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

Optimal. Leaf size=47 \[ \frac{2 a \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2}}+\frac{x}{b} \]

[Out]

x/b + (2*a*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b*Sqrt[a^2 + b^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0608682, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2735, 2660, 618, 206} \[ \frac{2 a \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2}}+\frac{x}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(a + b*Sinh[x]),x]

[Out]

x/b + (2*a*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b*Sqrt[a^2 + b^2])

Rule 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sinh (x)}{a+b \sinh (x)} \, dx &=\frac{x}{b}-\frac{a \int \frac{1}{a+b \sinh (x)} \, dx}{b}\\ &=\frac{x}{b}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{b}\\ &=\frac{x}{b}+\frac{(4 a) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac{x}{2}\right )\right )}{b}\\ &=\frac{x}{b}+\frac{2 a \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b \sqrt{a^2+b^2}}\\ \end{align*}

Mathematica [A]  time = 0.050002, size = 52, normalized size = 1.11 \[ \frac{x-\frac{2 a \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(a + b*Sinh[x]),x]

[Out]

(x - (2*a*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2])/b

________________________________________________________________________________________

Maple [A]  time = 0.015, size = 63, normalized size = 1.3 \begin{align*}{\frac{1}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-2\,{\frac{a}{b\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.02995, size = 367, normalized size = 7.81 \begin{align*} \frac{\sqrt{a^{2} + b^{2}} a \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) +{\left (a^{2} + b^{2}\right )} x}{a^{2} b + b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(a^2 + b^2)*a*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*si
nh(x) + 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x)
+ a)*sinh(x) - b)) + (a^2 + b^2)*x)/(a^2*b + b^3)

________________________________________________________________________________________

Sympy [A]  time = 96.1744, size = 296, normalized size = 6.3 \begin{align*} \begin{cases} \tilde{\infty } x & \text{for}\: a = 0 \wedge b = 0 \\\frac{\cosh{\left (x \right )}}{a} & \text{for}\: b = 0 \\\frac{x}{b} & \text{for}\: a = 0 \\- \frac{b^{2} x \tanh{\left (\frac{x}{2} \right )}}{- b^{3} \tanh{\left (\frac{x}{2} \right )} + i b^{2} \sqrt{b^{2}}} + \frac{i b x \sqrt{b^{2}}}{- b^{3} \tanh{\left (\frac{x}{2} \right )} + i b^{2} \sqrt{b^{2}}} - \frac{2 i b \sqrt{b^{2}} \tanh{\left (\frac{x}{2} \right )}}{- b^{3} \tanh{\left (\frac{x}{2} \right )} + i b^{2} \sqrt{b^{2}}} & \text{for}\: a = - \sqrt{- b^{2}} \\\frac{b^{2} x \tanh{\left (\frac{x}{2} \right )}}{b^{3} \tanh{\left (\frac{x}{2} \right )} + i b^{2} \sqrt{b^{2}}} + \frac{i b x \sqrt{b^{2}}}{b^{3} \tanh{\left (\frac{x}{2} \right )} + i b^{2} \sqrt{b^{2}}} - \frac{2 i b \sqrt{b^{2}} \tanh{\left (\frac{x}{2} \right )}}{b^{3} \tanh{\left (\frac{x}{2} \right )} + i b^{2} \sqrt{b^{2}}} & \text{for}\: a = \sqrt{- b^{2}} \\\frac{a \log{\left (\tanh{\left (\frac{x}{2} \right )} - \frac{b}{a} - \frac{\sqrt{a^{2} + b^{2}}}{a} \right )}}{b \sqrt{a^{2} + b^{2}}} - \frac{a \log{\left (\tanh{\left (\frac{x}{2} \right )} - \frac{b}{a} + \frac{\sqrt{a^{2} + b^{2}}}{a} \right )}}{b \sqrt{a^{2} + b^{2}}} + \frac{x}{b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (cosh(x)/a, Eq(b, 0)), (x/b, Eq(a, 0)), (-b**2*x*tanh(x/2)/(-b**3*tanh
(x/2) + I*b**2*sqrt(b**2)) + I*b*x*sqrt(b**2)/(-b**3*tanh(x/2) + I*b**2*sqrt(b**2)) - 2*I*b*sqrt(b**2)*tanh(x/
2)/(-b**3*tanh(x/2) + I*b**2*sqrt(b**2)), Eq(a, -sqrt(-b**2))), (b**2*x*tanh(x/2)/(b**3*tanh(x/2) + I*b**2*sqr
t(b**2)) + I*b*x*sqrt(b**2)/(b**3*tanh(x/2) + I*b**2*sqrt(b**2)) - 2*I*b*sqrt(b**2)*tanh(x/2)/(b**3*tanh(x/2)
+ I*b**2*sqrt(b**2)), Eq(a, sqrt(-b**2))), (a*log(tanh(x/2) - b/a - sqrt(a**2 + b**2)/a)/(b*sqrt(a**2 + b**2))
- a*log(tanh(x/2) - b/a + sqrt(a**2 + b**2)/a)/(b*sqrt(a**2 + b**2)) + x/b, True))

________________________________________________________________________________________

Giac [A]  time = 1.36999, size = 90, normalized size = 1.91 \begin{align*} -\frac{a \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} b} + \frac{x}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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