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

Optimal. Leaf size=25 \[ -\frac {2}{3} i \tanh (x)-\frac {i \text {sech}(x)}{3 (\sinh (x)+i)} \]

[Out]

-1/3*I*sech(x)/(I+sinh(x))-2/3*I*tanh(x)

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2672, 3767, 8} \[ -\frac {2}{3} i \tanh (x)-\frac {i \text {sech}(x)}{3 (\sinh (x)+i)} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(I + Sinh[x]),x]

[Out]

((-I/3)*Sech[x])/(I + Sinh[x]) - ((2*I)/3)*Tanh[x]

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

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.88 \[ -\frac {1}{3} i \left (2 \tanh (x)+\frac {\text {sech}(x)}{\sinh (x)+i}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(I + Sinh[x]),x]

[Out]

(-1/3*I)*(Sech[x]/(I + Sinh[x]) + 2*Tanh[x])

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fricas [A]  time = 0.40, size = 28, normalized size = 1.12 \[ -\frac {8 \, e^{x} + 4 i}{3 \, e^{\left (4 \, x\right )} + 6 i \, e^{\left (3 \, x\right )} + 6 i \, e^{x} - 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8*e^x + 4*I)/(3*e^(4*x) + 6*I*e^(3*x) + 6*I*e^x - 3)

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giac [A]  time = 0.14, size = 29, normalized size = 1.16 \[ \frac {1}{2 \, {\left (e^{x} - i\right )}} - \frac {3 \, e^{\left (2 \, x\right )} + 12 i \, e^{x} - 5}{6 \, {\left (e^{x} + i\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(e^x - I) - 1/6*(3*e^(2*x) + 12*I*e^x - 5)/(e^x + I)^3

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maple [B]  time = 0.05, size = 49, normalized size = 1.96 \[ -\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}+\frac {2 i}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {3 i}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 53, normalized size = 2.12 \[ -\frac {8 \, e^{\left (-x\right )}}{-6 i \, e^{\left (-x\right )} - 6 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} - 3} + \frac {4 i}{-6 i \, e^{\left (-x\right )} - 6 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} - 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*e^(-x)/(-6*I*e^(-x) - 6*I*e^(-3*x) + 3*e^(-4*x) - 3) + 4*I/(-6*I*e^(-x) - 6*I*e^(-3*x) + 3*e^(-4*x) - 3)

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mupad [B]  time = 0.58, size = 63, normalized size = 2.52 \[ -\frac {8\,{\mathrm {e}}^x}{3\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^3}-\frac {8\,{\mathrm {e}}^x\,\left ({\mathrm {e}}^{2\,x}-1\right )}{3\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^3}+\frac {{\mathrm {e}}^{2\,x}\,16{}\mathrm {i}}{3\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^3}-\frac {\left ({\mathrm {e}}^{2\,x}-1\right )\,4{}\mathrm {i}}{3\,{\left ({\mathrm {e}}^{2\,x}+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(sinh(x) + 1i)),x)

[Out]

(exp(2*x)*16i)/(3*(exp(2*x) + 1)^3) - (8*exp(x))/(3*(exp(2*x) + 1)^3) - ((exp(2*x) - 1)*4i)/(3*(exp(2*x) + 1)^
3) - (8*exp(x)*(exp(2*x) - 1))/(3*(exp(2*x) + 1)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)**2/(sinh(x) + I), x)

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