3.627 \(\int (\text{sech}(x)+i \tanh (x))^2 \, dx\)

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 = 0.0751763, antiderivative size = 20, normalized size of antiderivative = 1., 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 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]

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 8

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

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*}

Mathematica [A]  time = 0.0048836, size = 14, normalized size = 0.7 \[ -x+2 \tanh (x)-2 i \text{sech}(x) \]

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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Maple [A]  time = 0.015, size = 26, normalized size = 1.3 \begin{align*} 2\,\tanh \left ( x \right ) +2\,i \left ({\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{\cosh \left ( x \right ) }}-\cosh \left ( x \right ) \right ) -x \end{align*}

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*(sinh(x)^2/cosh(x)-cosh(x))-x

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Maxima [A]  time = 1.04959, size = 34, normalized size = 1.7 \begin{align*} -x - \frac{4 i}{e^{\left (-x\right )} + e^{x}} + \frac{4}{e^{\left (-2 \, x\right )} + 1} \end{align*}

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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Fricas [A]  time = 2.29544, size = 43, normalized size = 2.15 \begin{align*} -\frac{x e^{x} + i \, x + 4 i}{e^{x} + i} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (i \tanh{\left (x \right )} + \operatorname{sech}{\left (x \right )}\right )^{2}\, dx \end{align*}

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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Giac [A]  time = 1.11273, size = 16, normalized size = 0.8 \begin{align*} -x - \frac{4 i}{e^{x} + i} \end{align*}

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)