3.180 \(\int \frac {\text {sech}^4(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=49 \[ \frac {4 \tanh ^3(x)}{21}-\frac {4 \tanh (x)}{7}-\frac {\text {sech}^3(x)}{7 (\sinh (x)+i)}-\frac {i \text {sech}^3(x)}{7 (\sinh (x)+i)^2} \]

[Out]

-1/7*I*sech(x)^3/(I+sinh(x))^2-1/7*sech(x)^3/(I+sinh(x))-4/7*tanh(x)+4/21*tanh(x)^3

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Rubi [A]  time = 0.08, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2672, 3767} \[ \frac {4 \tanh ^3(x)}{21}-\frac {4 \tanh (x)}{7}-\frac {\text {sech}^3(x)}{7 (\sinh (x)+i)}-\frac {i \text {sech}^3(x)}{7 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(I + Sinh[x])^2,x]

[Out]

((-I/7)*Sech[x]^3)/(I + Sinh[x])^2 - Sech[x]^3/(7*(I + Sinh[x])) - (4*Tanh[x])/7 + (4*Tanh[x]^3)/21

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))^2} \, dx &=-\frac {i \text {sech}^3(x)}{7 (i+\sinh (x))^2}-\frac {5}{7} i \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx\\ &=-\frac {i \text {sech}^3(x)}{7 (i+\sinh (x))^2}-\frac {\text {sech}^3(x)}{7 (i+\sinh (x))}-\frac {4}{7} \int \text {sech}^4(x) \, dx\\ &=-\frac {i \text {sech}^3(x)}{7 (i+\sinh (x))^2}-\frac {\text {sech}^3(x)}{7 (i+\sinh (x))}-\frac {4}{7} i \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-i \tanh (x)\right )\\ &=-\frac {i \text {sech}^3(x)}{7 (i+\sinh (x))^2}-\frac {\text {sech}^3(x)}{7 (i+\sinh (x))}-\frac {4 \tanh (x)}{7}+\frac {4 \tanh ^3(x)}{21}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 0.96 \[ -\frac {\text {sech}^3(x) (-14 \sinh (x)-3 \sinh (3 x)+\sinh (5 x)+8 i \cosh (2 x)+4 i \cosh (4 x))}{42 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^4/(I + Sinh[x])^2,x]

[Out]

-1/42*(Sech[x]^3*((8*I)*Cosh[2*x] + (4*I)*Cosh[4*x] - 14*Sinh[x] - 3*Sinh[3*x] + Sinh[5*x]))/(I + Sinh[x])^2

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fricas [B]  time = 0.59, size = 82, normalized size = 1.67 \[ -\frac {224 \, e^{\left (4 \, x\right )} + 128 i \, e^{\left (3 \, x\right )} + 48 \, e^{\left (2 \, x\right )} + 64 i \, e^{x} - 16}{21 \, e^{\left (10 \, x\right )} + 84 i \, e^{\left (9 \, x\right )} - 63 \, e^{\left (8 \, x\right )} + 168 i \, e^{\left (7 \, x\right )} - 294 \, e^{\left (6 \, x\right )} - 294 \, e^{\left (4 \, x\right )} - 168 i \, e^{\left (3 \, x\right )} - 63 \, e^{\left (2 \, x\right )} - 84 i \, e^{x} + 21} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

-(224*e^(4*x) + 128*I*e^(3*x) + 48*e^(2*x) + 64*I*e^x - 16)/(21*e^(10*x) + 84*I*e^(9*x) - 63*e^(8*x) + 168*I*e
^(7*x) - 294*e^(6*x) - 294*e^(4*x) - 168*I*e^(3*x) - 63*e^(2*x) - 84*I*e^x + 21)

