3.2.92 \(\int \frac {\sinh (c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx\) [192]

Optimal. Leaf size=32 \[ \text {Int}\left (\frac {\sinh (c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))},x\right ) \]

[Out]

Unintegrable(sinh(d*x+c)/(f*x+e)^2/(a+I*a*sinh(d*x+c)),x)

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Rubi [A]
time = 0.04, 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 = {} \begin {gather*} \int \frac {\sinh (c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {\sinh (c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx &=\int \frac {\sinh (c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx\\ \end {align*}

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Mathematica [A]
time = 32.80, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh (c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [A]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\sinh \left (d x +c \right )}{\left (f x +e \right )^{2} \left (a +i a \sinh \left (d x +c \right )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-I*a*d*f^2*x^2 - 2*I*a*d*f*x*e - I*a*d*e^2 + (a*d*f^2*x^2 + 2*a*d*f*x*e + a*d*e^2)*e^(d*x + c))*integral(-(d
*f*x + d*e - (-I*d*f*x - I*d*e)*e^(d*x + c) + 4*f)/(-I*a*d*f^3*x^3 - 3*I*a*d*f^2*x^2*e - 3*I*a*d*f*x*e^2 - I*a
*d*e^3 + (a*d*f^3*x^3 + 3*a*d*f^2*x^2*e + 3*a*d*f*x*e^2 + a*d*e^3)*e^(d*x + c)), x) - 2)/(-I*a*d*f^2*x^2 - 2*I
*a*d*f*x*e - I*a*d*e^2 + (a*d*f^2*x^2 + 2*a*d*f*x*e + a*d*e^2)*e^(d*x + c))

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Sympy [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {2}{- i a d e^{2} - 2 i a d e f x - i a d f^{2} x^{2} + \left (a d e^{2} e^{c} + 2 a d e f x e^{c} + a d f^{2} x^{2} e^{c}\right ) e^{d x}} - \frac {i \left (\int \left (- \frac {4 i f}{e^{3} e^{c} e^{d x} - i e^{3} + 3 e^{2} f x e^{c} e^{d x} - 3 i e^{2} f x + 3 e f^{2} x^{2} e^{c} e^{d x} - 3 i e f^{2} x^{2} + f^{3} x^{3} e^{c} e^{d x} - i f^{3} x^{3}}\right )\, dx + \int \left (- \frac {i d e}{e^{3} e^{c} e^{d x} - i e^{3} + 3 e^{2} f x e^{c} e^{d x} - 3 i e^{2} f x + 3 e f^{2} x^{2} e^{c} e^{d x} - 3 i e f^{2} x^{2} + f^{3} x^{3} e^{c} e^{d x} - i f^{3} x^{3}}\right )\, dx + \int \left (- \frac {i d f x}{e^{3} e^{c} e^{d x} - i e^{3} + 3 e^{2} f x e^{c} e^{d x} - 3 i e^{2} f x + 3 e f^{2} x^{2} e^{c} e^{d x} - 3 i e f^{2} x^{2} + f^{3} x^{3} e^{c} e^{d x} - i f^{3} x^{3}}\right )\, dx + \int \frac {d e e^{c} e^{d x}}{e^{3} e^{c} e^{d x} - i e^{3} + 3 e^{2} f x e^{c} e^{d x} - 3 i e^{2} f x + 3 e f^{2} x^{2} e^{c} e^{d x} - 3 i e f^{2} x^{2} + f^{3} x^{3} e^{c} e^{d x} - i f^{3} x^{3}}\, dx + \int \frac {d f x e^{c} e^{d x}}{e^{3} e^{c} e^{d x} - i e^{3} + 3 e^{2} f x e^{c} e^{d x} - 3 i e^{2} f x + 3 e f^{2} x^{2} e^{c} e^{d x} - 3 i e f^{2} x^{2} + f^{3} x^{3} e^{c} e^{d x} - i f^{3} x^{3}}\, dx\right )}{a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(-I*a*d*e**2 - 2*I*a*d*e*f*x - I*a*d*f**2*x**2 + (a*d*e**2*exp(c) + 2*a*d*e*f*x*exp(c) + a*d*f**2*x**2*exp(
c))*exp(d*x)) - I*(Integral(-4*I*f/(e**3*exp(c)*exp(d*x) - I*e**3 + 3*e**2*f*x*exp(c)*exp(d*x) - 3*I*e**2*f*x
+ 3*e*f**2*x**2*exp(c)*exp(d*x) - 3*I*e*f**2*x**2 + f**3*x**3*exp(c)*exp(d*x) - I*f**3*x**3), x) + Integral(-I
*d*e/(e**3*exp(c)*exp(d*x) - I*e**3 + 3*e**2*f*x*exp(c)*exp(d*x) - 3*I*e**2*f*x + 3*e*f**2*x**2*exp(c)*exp(d*x
) - 3*I*e*f**2*x**2 + f**3*x**3*exp(c)*exp(d*x) - I*f**3*x**3), x) + Integral(-I*d*f*x/(e**3*exp(c)*exp(d*x) -
I*e**3 + 3*e**2*f*x*exp(c)*exp(d*x) - 3*I*e**2*f*x + 3*e*f**2*x**2*exp(c)*exp(d*x) - 3*I*e*f**2*x**2 + f**3*x
**3*exp(c)*exp(d*x) - I*f**3*x**3), x) + Integral(d*e*exp(c)*exp(d*x)/(e**3*exp(c)*exp(d*x) - I*e**3 + 3*e**2*
f*x*exp(c)*exp(d*x) - 3*I*e**2*f*x + 3*e*f**2*x**2*exp(c)*exp(d*x) - 3*I*e*f**2*x**2 + f**3*x**3*exp(c)*exp(d*
x) - I*f**3*x**3), x) + Integral(d*f*x*exp(c)*exp(d*x)/(e**3*exp(c)*exp(d*x) - I*e**3 + 3*e**2*f*x*exp(c)*exp(
d*x) - 3*I*e**2*f*x + 3*e*f**2*x**2*exp(c)*exp(d*x) - 3*I*e*f**2*x**2 + f**3*x**3*exp(c)*exp(d*x) - I*f**3*x**
3), x))/(a*d)

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Giac [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\mathrm {sinh}\left (c+d\,x\right )}{{\left (e+f\,x\right )}^2\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)/((e + f*x)^2*(a + a*sinh(c + d*x)*1i)),x)

[Out]

int(sinh(c + d*x)/((e + f*x)^2*(a + a*sinh(c + d*x)*1i)), x)

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