3.836 \(\int \frac{a+b \cosh (x)}{b^2+2 a b \cosh (x)+a^2 \cosh ^2(x)} \, dx\)

Optimal. Leaf size=11 \[ \frac{\sinh (x)}{a \cosh (x)+b} \]

[Out]

Sinh[x]/(b + a*Cosh[x])

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Rubi [A]  time = 0.086682, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3289, 2754, 8} \[ \frac{\sinh (x)}{a \cosh (x)+b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[x])/(b^2 + 2*a*b*Cosh[x] + a^2*Cosh[x]^2),x]

[Out]

Sinh[x]/(b + a*Cosh[x])

Rule 3289

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + cos[(d_.) + (e_.)*(x_)]^2*(c_.) + (a_))^(n_)*(cos[(d_.) + (e_.)*(x_)]*(B_
.) + (A_)), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Cos[d + e*x])*(b + 2*c*Cos[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{a+b \cosh (x)}{b^2+2 a b \cosh (x)+a^2 \cosh ^2(x)} \, dx &=\left (4 a^2\right ) \int \frac{a+b \cosh (x)}{\left (2 a b+2 a^2 \cosh (x)\right )^2} \, dx\\ &=\frac{\sinh (x)}{b+a \cosh (x)}+\frac{\int 0 \, dx}{a^2-b^2}\\ &=\frac{\sinh (x)}{b+a \cosh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0577431, size = 11, normalized size = 1. \[ \frac{\sinh (x)}{a \cosh (x)+b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sinh[x]/(b + a*Cosh[x])

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Maple [B]  time = 0.026, size = 29, normalized size = 2.6 \begin{align*} 2\,{\frac{\tanh \left ( x/2 \right ) }{a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}- \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+a+b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cosh(x))/(b^2+2*a*b*cosh(x)+a^2*cosh(x)^2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.53732, size = 159, normalized size = 14.45 \begin{align*} -\frac{2 \,{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + a^{2} + 2 \,{\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.13539, size = 35, normalized size = 3.18 \begin{align*} -\frac{2 \,{\left (b e^{x} + a\right )}}{{\left (a e^{\left (2 \, x\right )} + 2 \, b e^{x} + a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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