3.830 \(\int \frac{a+b \sinh (x)}{b^2-2 a b \sinh (x)+a^2 \sinh ^2(x)} \, dx\)

Optimal. Leaf size=12 \[ \frac{\cosh (x)}{b-a \sinh (x)} \]

[Out]

Cosh[x]/(b - a*Sinh[x])

________________________________________________________________________________________

Rubi [A]  time = 0.0906189, antiderivative size = 12, 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 = {3288, 2754, 8} \[ \frac{\cosh (x)}{b-a \sinh (x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cosh[x]/(b - a*Sinh[x])

Rule 3288

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Sin[d + e*x])*(b + 2*c*Sin[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 \sinh (x)}{b^2-2 a b \sinh (x)+a^2 \sinh ^2(x)} \, dx &=-\left (\left (4 a^2\right ) \int \frac{a+b \sinh (x)}{\left (2 i a b-2 i a^2 \sinh (x)\right )^2} \, dx\right )\\ &=\frac{\cosh (x)}{b-a \sinh (x)}-\frac{\int 0 \, dx}{a^2+b^2}\\ &=\frac{\cosh (x)}{b-a \sinh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0337063, size = 14, normalized size = 1.17 \[ -\frac{\cosh (x)}{a \sinh (x)-b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cosh[x]/(-b + a*Sinh[x]))

________________________________________________________________________________________

Maple [B]  time = 0.046, size = 36, normalized size = 3. \begin{align*} -2\,{\frac{1}{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+2\,a\tanh \left ( x/2 \right ) -b} \left ( -{\frac{a\tanh \left ( x/2 \right ) }{b}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*sinh(x))/(b^2-2*a*b*sinh(x)+a^2*sinh(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.74691, size = 159, normalized size = 13.25 \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*sinh(x))/(b^2-2*a*b*sinh(x)+a^2*sinh(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))

________________________________________________________________________________________

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*sinh(x))/(b**2-2*a*b*sinh(x)+a**2*sinh(x)**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.19291, size = 38, normalized size = 3.17 \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*sinh(x))/(b^2-2*a*b*sinh(x)+a^2*sinh(x)^2),x, algorithm="giac")

[Out]

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