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

Optimal. Leaf size=42 \[ \frac{\tan ^{-1}(\sinh (c+d x))}{2 a d}+\frac{i}{2 d (a+i a \sinh (c+d x))} \]

[Out]

ArcTan[Sinh[c + d*x]]/(2*a*d) + (I/2)/(d*(a + I*a*Sinh[c + d*x]))

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Rubi [A]  time = 0.0566407, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2667, 44, 206} \[ \frac{\tan ^{-1}(\sinh (c+d x))}{2 a d}+\frac{i}{2 d (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[Sinh[c + d*x]]/(2*a*d) + (I/2)/(d*(a + I*a*Sinh[c + d*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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0480522, size = 30, normalized size = 0.71 \[ \frac{\tan ^{-1}(\sinh (c+d x))+\frac{1}{\sinh (c+d x)-i}}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[Sinh[c + d*x]] + (-I + Sinh[c + d*x])^(-1))/(2*a*d)

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Maple [B]  time = 0.046, size = 91, normalized size = 2.2 \begin{align*}{\frac{{\frac{i}{2}}}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +i \right ) }-{\frac{i}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{{\frac{i}{2}}}{da}\ln \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I/d/a*ln(tanh(1/2*d*x+1/2*c)+I)-I/d/a/(-I+tanh(1/2*d*x+1/2*c))^2-1/2*I/d/a*ln(-I+tanh(1/2*d*x+1/2*c))-1/d/
a/(-I+tanh(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.16804, size = 117, normalized size = 2.79 \begin{align*} -\frac{2 \, e^{\left (-d x - c\right )}}{{\left (4 i \, a e^{\left (-d x - c\right )} + 2 \, a e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a\right )} d} - \frac{i \, \log \left (e^{\left (-d x - c\right )} + i\right )}{2 \, a d} + \frac{i \, \log \left (i \, e^{\left (-d x - c\right )} + 1\right )}{2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.12009, size = 267, normalized size = 6.36 \begin{align*} \frac{{\left (i \, e^{\left (2 \, d x + 2 \, c\right )} + 2 \, e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} + i\right ) +{\left (-i \, e^{\left (2 \, d x + 2 \, c\right )} - 2 \, e^{\left (d x + c\right )} + i\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 2 \, e^{\left (d x + c\right )}}{2 \, a d e^{\left (2 \, d x + 2 \, c\right )} - 4 i \, a d e^{\left (d x + c\right )} - 2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((I*e^(2*d*x + 2*c) + 2*e^(d*x + c) - I)*log(e^(d*x + c) + I) + (-I*e^(2*d*x + 2*c) - 2*e^(d*x + c) + I)*log(e
^(d*x + c) - I) + 2*e^(d*x + c))/(2*a*d*e^(2*d*x + 2*c) - 4*I*a*d*e^(d*x + c) - 2*a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{sech}{\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)/(a+I*a*sinh(d*x+c)),x)

[Out]

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

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Giac [B]  time = 1.15309, size = 147, normalized size = 3.5 \begin{align*} -\frac{i \, \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} - 2 i\right )}{4 \, a d} + \frac{i \, \log \left (i \, e^{\left (d x + c\right )} - i \, e^{\left (-d x - c\right )} - 2\right )}{4 \, a d} - \frac{-i \, e^{\left (d x + c\right )} + i \, e^{\left (-d x - c\right )} - 6}{4 \, a d{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} - 2 i\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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