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

Optimal. Leaf size=34 \[ \text {Int}\left (\frac {\sinh ^2(c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, 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 = {} \[ \int \frac {\sinh ^2(c+d x)}{(e+f x)^2 (a+i a \sinh (c+d x))} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 0.55, size = 0, normalized size = 0.00 \[ \frac {{\left (-i \, a d f^{2} x^{2} - 2 i \, a d e f x - i \, a d e^{2} + {\left (a d f^{2} x^{2} + 2 \, a d e f x + a d e^{2}\right )} e^{\left (d x + c\right )}\right )} {\rm integral}\left (\frac {d f x + d e + {\left (-i \, d f x - i \, d e\right )} e^{\left (3 \, d x + 3 \, c\right )} + {\left (d f x + d e\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-i \, d f x - i \, d e - 8 i \, f\right )} e^{\left (d x + c\right )}}{2 \, {\left (a d f^{3} x^{3} + 3 \, a d e f^{2} x^{2} + 3 \, a d e^{2} f x + a d e^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-2 i \, a d f^{3} x^{3} - 6 i \, a d e f^{2} x^{2} - 6 i \, a d e^{2} f x - 2 i \, a d e^{3}\right )} e^{\left (d x + c\right )}}, x\right ) - 2 i}{-i \, a d f^{2} x^{2} - 2 i \, a d e f x - i \, a d e^{2} + {\left (a d f^{2} x^{2} + 2 \, a d e f x + a d e^{2}\right )} e^{\left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^2/(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*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))*integral((d*
f*x + d*e + (-I*d*f*x - I*d*e)*e^(3*d*x + 3*c) + (d*f*x + d*e)*e^(2*d*x + 2*c) + (-I*d*f*x - I*d*e - 8*I*f)*e^
(d*x + c))/(2*(a*d*f^3*x^3 + 3*a*d*e*f^2*x^2 + 3*a*d*e^2*f*x + a*d*e^3)*e^(2*d*x + 2*c) + (-2*I*a*d*f^3*x^3 -
6*I*a*d*e*f^2*x^2 - 6*I*a*d*e^2*f*x - 2*I*a*d*e^3)*e^(d*x + c)), x) - 2*I)/(-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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}\left (d x +c \right )}{\left (f x +e \right )^{2} \left (a +i a \sinh \left (d x +c \right )\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sinh(d*x+c)^2/(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 \[ -4 i \, f \int \frac {1}{-i \, a d f^{3} x^{3} - 3 i \, a d e f^{2} x^{2} - 3 i \, a d e^{2} f x - i \, a d e^{3} + {\left (a d f^{3} x^{3} e^{c} + 3 \, a d e f^{2} x^{2} e^{c} + 3 \, a d e^{2} f x e^{c} + a d e^{3} e^{c}\right )} e^{\left (d x\right )}}\,{d x} + \frac {4 i \, d f x + 4 i \, d e - 4 \, {\left (d f x e^{c} + d e e^{c}\right )} e^{\left (d x\right )} - 8 i \, f}{4 \, {\left (-i \, a d f^{3} x^{2} - 2 i \, a d e f^{2} x - i \, a d e^{2} f + {\left (a d f^{3} x^{2} e^{c} + 2 \, a d e f^{2} x e^{c} + a d e^{2} f e^{c}\right )} e^{\left (d x\right )}\right )}} - \frac {i \, e^{\left (-c + \frac {d e}{f}\right )} E_{2}\left (\frac {{\left (f x + e\right )} d}{f}\right )}{2 \, {\left (f x + e\right )} a f} + \frac {i \, e^{\left (c - \frac {d e}{f}\right )} E_{2}\left (-\frac {{\left (f x + e\right )} d}{f}\right )}{2 \, {\left (f x + e\right )} a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*I*f*integrate(1/(-I*a*d*f^3*x^3 - 3*I*a*d*e*f^2*x^2 - 3*I*a*d*e^2*f*x - I*a*d*e^3 + (a*d*f^3*x^3*e^c + 3*a*
d*e*f^2*x^2*e^c + 3*a*d*e^2*f*x*e^c + a*d*e^3*e^c)*e^(d*x)), x) + 1/4*(4*I*d*f*x + 4*I*d*e - 4*(d*f*x*e^c + d*
e*e^c)*e^(d*x) - 8*I*f)/(-I*a*d*f^3*x^2 - 2*I*a*d*e*f^2*x - I*a*d*e^2*f + (a*d*f^3*x^2*e^c + 2*a*d*e*f^2*x*e^c
+ a*d*e^2*f*e^c)*e^(d*x)) - 1/2*I*e^(-c + d*e/f)*exp_integral_e(2, (f*x + e)*d/f)/((f*x + e)*a*f) + 1/2*I*e^(
c - d*e/f)*exp_integral_e(2, -(f*x + e)*d/f)/((f*x + e)*a*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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