3.640 \(\int \frac {1}{(\text {sech}(x)-i \tanh (x))^2} \, dx\)

Optimal. Leaf size=20 \[ -x-\frac {2 i \cosh (x)}{1-i \sinh (x)} \]

[Out]

-x-2*I*cosh(x)/(1-I*sinh(x))

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4391, 2680, 8} \[ -x-\frac {2 i \cosh (x)}{1-i \sinh (x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[x] - I*Tanh[x])^(-2),x]

[Out]

-x - ((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))^2} \, dx &=\int \frac {\cosh ^2(x)}{(1-i \sinh (x))^2} \, dx\\ &=-\frac {2 i \cosh (x)}{1-i \sinh (x)}-\int 1 \, dx\\ &=-x-\frac {2 i \cosh (x)}{1-i \sinh (x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 31, normalized size = 1.55 \[ -x+\frac {4 \sinh \left (\frac {x}{2}\right )}{\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x] - I*Tanh[x])^(-2),x]

[Out]

-x + (4*Sinh[x/2])/(Cosh[x/2] - I*Sinh[x/2])

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fricas [A]  time = 0.42, size = 17, normalized size = 0.85 \[ -\frac {x e^{x} + i \, x + 4 i}{e^{x} + i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))^2,x, algorithm="fricas")

[Out]

-(x*e^x + I*x + 4*I)/(e^x + I)

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giac [A]  time = 0.12, size = 12, normalized size = 0.60 \[ -x - \frac {4 i}{e^{x} + i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))^2,x, algorithm="giac")

[Out]

-x - 4*I/(e^x + I)

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maple [A]  time = 0.29, size = 29, normalized size = 1.45 \[ \frac {4}{\tanh \left (\frac {x}{2}\right )+i}+\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sech(x)-I*tanh(x))^2,x)

[Out]

4/(tanh(1/2*x)+I)+ln(tanh(1/2*x)-1)-ln(tanh(1/2*x)+1)

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maxima [A]  time = 0.50, size = 14, normalized size = 0.70 \[ -x - \frac {4 i}{e^{\left (-x\right )} - i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))^2,x, algorithm="maxima")

[Out]

-x - 4*I/(e^(-x) - I)

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mupad [B]  time = 1.51, size = 14, normalized size = 0.70 \[ -x-\frac {4{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tanh(x)*1i - 1/cosh(x))^2,x)

[Out]

- x - 4i/(exp(x) + 1i)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- i \tanh {\relax (x )} + \operatorname {sech}{\relax (x )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))**2,x)

[Out]

Integral((-I*tanh(x) + sech(x))**(-2), x)

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