3.632 \(\int \frac {1}{(\text {sech}(x)+i \tanh (x))^4} \, dx\)

Optimal. Leaf size=38 \[ x+\frac {2 i \cosh ^3(x)}{3 (1+i \sinh (x))^3}-\frac {2 i \cosh (x)}{1+i \sinh (x)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4391, 2680, 8} \[ x+\frac {2 i \cosh ^3(x)}{3 (1+i \sinh (x))^3}-\frac {2 i \cosh (x)}{1+i \sinh (x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[x] + I*Tanh[x])^(-4),x]

[Out]

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

Rule 8

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

Rule 2680

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

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 75, normalized size = 1.97 \[ \frac {3 (3 x+8 i) \cosh \left (\frac {x}{2}\right )-(3 x+16 i) \cosh \left (\frac {3 x}{2}\right )+6 i \sinh \left (\frac {x}{2}\right ) (2 x+x \cosh (x)+4 i)}{6 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x] + I*Tanh[x])^(-4),x]

[Out]

(3*(8*I + 3*x)*Cosh[x/2] - (16*I + 3*x)*Cosh[(3*x)/2] + (6*I)*(4*I + 2*x + x*Cosh[x])*Sinh[x/2])/(6*(Cosh[x/2]
+ I*Sinh[x/2])^3)

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fricas [A]  time = 0.42, size = 52, normalized size = 1.37 \[ \frac {3 \, x e^{\left (3 \, x\right )} + {\left (-9 i \, x - 24 i\right )} e^{\left (2 \, x\right )} - 3 \, {\left (3 \, x + 8\right )} e^{x} + 3 i \, x + 16 i}{3 \, e^{\left (3 \, x\right )} - 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} + 3 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))^4,x, algorithm="fricas")

[Out]

(3*x*e^(3*x) + (-9*I*x - 24*I)*e^(2*x) - 3*(3*x + 8)*e^x + 3*I*x + 16*I)/(3*e^(3*x) - 9*I*e^(2*x) - 9*e^x + 3*
I)

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giac [A]  time = 0.14, size = 22, normalized size = 0.58 \[ x - \frac {24 i \, e^{\left (2 \, x\right )} + 24 \, e^{x} - 16 i}{3 \, {\left (e^{x} - i\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))^4,x, algorithm="giac")

[Out]

x - 1/3*(24*I*e^(2*x) + 24*e^x - 16*I)/(e^x - I)^3

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maple [A]  time = 0.36, size = 41, normalized size = 1.08 \[ -\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )+\frac {8 i}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}-\frac {16}{3 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sech(x)+I*tanh(x))^4,x)

[Out]

-ln(tanh(1/2*x)-1)+ln(tanh(1/2*x)+1)+8*I/(tanh(1/2*x)-I)^2-16/3/(tanh(1/2*x)-I)^3

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maxima [A]  time = 0.42, size = 40, normalized size = 1.05 \[ x - \frac {24 \, e^{\left (-x\right )} - 24 i \, e^{\left (-2 \, x\right )} + 16 i}{9 \, e^{\left (-x\right )} - 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} + 3 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))^4,x, algorithm="maxima")

[Out]

x - (24*e^(-x) - 24*I*e^(-2*x) + 16*I)/(9*e^(-x) - 9*I*e^(-2*x) - 3*e^(-3*x) + 3*I)

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mupad [B]  time = 0.17, size = 67, normalized size = 1.76 \[ x+\frac {\frac {{\mathrm {e}}^{2\,x}\,8{}\mathrm {i}}{3}-\frac {8}{3}{}\mathrm {i}}{{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x-\mathrm {i}}-\frac {8{}\mathrm {i}}{3\,\left ({\mathrm {e}}^x-\mathrm {i}\right )}+\frac {{\mathrm {e}}^x\,8{}\mathrm {i}}{3\,\left (1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tanh(x)*1i + 1/cosh(x))^4,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (i \tanh {\relax (x )} + \operatorname {sech}{\relax (x )}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))**4,x)

[Out]

Integral((I*tanh(x) + sech(x))**(-4), x)

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