3.7.42 \(\int \frac {1}{(\text {sech}(x)-i \tanh (x))^4} \, dx\) [642]

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 = 0.05, 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 = {4476, 2759, 8} \begin {gather*} x-\frac {2 i \cosh ^3(x)}{3 (1-i \sinh (x))^3}+\frac {2 i \cosh (x)}{1-i \sinh (x)} \end {gather*}

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 2759

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 4476

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 = 0.01, size = 74, normalized size = 1.95 \begin {gather*} \frac {3 (-8 i+3 x) \cosh \left (\frac {x}{2}\right )+(16 i-3 x) \cosh \left (\frac {3 x}{2}\right )-6 i (-4 i+2 x+x \cosh (x)) \sinh \left (\frac {x}{2}\right )}{6 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right )^3} \end {gather*}

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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Maple [A]
time = 1.90, size = 41, normalized size = 1.08

method	result	size
risch	\(x +\frac {8 i \left (3 i {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x}-2\right )}{3 \left ({\mathrm e}^{x}+i\right )^{3}}\)	\(26\)
default	\(\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\frac {8 i}{\left (i+\tanh \left (\frac {x}{2}\right )\right )^{2}}-\frac {16}{3 \left (i+\tanh \left (\frac {x}{2}\right )\right )^{3}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sech(x)-I*tanh(x))^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 40, normalized size = 1.05 \begin {gather*} x - \frac {8 \, {\left (3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - 2 i\right )}}{3 \, {\left (3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 52, normalized size = 1.37 \begin {gather*} \frac {3 \, x e^{\left (3 \, x\right )} - 3 \, {\left (-3 i \, x - 8 i\right )} e^{\left (2 \, x\right )} - 3 \, {\left (3 \, x + 8\right )} e^{x} - 3 i \, x - 16 i}{3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (- i \tanh {\left (x \right )} + \operatorname {sech}{\left (x \right )}\right )^{4}}\, dx \end {gather*}

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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Giac [A]
time = 0.41, size = 22, normalized size = 0.58 \begin {gather*} x - \frac {8 \, {\left (-3 i \, e^{\left (2 \, x\right )} + 3 \, e^{x} + 2 i\right )}}{3 \, {\left (e^{x} + i\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.61, size = 65, normalized size = 1.71 \begin {gather*} 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 {{\mathrm {e}}^x\,8{}\mathrm {i}}{3\,\left ({\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}+\frac {8{}\mathrm {i}}{3\,\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )} \end {gather*}

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

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