\(\int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx\) [55]

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=\frac {\cosh (x)}{a}-\frac {\log (1+\cosh (x))}{a} \]

[Out]

cosh(x)/a-ln(1+cosh(x))/a

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3957, 2912, 12, 45} \[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=\frac {\cosh (x)}{a}-\frac {\log (\cosh (x)+1)}{a} \]

[In]

Int[Sinh[x]/(a + a*Sech[x]),x]

[Out]

Cosh[x]/a - Log[1 + Cosh[x]]/a

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cosh (x) \sinh (x)}{-a-a \cosh (x)} \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {x}{a (-a+x)} \, dx,x,-a \cosh (x)\right )}{a} \\ & = -\frac {\text {Subst}\left (\int \frac {x}{-a+x} \, dx,x,-a \cosh (x)\right )}{a^2} \\ & = -\frac {\text {Subst}\left (\int \left (1-\frac {a}{a-x}\right ) \, dx,x,-a \cosh (x)\right )}{a^2} \\ & = \frac {\cosh (x)}{a}-\frac {\log (1+\cosh (x))}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=\frac {\cosh (x)-2 \log \left (\cosh \left (\frac {x}{2}\right )\right )}{a} \]

[In]

Integrate[Sinh[x]/(a + a*Sech[x]),x]

[Out]

(Cosh[x] - 2*Log[Cosh[x/2]])/a

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(-\frac {\ln \left (1+\operatorname {sech}\left (x \right )\right )-\frac {1}{\operatorname {sech}\left (x \right )}-\ln \left (\operatorname {sech}\left (x \right )\right )}{a}\)	\(23\)
default	\(-\frac {\ln \left (1+\operatorname {sech}\left (x \right )\right )-\frac {1}{\operatorname {sech}\left (x \right )}-\ln \left (\operatorname {sech}\left (x \right )\right )}{a}\)	\(23\)
risch	\(\frac {x}{a}+\frac {{\mathrm e}^{x}}{2 a}+\frac {{\mathrm e}^{-x}}{2 a}-\frac {2 \ln \left ({\mathrm e}^{x}+1\right )}{a}\)	\(33\)

[In]

int(sinh(x)/(a+a*sech(x)),x,method=_RETURNVERBOSE)

[Out]

-1/a*(ln(1+sech(x))-1/sech(x)-ln(sech(x)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (17) = 34\).

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.94 \[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=\frac {2 \, x \cosh \left (x\right ) + \cosh \left (x\right )^{2} - 4 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + 2 \, {\left (x + \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}{2 \, {\left (a \cosh \left (x\right ) + a \sinh \left (x\right )\right )}} \]

[In]

integrate(sinh(x)/(a+a*sech(x)),x, algorithm="fricas")

[Out]

1/2*(2*x*cosh(x) + cosh(x)^2 - 4*(cosh(x) + sinh(x))*log(cosh(x) + sinh(x) + 1) + 2*(x + cosh(x))*sinh(x) + si
nh(x)^2 + 1)/(a*cosh(x) + a*sinh(x))

Sympy [F]

\[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=\frac {\int \frac {\sinh {\left (x \right )}}{\operatorname {sech}{\left (x \right )} + 1}\, dx}{a} \]

[In]

integrate(sinh(x)/(a+a*sech(x)),x)

[Out]

Integral(sinh(x)/(sech(x) + 1), x)/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.06 \[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=-\frac {x}{a} + \frac {e^{\left (-x\right )}}{2 \, a} + \frac {e^{x}}{2 \, a} - \frac {2 \, \log \left (e^{\left (-x\right )} + 1\right )}{a} \]

[In]

integrate(sinh(x)/(a+a*sech(x)),x, algorithm="maxima")

[Out]

-x/a + 1/2*e^(-x)/a + 1/2*e^x/a - 2*log(e^(-x) + 1)/a

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.88 \[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a} + \frac {e^{\left (-x\right )}}{2 \, a} + \frac {e^{x}}{2 \, a} - \frac {2 \, \log \left (e^{x} + 1\right )}{a} \]

[In]

integrate(sinh(x)/(a+a*sech(x)),x, algorithm="giac")

[Out]

x/a + 1/2*e^(-x)/a + 1/2*e^x/a - 2*log(e^x + 1)/a

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {\sinh (x)}{a+a \text {sech}(x)} \, dx=-\frac {\ln \left (\mathrm {cosh}\left (x\right )+1\right )-\mathrm {cosh}\left (x\right )}{a} \]

[In]

int(sinh(x)/(a + a/cosh(x)),x)

[Out]

-(log(cosh(x) + 1) - cosh(x))/a