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

Optimal. Leaf size=32 \[ \text {Int}\left (\frac {\sinh (c+d x)}{(e+f x) (a+i a \sinh (c+d x))},x\right ) \]

[Out]

Unintegrable(sinh(d*x+c)/(f*x+e)/(a+I*a*sinh(d*x+c)),x)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\sinh (c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 48.57, size = 0, normalized size = 0.00 \[ \int \frac {\sinh (c+d x)}{(e+f x) (a+i a \sinh (c+d x))} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 0, normalized size = 0.00 \[ \frac {{\left (-i \, a d f x - i \, a d e + {\left (a d f x + a d e\right )} e^{\left (d x + c\right )}\right )} {\rm integral}\left (-\frac {d f x + d e - {\left (-i \, d f x - i \, d e\right )} e^{\left (d x + c\right )} + 2 \, f}{-i \, a d f^{2} x^{2} - 2 i \, a d e f x - i \, a d e^{2} + {\left (a d f^{2} x^{2} + 2 \, a d e f x + a d e^{2}\right )} e^{\left (d x + c\right )}}, x\right ) - 2}{-i \, a d f x - i \, a d e + {\left (a d f x + a d e\right )} e^{\left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (d x + c\right )}{{\left (f x + e\right )} {\left (i \, a \sinh \left (d x + c\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (d x +c \right )}{\left (f x +e \right ) \left (a +i a \sinh \left (d x +c \right )\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -2 \, f \int \frac {1}{-i \, a d f^{2} x^{2} - 2 i \, a d e f x - i \, a d e^{2} + {\left (a d f^{2} x^{2} e^{c} + 2 \, a d e f x e^{c} + a d e^{2} e^{c}\right )} e^{\left (d x\right )}}\,{d x} - \frac {2}{-i \, a d f x - i \, a d e + {\left (a d f x e^{c} + a d e e^{c}\right )} e^{\left (d x\right )}} - \frac {i \, \log \left (f x + e\right )}{a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [A]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {sinh}\left (c+d\,x\right )}{\left (e+f\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)/((e + f*x)*(a + a*sinh(c + d*x)*1i)),x)

[Out]

int(sinh(c + d*x)/((e + f*x)*(a + a*sinh(c + d*x)*1i)), x)

________________________________________________________________________________________

sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {2 i e^{c}}{- a d e e^{c} - a d f x e^{c} + \left (i a d e + i a d f x\right ) e^{- d x}} - \frac {i \left (\int \left (- \frac {i d e}{e^{2} e^{c} e^{d x} - i e^{2} + 2 e f x e^{c} e^{d x} - 2 i e f x + f^{2} x^{2} e^{c} e^{d x} - i f^{2} x^{2}}\right )\, dx + \int \left (- \frac {2 f e^{c} e^{d x}}{e^{2} e^{c} e^{d x} - i e^{2} + 2 e f x e^{c} e^{d x} - 2 i e f x + f^{2} x^{2} e^{c} e^{d x} - i f^{2} x^{2}}\right )\, dx + \int \left (- \frac {i d f x}{e^{2} e^{c} e^{d x} - i e^{2} + 2 e f x e^{c} e^{d x} - 2 i e f x + f^{2} x^{2} e^{c} e^{d x} - i f^{2} x^{2}}\right )\, dx + \int \frac {d e e^{c} e^{d x}}{e^{2} e^{c} e^{d x} - i e^{2} + 2 e f x e^{c} e^{d x} - 2 i e f x + f^{2} x^{2} e^{c} e^{d x} - i f^{2} x^{2}}\, dx + \int \frac {d f x e^{c} e^{d x}}{e^{2} e^{c} e^{d x} - i e^{2} + 2 e f x e^{c} e^{d x} - 2 i e f x + f^{2} x^{2} e^{c} e^{d x} - i f^{2} x^{2}}\, dx\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*exp(c)/(-a*d*e*exp(c) - a*d*f*x*exp(c) + (I*a*d*e + I*a*d*f*x)*exp(-d*x)) - I*(Integral(-I*d*e/(e**2*exp(
c)*exp(d*x) - I*e**2 + 2*e*f*x*exp(c)*exp(d*x) - 2*I*e*f*x + f**2*x**2*exp(c)*exp(d*x) - I*f**2*x**2), x) + In
tegral(-2*f*exp(c)*exp(d*x)/(e**2*exp(c)*exp(d*x) - I*e**2 + 2*e*f*x*exp(c)*exp(d*x) - 2*I*e*f*x + f**2*x**2*e
xp(c)*exp(d*x) - I*f**2*x**2), x) + Integral(-I*d*f*x/(e**2*exp(c)*exp(d*x) - I*e**2 + 2*e*f*x*exp(c)*exp(d*x)
- 2*I*e*f*x + f**2*x**2*exp(c)*exp(d*x) - I*f**2*x**2), x) + Integral(d*e*exp(c)*exp(d*x)/(e**2*exp(c)*exp(d*
x) - I*e**2 + 2*e*f*x*exp(c)*exp(d*x) - 2*I*e*f*x + f**2*x**2*exp(c)*exp(d*x) - I*f**2*x**2), x) + Integral(d*
f*x*exp(c)*exp(d*x)/(e**2*exp(c)*exp(d*x) - I*e**2 + 2*e*f*x*exp(c)*exp(d*x) - 2*I*e*f*x + f**2*x**2*exp(c)*ex
p(d*x) - I*f**2*x**2), x))/(a*d)

________________________________________________________________________________________