3.691 \(\int \frac {\cosh (x)}{a \cosh (x)+b \sinh (x)} \, dx\)

Optimal. Leaf size=39 \[ \frac {a x}{a^2-b^2}-\frac {b \log (a \cosh (x)+b \sinh (x))}{a^2-b^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3098, 3133} \[ \frac {a x}{a^2-b^2}-\frac {b \log (a \cosh (x)+b \sinh (x))}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(a*Cosh[x] + b*Sinh[x]),x]

[Out]

(a*x)/(a^2 - b^2) - (b*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2)

Rule 3098

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

\begin {align*} \int \frac {\cosh (x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac {a x}{a^2-b^2}-\frac {(i b) \int \frac {-i b \cosh (x)-i a \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=\frac {a x}{a^2-b^2}-\frac {b \log (a \cosh (x)+b \sinh (x))}{a^2-b^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 29, normalized size = 0.74 \[ \frac {a x-b \log (a \cosh (x)+b \sinh (x))}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x - b*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2)

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fricas [A]  time = 0.43, size = 42, normalized size = 1.08 \[ \frac {{\left (a + b\right )} x - b \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} - b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 43, normalized size = 1.10 \[ -\frac {b \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{2} - b^{2}} + \frac {x}{a - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 71, normalized size = 1.82 \[ -\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b +2 a}-\frac {b \ln \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}{\left (a -b \right ) \left (a +b \right )}+\frac {2 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a -2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(2*b+2*a)*ln(tanh(1/2*x)-1)-b/(a-b)/(a+b)*ln(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)+2/(2*a-2*b)*ln(tanh(1/2*x)+
1)

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maxima [A]  time = 0.31, size = 41, normalized size = 1.05 \[ -\frac {b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{2} - b^{2}} + \frac {x}{a + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.52, size = 29, normalized size = 0.74 \[ \frac {a\,x-b\,\ln \left (a\,\mathrm {cosh}\relax (x)+b\,\mathrm {sinh}\relax (x)\right )}{a^2-b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.58, size = 150, normalized size = 3.85 \[ \begin {cases} \tilde {\infty } \log {\left (\sinh {\relax (x )} \right )} & \text {for}\: a = 0 \wedge b = 0 \\\frac {\log {\left (\sinh {\relax (x )} \right )}}{b} & \text {for}\: a = 0 \\\frac {x \sinh {\relax (x )}}{- 2 b \sinh {\relax (x )} + 2 b \cosh {\relax (x )}} - \frac {x \cosh {\relax (x )}}{- 2 b \sinh {\relax (x )} + 2 b \cosh {\relax (x )}} - \frac {\cosh {\relax (x )}}{- 2 b \sinh {\relax (x )} + 2 b \cosh {\relax (x )}} & \text {for}\: a = - b \\\frac {x \sinh {\relax (x )}}{2 b \sinh {\relax (x )} + 2 b \cosh {\relax (x )}} + \frac {x \cosh {\relax (x )}}{2 b \sinh {\relax (x )} + 2 b \cosh {\relax (x )}} - \frac {\cosh {\relax (x )}}{2 b \sinh {\relax (x )} + 2 b \cosh {\relax (x )}} & \text {for}\: a = b \\\frac {a x}{a^{2} - b^{2}} - \frac {b \log {\left (\cosh {\relax (x )} + \frac {b \sinh {\relax (x )}}{a} \right )}}{a^{2} - b^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*log(sinh(x)), Eq(a, 0) & Eq(b, 0)), (log(sinh(x))/b, Eq(a, 0)), (x*sinh(x)/(-2*b*sinh(x) + 2*b*
cosh(x)) - x*cosh(x)/(-2*b*sinh(x) + 2*b*cosh(x)) - cosh(x)/(-2*b*sinh(x) + 2*b*cosh(x)), Eq(a, -b)), (x*sinh(
x)/(2*b*sinh(x) + 2*b*cosh(x)) + x*cosh(x)/(2*b*sinh(x) + 2*b*cosh(x)) - cosh(x)/(2*b*sinh(x) + 2*b*cosh(x)),
Eq(a, b)), (a*x/(a**2 - b**2) - b*log(cosh(x) + b*sinh(x)/a)/(a**2 - b**2), True))

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