left (-x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} + \frac{32 i \, e^{\left (-2 \, x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} - \frac{96 \, e^{\left (-3 \, x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} + \frac{16 i}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(I+sinh(x)),x, algorithm="maxima")

[Out]

-32*e^(-x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-
8*x) - 15) + 32*I*e^(-2*x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-6*x) - 30*I*e^
(-7*x) + 15*e^(-8*x) - 15) - 96*e^(-3*x)/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) + 30*e^(-
6*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15) + 16*I/(-30*I*e^(-x) - 30*e^(-2*x) - 90*I*e^(-3*x) - 90*I*e^(-5*x) +
30*e^(-6*x) - 30*I*e^(-7*x) + 15*e^(-8*x) - 15)

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Fricas [B]  time = 1.78141, size = 197, normalized size = 5.32 \begin{align*} -\frac{96 \, e^{\left (3 \, x\right )} + 32 i \, e^{\left (2 \, x\right )} + 32 \, e^{x} + 16 i}{15 \, e^{\left (8 \, x\right )} + 30 i \, e^{\left (7 \, x\right )} + 30 \, e^{\left (6 \, x\right )} + 90 i \, e^{\left (5 \, x\right )} + 90 i \, e^{\left (3 \, x\right )} - 30 \, e^{\left (2 \, x\right )} + 30 i \, e^{x} - 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(I+sinh(x)),x, algorithm="fricas")

[Out]

-(96*e^(3*x) + 32*I*e^(2*x) + 32*e^x + 16*I)/(15*e^(8*x) + 30*I*e^(7*x) + 30*e^(6*x) + 90*I*e^(5*x) + 90*I*e^(
3*x) - 30*e^(2*x) + 30*I*e^x - 15)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**4/(I+sinh(x)),x)

[Out]

Timed out

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Giac [B]  time = 1.27, size = 72, normalized size = 1.95 \begin{align*} \frac{9 \, e^{\left (2 \, x\right )} - 24 i \, e^{x} - 11}{24 \,{\left (e^{x} - i\right )}^{3}} - \frac{45 \, e^{\left (4 \, x\right )} + 240 i \, e^{\left (3 \, x\right )} - 490 \, e^{\left (2 \, x\right )} - 320 i \, e^{x} + 73}{120 \,{\left (e^{x} + i\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(I+sinh(x)),x, algorithm="giac")

[Out]

1/24*(9*e^(2*x) - 24*I*e^x - 11)/(e^x - I)^3 - 1/120*(45*e^(4*x) + 240*I*e^(3*x) - 490*e^(2*x) - 320*I*e^x + 7
3)/(e^x + I)^5