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

Optimal. Leaf size=35 \[ -\frac{\cosh (c+d x)}{d (a+i a \sinh (c+d x))}-\frac{i x}{a} \]

[Out]

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

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Rubi [A]  time = 0.0438023, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2735, 2648} \[ -\frac{\cosh (c+d x)}{d (a+i a \sinh (c+d x))}-\frac{i x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac{i x}{a}+i \int \frac{1}{a+i a \sinh (c+d x)} \, dx\\ &=-\frac{i x}{a}-\frac{\cosh (c+d x)}{d (a+i a \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.206753, size = 61, normalized size = 1.74 \[ \frac{i \cosh (c+d x) \left (1-\frac{\sinh ^{-1}(\sinh (c+d x)) (\sinh (c+d x)-i)}{\sqrt{\cosh ^2(c+d x)}}\right )}{a d (\sinh (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 67, normalized size = 1.9 \begin{align*}{\frac{-i}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{2\,i}{da} \left ( -i+\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{i}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1521, size = 49, normalized size = 1.4 \begin{align*} -\frac{i \,{\left (d x + c\right )}}{a d} - \frac{2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.32153, size = 78, normalized size = 2.23 \begin{align*} \frac{-i \, d x e^{\left (d x + c\right )} - d x - 2}{a d e^{\left (d x + c\right )} - i \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.352585, size = 24, normalized size = 0.69 \begin{align*} - \frac{i x}{a} - \frac{2 e^{- c}}{a d \left (e^{d x} - i e^{- c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x/a - 2*exp(-c)/(a*d*(exp(d*x) - I*exp(-c)))

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Giac [A]  time = 1.23488, size = 46, normalized size = 1.31 \begin{align*} -\frac{i \,{\left (d x + c\right )}}{a d} - \frac{2 i}{a d{\left (i \, e^{\left (d x + c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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