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]

-1/5*I*sech(x)^3/(I+sinh(x))-4/5*I*tanh(x)+4/15*I*tanh(x)^3

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.07, 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]

(-1/15*I)*((3*Sech[x]^3)/(I + Sinh[x]) + 12*Sech[x]^2*Tanh[x] + 8*Tanh[x]^3)

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fricas [B]  time = 0.44, size = 64, normalized size = 1.73 \[ -\frac {96 \, e^{\left (3 \, x\right )} + 32 i \, e^{\left (2 \, x\right )} + 32 \, e^{x} + 16 i}{15 \, e^{\left (8 \, x\right )} + 30 i \, e^{\left (7 \, x\right )} + 30 \, e^{\left (6 \, x\right )} + 90 i \, e^{\left (5 \, x\right )} + 90 i \, e^{\left (3 \, x\right )} - 30 \, e^{\left (2 \, x\right )} + 30 i \, e^{x} - 15} \]

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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giac [B]  time = 0.19, size = 53, normalized size = 1.43 \[ \frac {9 \, e^{\left (2 \, x\right )} - 24 i \, e^{x} - 11}{24 \, {\left (e^{x} - i\right )}^{3}} - \frac {45 \, e^{\left (4 \, x\right )} + 240 i \, e^{\left (3 \, x\right )} - 490 \, e^{\left (2 \, x\right )} - 320 i \, e^{x} + 73}{120 \, {\left (e^{x} + i\right )}^{5}} \]

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

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maple [B]  time = 0.06, size = 93, normalized size = 2.51 \[ -\frac {2 i}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}+\frac {5 i}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {11 i}{8 \left (\tanh \left (\frac {x}{2}\right )+i\right )}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}-\frac {3}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}+\frac {i}{6 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}-\frac {5 i}{8 \left (\tanh \left (\frac {x}{2}\right )-i\right )}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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+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

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maxima [B]  time = 0.31, size = 205, normalized size = 5.54 \[ -\frac {32 \, e^{\left (-x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} + \frac {32 i \, e^{\left (-2 \, x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} - \frac {96 \, e^{\left (-3 \, x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} + \frac {16 i}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} \]

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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mupad [B]  time = 1.01, size = 231, normalized size = 6.24 \[ -\frac {1}{6\,\left ({\mathrm {e}}^{2\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x-\mathrm {i}\right )}-\frac {\frac {3\,{\mathrm {e}}^x}{40}+\frac {1}{8}{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}}-\frac {\frac {3\,{\mathrm {e}}^{2\,x}}{40}-\frac {5}{24}+\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{4}}{{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}+{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x-\mathrm {i}}+\frac {1{}\mathrm {i}}{4\,\left (1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}+\frac {3}{8\,\left ({\mathrm {e}}^x-\mathrm {i}\right )}-\frac {3}{40\,\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}-\frac {\frac {{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}}{8}+\frac {3\,{\mathrm {e}}^{3\,x}}{40}-\frac {5\,{\mathrm {e}}^x}{8}-\frac {1}{8}{}\mathrm {i}}{{\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1+{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}-{\mathrm {e}}^x\,4{}\mathrm {i}}-\frac {\frac {3\,{\mathrm {e}}^{4\,x}}{40}-\frac {5\,{\mathrm {e}}^{2\,x}}{4}+\frac {3}{40}+\frac {{\mathrm {e}}^{3\,x}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{2}}{{\mathrm {e}}^{5\,x}-10\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}\,5{}\mathrm {i}-{\mathrm {e}}^{2\,x}\,10{}\mathrm {i}+5\,{\mathrm {e}}^x+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^4*(sinh(x) + 1i)),x)

[Out]

1i/(4*(exp(x)*2i - exp(2*x) + 1)) - ((3*exp(x))/40 + 1i/8)/(exp(2*x) + exp(x)*2i - 1) - ((3*exp(2*x))/40 + (ex
p(x)*1i)/4 - 5/24)/(exp(2*x)*3i + exp(3*x) - 3*exp(x) - 1i) - 1/(6*(exp(2*x)*3i - exp(3*x) + 3*exp(x) - 1i)) +
3/(8*(exp(x) - 1i)) - 3/(40*(exp(x) + 1i)) - ((exp(2*x)*3i)/8 + (3*exp(3*x))/40 - (5*exp(x))/8 - 1i/8)/(exp(3
*x)*4i - 6*exp(2*x) + exp(4*x) - exp(x)*4i + 1) - ((exp(3*x)*1i)/2 - (5*exp(2*x))/4 + (3*exp(4*x))/40 - (exp(x
)*1i)/2 + 3/40)/(exp(4*x)*5i - 10*exp(3*x) - exp(2*x)*10i + exp(5*x) + 5*exp(x) + 1i)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{4}{\relax (x )}}{\sinh {\relax (x )} + i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)**4/(sinh(x) + I), x)

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