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

Optimal. Leaf size=13 \[ -i \log (-\sinh (x)+i) \]

[Out]

-I*ln(I-sinh(x))

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Rubi [A]  time = 0.03, antiderivative size = 13, 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 = {3159, 2667, 31} \[ -i \log (-\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[(Sech[x] + I*Tanh[x])^(-1),x]

[Out]

(-I)*Log[I - Sinh[x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2667

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

Rule 3159

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Int[Cos[d + e*x
]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

Rubi steps

\begin {align*} \int \frac {1}{\text {sech}(x)+i \tanh (x)} \, dx &=\int \frac {\cosh (x)}{1+i \sinh (x)} \, dx\\ &=-\left (i \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,i \sinh (x)\right )\right )\\ &=-i \log (i-\sinh (x))\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.31 \[ 2 \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )-i \log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x] + I*Tanh[x])^(-1),x]

[Out]

2*ArcTan[Tanh[x/2]] - I*Log[Cosh[x]]

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fricas [A]  time = 0.43, size = 11, normalized size = 0.85 \[ i \, x - 2 i \, \log \left (e^{x} - i\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x - 2*I*log(e^x - I)

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giac [A]  time = 0.14, size = 13, normalized size = 1.00 \[ i \, x - 2 i \, \log \left (i \, e^{x} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x - 2*I*log(I*e^x + 1)

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maple [B]  time = 0.24, size = 33, normalized size = 2.54 \[ i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-2 i \ln \left (\tanh \left (\frac {x}{2}\right )-i\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 15, normalized size = 1.15 \[ -i \, x - 2 i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x - 2*I*log(I*e^(-x) - 1)

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mupad [B]  time = 1.56, size = 14, normalized size = 1.08 \[ x\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tanh(x)*1i + 1/cosh(x)),x)

[Out]

x*1i - log(exp(x) - 1i)*2i

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sympy [B]  time = 0.21, size = 22, normalized size = 1.69 \[ - i x - i \log {\left (i \tanh {\relax (x )} + \operatorname {sech}{\relax (x )} \right )} + i \log {\left (\tanh {\relax (x )} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x - I*log(I*tanh(x) + sech(x)) + I*log(tanh(x) + 1)

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