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

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 = 0.07, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4391, 2670, 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 2670

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a/g)^
(2*m), Int[(g*Cos[e + f*x])^(2*m + p)/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 -
b^2, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]

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 (\text {sech}(x)-i \tanh (x))^2 \, dx &=\int \text {sech}^2(x) (1-i \sinh (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 = 0.01, size = 14, normalized size = 0.70 \[ -x+2 \tanh (x)+2 i \text {sech}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x] - I*Tanh[x])^2,x]

[Out]

-x + (2*I)*Sech[x] + 2*Tanh[x]

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

[Out]

-x + 4*I/(e^x - I)

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maple [A]  time = 0.35, size = 16, normalized size = 0.80 \[ 2 \tanh \relax (x )+\frac {2 i}{\cosh \relax (x )}-x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*tanh(x)+2*I/cosh(x)-x

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maxima [A]  time = 0.41, size = 25, normalized size = 1.25 \[ -x + \frac {4 i}{e^{\left (-x\right )} + e^{x}} + \frac {4}{e^{\left (-2 \, x\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 4*I/(e^(-x) + e^x) + 4/(e^(-2*x) + 1)

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mupad [B]  time = 1.56, size = 14, normalized size = 0.70 \[ -x+\frac {4{}\mathrm {i}}{{\mathrm {e}}^x-\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x)*1i - 1/cosh(x))^2,x)

[Out]

4i/(exp(x) - 1i) - x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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