3.410 \(\int \frac{\sinh ^2(c+d x) \tanh (c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx\)

Optimal. Leaf size=36 \[ \text{Unintegrable}\left (\frac{\sinh ^2(c+d x) \tanh (c+d x)}{(e+f x) (a+b \sinh (c+d x))},x\right ) \]

[Out]

Unintegrable[(Sinh[c + d*x]^2*Tanh[c + d*x])/((e + f*x)*(a + b*Sinh[c + d*x])), x]

________________________________________________________________________________________

Rubi [A]  time = 0.0891297, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\sinh ^2(c+d x) \tanh (c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [F]  time = 180.002, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

Maple [A]  time = 0.934, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}\tanh \left ( dx+c \right ) }{ \left ( fx+e \right ) \left ( a+b\sinh \left ( dx+c \right ) \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{e^{\left (-c + \frac{d e}{f}\right )} E_{1}\left (\frac{{\left (f x + e\right )} d}{f}\right )}{2 \, b f} - \frac{e^{\left (c - \frac{d e}{f}\right )} E_{1}\left (-\frac{{\left (f x + e\right )} d}{f}\right )}{2 \, b f} - \frac{a \log \left (f x + e\right )}{b^{2} f} + \frac{1}{4} \, \int -\frac{8 \,{\left (a^{4} e^{\left (d x + c\right )} - a^{3} b\right )}}{a^{2} b^{3} e + b^{5} e +{\left (a^{2} b^{3} f + b^{5} f\right )} x -{\left (a^{2} b^{3} e e^{\left (2 \, c\right )} + b^{5} e e^{\left (2 \, c\right )} +{\left (a^{2} b^{3} f e^{\left (2 \, c\right )} + b^{5} f e^{\left (2 \, c\right )}\right )} x\right )} e^{\left (2 \, d x\right )} - 2 \,{\left (a^{3} b^{2} e e^{c} + a b^{4} e e^{c} +{\left (a^{3} b^{2} f e^{c} + a b^{4} f e^{c}\right )} x\right )} e^{\left (d x\right )}}\,{d x} - \frac{1}{4} \, \int \frac{8 \,{\left (b e^{\left (d x + c\right )} - a\right )}}{a^{2} e + b^{2} e +{\left (a^{2} f + b^{2} f\right )} x +{\left (a^{2} e e^{\left (2 \, c\right )} + b^{2} e e^{\left (2 \, c\right )} +{\left (a^{2} f e^{\left (2 \, c\right )} + b^{2} f e^{\left (2 \, c\right )}\right )} x\right )} e^{\left (2 \, d x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^(-c + d*e/f)*exp_integral_e(1, (f*x + e)*d/f)/(b*f) - 1/2*e^(c - d*e/f)*exp_integral_e(1, -(f*x + e)*d/
f)/(b*f) - a*log(f*x + e)/(b^2*f) + 1/4*integrate(-8*(a^4*e^(d*x + c) - a^3*b)/(a^2*b^3*e + b^5*e + (a^2*b^3*f
+ b^5*f)*x - (a^2*b^3*e*e^(2*c) + b^5*e*e^(2*c) + (a^2*b^3*f*e^(2*c) + b^5*f*e^(2*c))*x)*e^(2*d*x) - 2*(a^3*b
^2*e*e^c + a*b^4*e*e^c + (a^3*b^2*f*e^c + a*b^4*f*e^c)*x)*e^(d*x)), x) - 1/4*integrate(8*(b*e^(d*x + c) - a)/(
a^2*e + b^2*e + (a^2*f + b^2*f)*x + (a^2*e*e^(2*c) + b^2*e*e^(2*c) + (a^2*f*e^(2*c) + b^2*f*e^(2*c))*x)*e^(2*d
*x)), x)

________________________________________________________________________________________

Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sinh \left (d x + c\right )^{2} \tanh \left (d x + c\right )}{a f x + a e +{\left (b f x + b e\right )} \sinh \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sinh(d*x + c)^2*tanh(d*x + c)/(a*f*x + a*e + (b*f*x + b*e)*sinh(d*x + c)), x)

________________________________________________________________________________________

Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{2}{\left (c + d x \right )} \tanh{\left (c + d x \right )}}{\left (a + b \sinh{\left (c + d x \right )}\right ) \left (e + f x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(c + d*x)**2*tanh(c + d*x)/((a + b*sinh(c + d*x))*(e + f*x)), x)

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out