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giac [A]  time = 0.23, size = 65, normalized size = 1.33 \[ -\frac {6 i \, e^{\left (2 \, x\right )} + 15 \, e^{x} - 7 i}{24 \, {\left (e^{x} - i\right )}^{3}} - \frac {-42 i \, e^{\left (6 \, x\right )} + 315 \, e^{\left (5 \, x\right )} + 1015 i \, e^{\left (4 \, x\right )} - 1750 \, e^{\left (3 \, x\right )} - 1344 i \, e^{\left (2 \, x\right )} + 511 \, e^{x} + 79 i}{168 \, {\left (e^{x} + i\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-1/24*(6*I*e^(2*x) + 15*e^x - 7*I)/(e^x - I)^3 - 1/168*(-42*I*e^(6*x) + 315*e^(5*x) + 1015*I*e^(4*x) - 1750*e^
(3*x) - 1344*I*e^(2*x) + 511*e^x + 79*I)/(e^x + I)^7

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maple [B]  time = 0.09, size = 116, normalized size = 2.37 \[ \frac {2 i}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{6}}-\frac {5 i}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}+\frac {23 i}{8 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}+\frac {4}{7 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{7}}-\frac {4}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}+\frac {55}{12 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {13}{8 \left (\tanh \left (\frac {x}{2}\right )+i\right )}-\frac {i}{8 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}+\frac {1}{12 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}-\frac {3}{8 \left (\tanh \left (\frac {x}{2}\right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^4/(I+sinh(x))^2,x)

[Out]

2*I/(tanh(1/2*x)+I)^6-5*I/(tanh(1/2*x)+I)^4+23/8*I/(tanh(1/2*x)+I)^2+4/7/(tanh(1/2*x)+I)^7-4/(tanh(1/2*x)+I)^5
+55/12/(tanh(1/2*x)+I)^3-13/8/(tanh(1/2*x)+I)-1/8*I/(tanh(1/2*x)-I)^2+1/12/(tanh(1/2*x)-I)^3-3/8/(tanh(1/2*x)-
I)

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maxima [B]  time = 0.34, size = 317, normalized size = 6.47 \[ -\frac {64 i \, e^{\left (-x\right )}}{84 i \, e^{\left (-x\right )} - 63 \, e^{\left (-2 \, x\right )} + 168 i \, e^{\left (-3 \, x\right )} - 294 \, e^{\left (-4 \, x\right )} - 294 \, e^{\left (-6 \, x\right )} - 168 i \, e^{\left (-7 \, x\right )} - 63 \, e^{\left (-8 \, x\right )} - 84 i \, e^{\left (-9 \, x\right )} + 21 \, e^{\left (-10 \, x\right )} + 21} + \frac {48 \, e^{\left (-2 \, x\right )}}{84 i \, e^{\left (-x\right )} - 63 \, e^{\left (-2 \, x\right )} + 168 i \, e^{\left (-3 \, x\right )} - 294 \, e^{\left (-4 \, x\right )} - 294 \, e^{\left (-6 \, x\right )} - 168 i \, e^{\left (-7 \, x\right )} - 63 \, e^{\left (-8 \, x\right )} - 84 i \, e^{\left (-9 \, x\right )} + 21 \, e^{\left (-10 \, x\right )} + 21} - \frac {128 i \, e^{\left (-3 \, x\right )}}{84 i \, e^{\left (-x\right )} - 63 \, e^{\left (-2 \, x\right )} + 168 i \, e^{\left (-3 \, x\right )} - 294 \, e^{\left (-4 \, x\right )} - 294 \, e^{\left (-6 \, x\right )} - 168 i \, e^{\left (-7 \, x\right )} - 63 \, e^{\left (-8 \, x\right )} - 84 i \, e^{\left (-9 \, x\right )} + 21 \, e^{\left (-10 \, x\right )} + 21} + \frac {224 \, e^{\left (-4 \, x\right )}}{84 i \, e^{\left (-x\right )} - 63 \, e^{\left (-2 \, x\right )} + 168 i \, e^{\left (-3 \, x\right )} - 294 \, e^{\left (-4 \, x\right )} - 294 \, e^{\left (-6 \, x\right )} - 168 i \, e^{\left (-7 \, x\right )} - 63 \, e^{\left (-8 \, x\right )} - 84 i \, e^{\left (-9 \, x\right )} + 21 \, e^{\left (-10 \, x\right )} + 21} - \frac {16}{84 i \, e^{\left (-x\right )} - 63 \, e^{\left (-2 \, x\right )} + 168 i \, e^{\left (-3 \, x\right )} - 294 \, e^{\left (-4 \, x\right )} - 294 \, e^{\left (-6 \, x\right )} - 168 i \, e^{\left (-7 \, x\right )} - 63 \, e^{\left (-8 \, x\right )} - 84 i \, e^{\left (-9 \, x\right )} + 21 \, e^{\left (-10 \, x\right )} + 21} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-64*I*e^(-x)/(84*I*e^(-x) - 63*e^(-2*x) + 168*I*e^(-3*x) - 294*e^(-4*x) - 294*e^(-6*x) - 168*I*e^(-7*x) - 63*e
^(-8*x) - 84*I*e^(-9*x) + 21*e^(-10*x) + 21) + 48*e^(-2*x)/(84*I*e^(-x) - 63*e^(-2*x) + 168*I*e^(-3*x) - 294*e
^(-4*x) - 294*e^(-6*x) - 168*I*e^(-7*x) - 63*e^(-8*x) - 84*I*e^(-9*x) + 21*e^(-10*x) + 21) - 128*I*e^(-3*x)/(8
4*I*e^(-x) - 63*e^(-2*x) + 168*I*e^(-3*x) - 294*e^(-4*x) - 294*e^(-6*x) - 168*I*e^(-7*x) - 63*e^(-8*x) - 84*I*
e^(-9*x) + 21*e^(-10*x) + 21) + 224*e^(-4*x)/(84*I*e^(-x) - 63*e^(-2*x) + 168*I*e^(-3*x) - 294*e^(-4*x) - 294*
e^(-6*x) - 168*I*e^(-7*x) - 63*e^(-8*x) - 84*I*e^(-9*x) + 21*e^(-10*x) + 21) - 16/(84*I*e^(-x) - 63*e^(-2*x) +
168*I*e^(-3*x) - 294*e^(-4*x) - 294*e^(-6*x) - 168*I*e^(-7*x) - 63*e^(-8*x) - 84*I*e^(-9*x) + 21*e^(-10*x) +
21)

