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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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 = 0.05, 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

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 134, normalized size = 2.85 \[ \frac {\sqrt {a^{2} + b^{2}} a \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) - b}\right ) + {\left (a^{2} + b^{2}\right )} x}{a^{2} b + b^{3}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 67, normalized size = 1.43 \[ -\frac {a \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b} + \frac {x}{b} \]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 63, normalized size = 1.34 \[ -\frac {2 a \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b \sqrt {a^{2}+b^{2}}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 65, normalized size = 1.38 \[ -\frac {a \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b} + \frac {x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.53, size = 99, normalized size = 2.11 \[ \frac {x}{b}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^2}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^x\right )}{b^2\,\sqrt {a^2+b^2}}\right )}{b\,\sqrt {a^2+b^2}}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^2}+\frac {2\,a\,\left (b-a\,{\mathrm {e}}^x\right )}{b^2\,\sqrt {a^2+b^2}}\right )}{b\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(a + b*sinh(x)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 58.30, size = 252, normalized size = 5.36 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\\frac {\cosh {\relax (x )}}{a} & \text {for}\: b = 0 \\\frac {x}{b} & \text {for}\: a = 0 \\\frac {b x \tanh {\left (\frac {x}{2} \right )}}{b^{2} \tanh {\left (\frac {x}{2} \right )} - i b \sqrt {b^{2}}} - \frac {2 b}{b^{2} \tanh {\left (\frac {x}{2} \right )} - i b \sqrt {b^{2}}} - \frac {i x \sqrt {b^{2}}}{b^{2} \tanh {\left (\frac {x}{2} \right )} - i b \sqrt {b^{2}}} & \text {for}\: a = - \sqrt {- b^{2}} \\\frac {b x \tanh {\left (\frac {x}{2} \right )}}{b^{2} \tanh {\left (\frac {x}{2} \right )} + i b \sqrt {b^{2}}} - \frac {2 b}{b^{2} \tanh {\left (\frac {x}{2} \right )} + i b \sqrt {b^{2}}} + \frac {i x \sqrt {b^{2}}}{b^{2} \tanh {\left (\frac {x}{2} \right )} + i b \sqrt {b^{2}}} & \text {for}\: a = \sqrt {- b^{2}} \\\frac {a \log {\left (\tanh {\left (\frac {x}{2} \right )} - \frac {b}{a} - \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b \sqrt {a^{2} + b^{2}}} - \frac {a \log {\left (\tanh {\left (\frac {x}{2} \right )} - \frac {b}{a} + \frac {\sqrt {a^{2} + b^{2}}}{a} \right )}}{b \sqrt {a^{2} + b^{2}}} + \frac {x}{b} & \text {otherwise} \end {cases} \]

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*x*tanh(x/2)/(b**2*tanh(x/2)
- I*b*sqrt(b**2)) - 2*b/(b**2*tanh(x/2) - I*b*sqrt(b**2)) - I*x*sqrt(b**2)/(b**2*tanh(x/2) - I*b*sqrt(b**2)),
Eq(a, -sqrt(-b**2))), (b*x*tanh(x/2)/(b**2*tanh(x/2) + I*b*sqrt(b**2)) - 2*b/(b**2*tanh(x/2) + I*b*sqrt(b**2)
) + I*x*sqrt(b**2)/(b**2*tanh(x/2) + I*b*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, T
rue))

________________________________________________________________________________________