3.280 \(\int \frac{\text{sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=47 \[ \frac{2 \tanh (c+d x)}{3 a d}+\frac{i \text{sech}(c+d x)}{3 d (a+i a \sinh (c+d x))} \]

[Out]

((I/3)*Sech[c + d*x])/(d*(a + I*a*Sinh[c + d*x])) + (2*Tanh[c + d*x])/(3*a*d)

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Rubi [A]  time = 0.054996, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2672, 3767, 8} \[ \frac{2 \tanh (c+d x)}{3 a d}+\frac{i \text{sech}(c+d x)}{3 d (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^2/(a + I*a*Sinh[c + d*x]),x]

[Out]

((I/3)*Sech[c + d*x])/(d*(a + I*a*Sinh[c + d*x])) + (2*Tanh[c + d*x])/(3*a*d)

Rule 2672

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\text{sech}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=\frac{i \text{sech}(c+d x)}{3 d (a+i a \sinh (c+d x))}+\frac{2 \int \text{sech}^2(c+d x) \, dx}{3 a}\\ &=\frac{i \text{sech}(c+d x)}{3 d (a+i a \sinh (c+d x))}+\frac{(2 i) \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{3 a d}\\ &=\frac{i \text{sech}(c+d x)}{3 d (a+i a \sinh (c+d x))}+\frac{2 \tanh (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.0461272, size = 47, normalized size = 1. \[ \frac{\text{sech}(c+d x) (\cosh (2 (c+d x))-2 i \sinh (c+d x))}{3 a d (\sinh (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^2/(a + I*a*Sinh[c + d*x]),x]

[Out]

(Sech[c + d*x]*(Cosh[2*(c + d*x)] - (2*I)*Sinh[c + d*x]))/(3*a*d*(-I + Sinh[c + d*x]))

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Maple [A]  time = 0.049, size = 75, normalized size = 1.6 \begin{align*} 2\,{\frac{1}{da} \left ( 1/4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) +i \right ) ^{-1}-1/3\, \left ( -i+\tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{-3}+{\frac{i/2}{ \left ( -i+\tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}+3/4\, \left ( -i+\tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x)

[Out]

2/d/a*(1/4/(tanh(1/2*d*x+1/2*c)+I)-1/3/(-I+tanh(1/2*d*x+1/2*c))^3+1/2*I/(-I+tanh(1/2*d*x+1/2*c))^2+3/4/(-I+tan
h(1/2*d*x+1/2*c)))

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Maxima [B]  time = 1.18232, size = 140, normalized size = 2.98 \begin{align*} \frac{8 \, e^{\left (-d x - c\right )}}{{\left (6 \, a e^{\left (-d x - c\right )} + 6 \, a e^{\left (-3 \, d x - 3 \, c\right )} - 3 i \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 i \, a\right )} d} + \frac{4 i}{{\left (6 \, a e^{\left (-d x - c\right )} + 6 \, a e^{\left (-3 \, d x - 3 \, c\right )} - 3 i \, a e^{\left (-4 \, d x - 4 \, c\right )} + 3 i \, a\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

8*e^(-d*x - c)/((6*a*e^(-d*x - c) + 6*a*e^(-3*d*x - 3*c) - 3*I*a*e^(-4*d*x - 4*c) + 3*I*a)*d) + 4*I/((6*a*e^(-
d*x - c) + 6*a*e^(-3*d*x - 3*c) - 3*I*a*e^(-4*d*x - 4*c) + 3*I*a)*d)

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Fricas [A]  time = 2.07062, size = 144, normalized size = 3.06 \begin{align*} -\frac{4 \,{\left (-2 i \, e^{\left (d x + c\right )} - 1\right )}}{3 \, a d e^{\left (4 \, d x + 4 \, c\right )} - 6 i \, a d e^{\left (3 \, d x + 3 \, c\right )} - 6 i \, a d e^{\left (d x + c\right )} - 3 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-4*(-2*I*e^(d*x + c) - 1)/(3*a*d*e^(4*d*x + 4*c) - 6*I*a*d*e^(3*d*x + 3*c) - 6*I*a*d*e^(d*x + c) - 3*a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{sech}^{2}{\left (c + d x \right )}}{i \sinh{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)

[Out]

Integral(sech(c + d*x)**2/(I*sinh(c + d*x) + 1), x)/a

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Giac [A]  time = 1.15949, size = 81, normalized size = 1.72 \begin{align*} \frac{1}{2 \, a d{\left (i \, e^{\left (d x + c\right )} - 1\right )}} - \frac{-3 i \, e^{\left (2 \, d x + 2 \, c\right )} - 12 \, e^{\left (d x + c\right )} + 5 i}{6 \, a d{\left (e^{\left (d x + c\right )} - i\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

1/2/(a*d*(I*e^(d*x + c) - 1)) - 1/6*(-3*I*e^(2*d*x + 2*c) - 12*e^(d*x + c) + 5*I)/(a*d*(e^(d*x + c) - I)^3)