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=37 \[ \text {Int}\left (\frac {\sinh ^2(c+d x) \tanh (c+d x)}{(e+f x) (a+b \sinh (c+d x))},x\right ) \]

[Out]

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

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Rubi [A]  time = 0.09, 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 = {} \[ \int \frac {\sinh ^2(c+d x) \tanh (c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx \]

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*}

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

[Out]

$Aborted

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fricas [A]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sinh \left (d x + c\right )^{2} \tanh \left (d x + c\right )}{a f x + a e + {\left (b f x + b e\right )} \sinh \left (d x + c\right )}, x\right ) \]

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)

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

[Out]

Timed out

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maple [A]  time = 1.23, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sinh ^{2}\left (d x +c \right )\right ) \tanh \left (d x +c \right )}{\left (f x +e \right ) \left (a +b \sinh \left (d x +c \right )\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^2*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)

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {e^{\left (-c + \frac {d e}{f}\right )} E_{1}\left (\frac {{\left (f x + e\right )} d}{f}\right )}{2 \, b f} - \frac {e^{\left (c - \frac {d e}{f}\right )} E_{1}\left (-\frac {{\left (f x + e\right )} d}{f}\right )}{2 \, b f} - \frac {a \log \left (f x + e\right )}{b^{2} f} + \frac {1}{4} \, \int -\frac {8 \, {\left (a^{4} e^{\left (d x + c\right )} - a^{3} b\right )}}{a^{2} b^{3} e + b^{5} e + {\left (a^{2} b^{3} f + b^{5} f\right )} x - {\left (a^{2} b^{3} e e^{\left (2 \, c\right )} + b^{5} e e^{\left (2 \, c\right )} + {\left (a^{2} b^{3} f e^{\left (2 \, c\right )} + b^{5} f e^{\left (2 \, c\right )}\right )} x\right )} e^{\left (2 \, d x\right )} - 2 \, {\left (a^{3} b^{2} e e^{c} + a b^{4} e e^{c} + {\left (a^{3} b^{2} f e^{c} + a b^{4} f e^{c}\right )} x\right )} e^{\left (d x\right )}}\,{d x} - \frac {1}{4} \, \int \frac {8 \, {\left (b e^{\left (d x + c\right )} - a\right )}}{a^{2} e + b^{2} e + {\left (a^{2} f + b^{2} f\right )} x + {\left (a^{2} e e^{\left (2 \, c\right )} + b^{2} e e^{\left (2 \, c\right )} + {\left (a^{2} f e^{\left (2 \, c\right )} + b^{2} f e^{\left (2 \, c\right )}\right )} x\right )} e^{\left (2 \, d x\right )}}\,{d x} \]

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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