\(\int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx\) [158]

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=-\frac {x}{a}+\frac {\sinh (x)}{a} \]

[Out]

-x/a+sinh(x)/a

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2761, 8} \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=\frac {\sinh (x)}{a}-\frac {x}{a} \]

[In]

Int[Sinh[x]^2/(a + a*Cosh[x]),x]

[Out]

-(x/a) + Sinh[x]/a

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2761

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*((g*Cos[e
+ f*x])^(p - 1)/(b*f*(p - 1))), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {\sinh (x)}{a}-\frac {\int 1 \, dx}{a} \\ & = -\frac {x}{a}+\frac {\sinh (x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=\frac {2 \left (-\frac {x}{2}+\frac {\sinh (x)}{2}\right )}{a} \]

[In]

Integrate[Sinh[x]^2/(a + a*Cosh[x]),x]

[Out]

(2*(-1/2*x + Sinh[x]/2))/a

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.85

method	result	size
risch	\(-\frac {x}{a}+\frac {{\mathrm e}^{x}}{2 a}-\frac {{\mathrm e}^{-x}}{2 a}\)	\(24\)
default	\(\frac {-\frac {1}{\tanh \left (\frac {x}{2}\right )-1}+\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}\)	\(45\)

[In]

int(sinh(x)^2/(a+a*cosh(x)),x,method=_RETURNVERBOSE)

[Out]

-x/a+1/2/a*exp(x)-1/2/a*exp(-x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=-\frac {x - \sinh \left (x\right )}{a} \]

[In]

integrate(sinh(x)^2/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

-(x - sinh(x))/a

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (7) = 14\).

Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 3.54 \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=- \frac {x \tanh ^{2}{\left (\frac {x}{2} \right )}}{a \tanh ^{2}{\left (\frac {x}{2} \right )} - a} + \frac {x}{a \tanh ^{2}{\left (\frac {x}{2} \right )} - a} - \frac {2 \tanh {\left (\frac {x}{2} \right )}}{a \tanh ^{2}{\left (\frac {x}{2} \right )} - a} \]

[In]

integrate(sinh(x)**2/(a+a*cosh(x)),x)

[Out]

-x*tanh(x/2)**2/(a*tanh(x/2)**2 - a) + x/(a*tanh(x/2)**2 - a) - 2*tanh(x/2)/(a*tanh(x/2)**2 - a)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.77 \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=-\frac {x}{a} - \frac {e^{\left (-x\right )}}{2 \, a} + \frac {e^{x}}{2 \, a} \]

[In]

integrate(sinh(x)^2/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-x/a - 1/2*e^(-x)/a + 1/2*e^x/a

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=-\frac {2 \, x + e^{\left (-x\right )} - e^{x}}{2 \, a} \]

[In]

integrate(sinh(x)^2/(a+a*cosh(x)),x, algorithm="giac")

[Out]

-1/2*(2*x + e^(-x) - e^x)/a

Mupad [B] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.77 \[ \int \frac {\sinh ^2(x)}{a+a \cosh (x)} \, dx=\frac {{\mathrm {e}}^x}{2\,a}-\frac {x}{a}-\frac {{\mathrm {e}}^{-x}}{2\,a} \]

[In]

int(sinh(x)^2/(a + a*cosh(x)),x)

[Out]

exp(x)/(2*a) - x/a - exp(-x)/(2*a)