3.178 \(\int \frac {\text {sech}^2(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=37 \[ -\frac {2 \tanh (x)}{5}-\frac {\text {sech}(x)}{5 (\sinh (x)+i)}-\frac {i \text {sech}(x)}{5 (\sinh (x)+i)^2} \]

[Out]

-1/5*I*sech(x)/(I+sinh(x))^2-1/5*sech(x)/(I+sinh(x))-2/5*tanh(x)

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Rubi [A]  time = 0.07, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2672, 3767, 8} \[ -\frac {2 \tanh (x)}{5}-\frac {\text {sech}(x)}{5 (\sinh (x)+i)}-\frac {i \text {sech}(x)}{5 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(I + Sinh[x])^2,x]

[Out]

((-I/5)*Sech[x])/(I + Sinh[x])^2 - Sech[x]/(5*(I + Sinh[x])) - (2*Tanh[x])/5

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.84 \[ -\frac {\text {sech}(x) (-5 \sinh (x)+\sinh (3 x)+4 i \cosh (2 x))}{10 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(I + Sinh[x])^2,x]

[Out]

-1/10*(Sech[x]*((4*I)*Cosh[2*x] - 5*Sinh[x] + Sinh[3*x]))/(I + Sinh[x])^2

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fricas [A]  time = 0.64, size = 46, normalized size = 1.24 \[ -\frac {20 \, e^{\left (2 \, x\right )} + 16 i \, e^{x} - 4}{5 \, e^{\left (6 \, x\right )} + 20 i \, e^{\left (5 \, x\right )} - 25 \, e^{\left (4 \, x\right )} - 25 \, e^{\left (2 \, x\right )} - 20 i \, e^{x} + 5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

-(20*e^(2*x) + 16*I*e^x - 4)/(5*e^(6*x) + 20*I*e^(5*x) - 25*e^(4*x) - 25*e^(2*x) - 20*I*e^x + 5)

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giac [A]  time = 0.20, size = 41, normalized size = 1.11 \[ -\frac {i}{4 \, {\left (e^{x} - i\right )}} - \frac {-5 i \, e^{\left (4 \, x\right )} + 30 \, e^{\left (3 \, x\right )} + 80 i \, e^{\left (2 \, x\right )} - 50 \, e^{x} - 11 i}{20 \, {\left (e^{x} + i\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-1/4*I/(e^x - I) - 1/20*(-5*I*e^(4*x) + 30*e^(3*x) + 80*I*e^(2*x) - 50*e^x - 11*I)/(e^x + I)^5

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maple [B]  time = 0.06, size = 70, normalized size = 1.89 \[ -\frac {2 i}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}+\frac {5 i}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {4}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}+\frac {3}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {7}{4 \left (\tanh \left (\frac {x}{2}\right )+i\right )}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(I+sinh(x))^2,x)

[Out]

-2*I/(tanh(1/2*x)+I)^4+5/2*I/(tanh(1/2*x)+I)^2-4/5/(tanh(1/2*x)+I)^5+3/(tanh(1/2*x)+I)^3-7/4/(tanh(1/2*x)+I)-1
/4/(tanh(1/2*x)-I)

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maxima [B]  time = 0.32, size = 117, normalized size = 3.16 \[ -\frac {16 i \, e^{\left (-x\right )}}{20 i \, e^{\left (-x\right )} - 25 \, e^{\left (-2 \, x\right )} - 25 \, e^{\left (-4 \, x\right )} - 20 i \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-6 \, x\right )} + 5} + \frac {20 \, e^{\left (-2 \, x\right )}}{20 i \, e^{\left (-x\right )} - 25 \, e^{\left (-2 \, x\right )} - 25 \, e^{\left (-4 \, x\right )} - 20 i \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-6 \, x\right )} + 5} - \frac {4}{20 i \, e^{\left (-x\right )} - 25 \, e^{\left (-2 \, x\right )} - 25 \, e^{\left (-4 \, x\right )} - 20 i \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-6 \, x\right )} + 5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-16*I*e^(-x)/(20*I*e^(-x) - 25*e^(-2*x) - 25*e^(-4*x) - 20*I*e^(-5*x) + 5*e^(-6*x) + 5) + 20*e^(-2*x)/(20*I*e^
(-x) - 25*e^(-2*x) - 25*e^(-4*x) - 20*I*e^(-5*x) + 5*e^(-6*x) + 5) - 4/(20*I*e^(-x) - 25*e^(-2*x) - 25*e^(-4*x
) - 20*I*e^(-5*x) + 5*e^(-6*x) + 5)

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mupad [B]  time = 0.73, size = 109, normalized size = 2.95 \[ -\frac {16\,{\mathrm {e}}^x\,\left (4\,{\mathrm {e}}^{3\,x}-4\,{\mathrm {e}}^x\right )}{5\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^5}-\frac {\left (4\,{\mathrm {e}}^{2\,x}-\frac {4}{5}\right )\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1\right )}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^5}-\frac {{\mathrm {e}}^x\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1\right )\,16{}\mathrm {i}}{5\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^5}+\frac {\left (4\,{\mathrm {e}}^{3\,x}-4\,{\mathrm {e}}^x\right )\,\left (4\,{\mathrm {e}}^{2\,x}-\frac {4}{5}\right )\,1{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(sinh(x) + 1i)^2),x)

[Out]

((4*exp(3*x) - 4*exp(x))*(4*exp(2*x) - 4/5)*1i)/(exp(2*x) + 1)^5 - (exp(x)*(exp(4*x) - 6*exp(2*x) + 1)*16i)/(5
*(exp(2*x) + 1)^5) - (16*exp(x)*(4*exp(3*x) - 4*exp(x)))/(5*(exp(2*x) + 1)^5) - ((4*exp(2*x) - 4/5)*(exp(4*x)
- 6*exp(2*x) + 1))/(exp(2*x) + 1)^5

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{2}{\relax (x )}}{\left (\sinh {\relax (x )} + i\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(I+sinh(x))**2,x)

[Out]

Integral(sech(x)**2/(sinh(x) + I)**2, x)

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