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mupad [B]  time = 0.53, size = 139, normalized size = 2.84 \[ \frac {\left (4\,{\mathrm {e}}^{3\,x}-4\,{\mathrm {e}}^x\right )\,\left (\frac {16\,{\mathrm {e}}^{2\,x}}{7}+\frac {32\,{\mathrm {e}}^{4\,x}}{3}-\frac {16}{21}\right )\,1{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^7}-\frac {\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1\right )\,\left (\frac {16\,{\mathrm {e}}^{2\,x}}{7}+\frac {32\,{\mathrm {e}}^{4\,x}}{3}-\frac {16}{21}\right )}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^7}-\frac {\left (4\,{\mathrm {e}}^{3\,x}-4\,{\mathrm {e}}^x\right )\,\left (\frac {128\,{\mathrm {e}}^{3\,x}}{21}+\frac {64\,{\mathrm {e}}^x}{21}\right )}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^7}-\frac {\left (\frac {128\,{\mathrm {e}}^{3\,x}}{21}+\frac {64\,{\mathrm {e}}^x}{21}\right )\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1\right )\,1{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^4*(sinh(x) + 1i)^2),x)

[Out]

((4*exp(3*x) - 4*exp(x))*((16*exp(2*x))/7 + (32*exp(4*x))/3 - 16/21)*1i)/(exp(2*x) + 1)^7 - ((exp(4*x) - 6*exp
(2*x) + 1)*((16*exp(2*x))/7 + (32*exp(4*x))/3 - 16/21))/(exp(2*x) + 1)^7 - ((4*exp(3*x) - 4*exp(x))*((128*exp(
3*x))/21 + (64*exp(x))/21))/(exp(2*x) + 1)^7 - (((128*exp(3*x))/21 + (64*exp(x))/21)*(exp(4*x) - 6*exp(2*x) +
1)*1i)/(exp(2*x) + 1)^7

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{4}{\relax (x )}}{\left (\sinh {\relax (x )} + i\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**4/(I+sinh(x))**2,x)

[Out]

Integral(sech(x)**4/(sinh(x) + I)**2, x)

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