\(\int \frac {\cosh (x)}{a+b \sinh (x)} \, dx\) [193]

Optimal result

Integrand size = 11, antiderivative size = 11 \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\log (a+b \sinh (x))}{b} \]

[Out]

ln(a+b*sinh(x))/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2747, 31} \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\log (a+b \sinh (x))}{b} \]

[In]

Int[Cosh[x]/(a + b*Sinh[x]),x]

[Out]

Log[a + b*Sinh[x]]/b

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\log (a+b \sinh (x))}{b} \]

[In]

Integrate[Cosh[x]/(a + b*Sinh[x]),x]

[Out]

Log[a + b*Sinh[x]]/b

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\ln \left (a +b \sinh \left (x \right )\right )}{b}\)	\(12\)
default	\(\frac {\ln \left (a +b \sinh \left (x \right )\right )}{b}\)	\(12\)
risch	\(-\frac {x}{b}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{b}\)	\(27\)

[In]

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

[Out]

ln(a+b*sinh(x))/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (11) = 22\).

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.45 \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=-\frac {x - \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{b} \]

[In]

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

[Out]

-(x - log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))))/b

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=\begin {cases} \frac {\log {\left (\frac {a}{b} + \sinh {\left (x \right )} \right )}}{b} & \text {for}\: b \neq 0 \\\frac {\sinh {\left (x \right )}}{a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(a/b + sinh(x))/b, Ne(b, 0)), (sinh(x)/a, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\log \left (b \sinh \left (x\right ) + a\right )}{b} \]

[In]

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

[Out]

log(b*sinh(x) + a)/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 2.00 \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{b} \]

[In]

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

[Out]

log(abs(-b*(e^(-x) - e^x) + 2*a))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (x)}{a+b \sinh (x)} \, dx=\frac {\ln \left (a+b\,\mathrm {sinh}\left (x\right )\right )}{b} \]

[In]

int(cosh(x)/(a + b*sinh(x)),x)

[Out]

log(a + b*sinh(x))/b