3.4.0 ∫ sech(c+dx) tanh2(c+dx) (e+fx)(a+b sinh(c+dx)) dx [390]

Integrand size = 34, antiderivative size = 34 \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\text {Int}\left (\frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))},x\right ) \]

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

Rubi [N/A]

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx \]

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

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

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

Mathematica [N/A]

Time = 68.83 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx \]

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

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

Maple [N/A] (veriﬁed)

Time = 0.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00

\[\int \frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )^{2}}{\left (f x +e \right ) \left (a +b \sinh \left (d x +c \right )\right )}d x\]

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

Fricas [N/A]

Time = 4.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right ) \tanh \left (d x + c\right )^{2}}{{\left (f x + e\right )} {\left (b \sinh \left (d x + c\right ) + a\right )}} \,d x } \]

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

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

Sympy [N/A]

Time = 2.91 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\int \frac {\tanh ^{2}{\left (c + d x \right )} \operatorname {sech}{\left (c + d x \right )}}{\left (a + b \sinh {\left (c + d x \right )}\right ) \left (e + f x\right )}\, dx \]

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

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

Maxima [N/A]

Time = 1.94 (sec) , antiderivative size = 1102, normalized size of antiderivative = 32.41 \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right ) \tanh \left (d x + c\right )^{2}}{{\left (f x + e\right )} {\left (b \sinh \left (d x + c\right ) + a\right )}} \,d x } \]

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

(b*f - (a*d*f*x*e^(3*c) + (d*e - f)*a*e^(3*c))*e^(3*d*x) - (2*b*d*f*x*e^(2*c) + (2*d*e - f)*b*e^(2*c))*e^(2*d*

x) + (a*d*f*x*e^c + (d*e + f)*a*e^c)*e^(d*x))/(a^2*d^2*e^2 + b^2*d^2*e^2 + (a^2*d^2*f^2 + b^2*d^2*f^2)*x^2 + 2

*(a^2*d^2*e*f + b^2*d^2*e*f)*x + (a^2*d^2*e^2*e^(4*c) + b^2*d^2*e^2*e^(4*c) + (a^2*d^2*f^2*e^(4*c) + b^2*d^2*f

^2*e^(4*c))*x^2 + 2*(a^2*d^2*e*f*e^(4*c) + b^2*d^2*e*f*e^(4*c))*x)*e^(4*d*x) + 2*(a^2*d^2*e^2*e^(2*c) + b^2*d^

2*e^2*e^(2*c) + (a^2*d^2*f^2*e^(2*c) + b^2*d^2*f^2*e^(2*c))*x^2 + 2*(a^2*d^2*e*f*e^(2*c) + b^2*d^2*e*f*e^(2*c)

)*x)*e^(2*d*x)) + 2*integrate(1/2*(2*a^2*b*d^2*f^2*x^2 + 4*a^2*b*d^2*e*f*x + 2*b^3*f^2 + 2*(d^2*e^2 + f^2)*a^2

*b + ((d^2*e^2 + 2*f^2)*a^3*e^c - (d^2*e^2 - 2*f^2)*a*b^2*e^c + (a^3*d^2*f^2*e^c - a*b^2*d^2*f^2*e^c)*x^2 + 2*

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

3 + 2*a^2*b^2*d^2*f^3 + b^4*d^2*f^3)*x^3 + 3*(a^4*d^2*e*f^2 + 2*a^2*b^2*d^2*e*f^2 + b^4*d^2*e*f^2)*x^2 + 3*(a^

4*d^2*e^2*f + 2*a^2*b^2*d^2*e^2*f + b^4*d^2*e^2*f)*x + (a^4*d^2*e^3*e^(2*c) + 2*a^2*b^2*d^2*e^3*e^(2*c) + b^4*

d^2*e^3*e^(2*c) + (a^4*d^2*f^3*e^(2*c) + 2*a^2*b^2*d^2*f^3*e^(2*c) + b^4*d^2*f^3*e^(2*c))*x^3 + 3*(a^4*d^2*e*f

^2*e^(2*c) + 2*a^2*b^2*d^2*e*f^2*e^(2*c) + b^4*d^2*e*f^2*e^(2*c))*x^2 + 3*(a^4*d^2*e^2*f*e^(2*c) + 2*a^2*b^2*d

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

b*e + 2*a^2*b^3*e + b^5*e + (a^4*b*f + 2*a^2*b^3*f + b^5*f)*x - (a^4*b*e*e^(2*c) + 2*a^2*b^3*e*e^(2*c) + b^5*e

*e^(2*c) + (a^4*b*f*e^(2*c) + 2*a^2*b^3*f*e^(2*c) + b^5*f*e^(2*c))*x)*e^(2*d*x) - 2*(a^5*e*e^c + 2*a^3*b^2*e*e

^c + a*b^4*e*e^c + (a^5*f*e^c + 2*a^3*b^2*f*e^c + a*b^4*f*e^c)*x)*e^(d*x)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\text {Timed out} \]

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

Mupad [N/A]

Time = 3.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx=\int \frac {{\mathrm {tanh}\left (c+d\,x\right )}^2}{\mathrm {cosh}\left (c+d\,x\right )\,\left (e+f\,x\right )\,\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )} \,d x \]

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

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

\(\int \frac {\text {sech}(c+d x) \tanh ^2(c+d x)}{(e+f x) (a+b \sinh (c+d x))} \, dx\) [390]

Optimal result