3.2.59 \(\int \frac {\sinh (x)}{a+a \cosh (x)} \, dx\) [159]

Optimal. Leaf size=9 \[ \frac {\log (1+\cosh (x))}{a} \]

[Out]

ln(1+cosh(x))/a

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Rubi [A]
time = 0.02, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2746, 31} \begin {gather*} \frac {\log (\cosh (x)+1)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(a + a*Cosh[x]),x]

[Out]

Log[1 + Cosh[x]]/a

Rule 31

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

Rule 2746

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.33 \begin {gather*} \frac {2 \log \left (\cosh \left (\frac {x}{2}\right )\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(a + a*Cosh[x]),x]

[Out]

(2*Log[Cosh[x/2]])/a

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Maple [A]
time = 0.32, size = 12, normalized size = 1.33

method	result	size
derivativedivides	\(\frac {\ln \left (a +a \cosh \left (x \right )\right )}{a}\)	\(12\)
default	\(\frac {\ln \left (a +a \cosh \left (x \right )\right )}{a}\)	\(12\)
risch	\(-\frac {x}{a}+\frac {2 \ln \left ({\mathrm e}^{x}+1\right )}{a}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(a+a*cosh(x))/a

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Maxima [A]
time = 0.29, size = 11, normalized size = 1.22 \begin {gather*} \frac {\log \left (a \cosh \left (x\right ) + a\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*cosh(x) + a)/a

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Fricas [A]
time = 0.43, size = 16, normalized size = 1.78 \begin {gather*} -\frac {x - 2 \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - 2*log(cosh(x) + sinh(x) + 1))/a

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Sympy [A]
time = 0.05, size = 7, normalized size = 0.78 \begin {gather*} \frac {\log {\left (\cosh {\left (x \right )} + 1 \right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(cosh(x) + 1)/a

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Giac [A]
time = 0.42, size = 17, normalized size = 1.89 \begin {gather*} -\frac {x}{a} + \frac {2 \, \log \left (e^{x} + 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/a + 2*log(e^x + 1)/a

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Mupad [B]
time = 0.89, size = 9, normalized size = 1.00 \begin {gather*} \frac {\ln \left (\mathrm {cosh}\left (x\right )+1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(a + a*cosh(x)),x)

[Out]

log(cosh(x) + 1)/a

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