3.117 \(\int \frac {\cosh (x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=72 \[ \frac {a \sinh (x)}{a^2-b^2}-\frac {b \cosh (x)}{a^2-b^2}+\frac {a b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}} \]

[Out]

a*b*arctanh((a*cosh(x)+b*sinh(x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)-b*cosh(x)/(a^2-b^2)+a*sinh(x)/(a^2-b^2)

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Rubi [A]  time = 0.11, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {3518, 3109, 2637, 2638, 3074, 206} \[ \frac {a \sinh (x)}{a^2-b^2}-\frac {b \cosh (x)}{a^2-b^2}+\frac {a b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(a + b*Coth[x]),x]

[Out]

(a*b*ArcTanh[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - (b*Cosh[x])/(a^2 - b^2) + (a*Sinh[x
])/(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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3074

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

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {\cosh (x)}{a+b \coth (x)} \, dx &=-\left (i \int \frac {\cosh (x) \sinh (x)}{-i b \cosh (x)-i a \sinh (x)} \, dx\right )\\ &=\frac {a \int \cosh (x) \, dx}{a^2-b^2}-\frac {b \int \sinh (x) \, dx}{a^2-b^2}+\frac {(i a b) \int \frac {1}{-i b \cosh (x)-i a \sinh (x)} \, dx}{a^2-b^2}\\ &=-\frac {b \cosh (x)}{a^2-b^2}+\frac {a \sinh (x)}{a^2-b^2}-\frac {(a b) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,-a \cosh (x)-b \sinh (x)\right )}{a^2-b^2}\\ &=\frac {a b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {b \cosh (x)}{a^2-b^2}+\frac {a \sinh (x)}{a^2-b^2}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 79, normalized size = 1.10 \[ \frac {a \sinh (x)}{a^2-b^2}+\frac {b \cosh (x)}{b^2-a^2}+\frac {2 a b \tan ^{-1}\left (\frac {a+b \tanh \left (\frac {x}{2}\right )}{\sqrt {b-a} \sqrt {a+b}}\right )}{(b-a)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/(a + b*Coth[x]),x]

[Out]

(2*a*b*ArcTan[(a + b*Tanh[x/2])/(Sqrt[-a + b]*Sqrt[a + b])])/((-a + b)^(3/2)*(a + b)^(3/2)) + (b*Cosh[x])/(-a^
2 + b^2) + (a*Sinh[x])/(a^2 - b^2)

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fricas [B]  time = 0.42, size = 431, normalized size = 5.99 \[ \left [-\frac {a^{3} + a^{2} b - a b^{2} - b^{3} - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \relax (x)^{2} + 2 \, {\left (a b \cosh \relax (x) + a b \sinh \relax (x)\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} + a - b}{{\left (a + b\right )} \cosh \relax (x)^{2} + 2 \, {\left (a + b\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a + b\right )} \sinh \relax (x)^{2} - a + b}\right )}{2 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)\right )}}, -\frac {a^{3} + a^{2} b - a b^{2} - b^{3} - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x) \sinh \relax (x) - {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \relax (x)^{2} + 4 \, {\left (a b \cosh \relax (x) + a b \sinh \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (\frac {\sqrt {-a^{2} + b^{2}}}{{\left (a + b\right )} \cosh \relax (x) + {\left (a + b\right )} \sinh \relax (x)}\right )}{2 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a^3 + a^2*b - a*b^2 - b^3 - (a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^2 - 2*(a^3 - a^2*b - a*b^2 + b^3)*cosh(
x)*sinh(x) - (a^3 - a^2*b - a*b^2 + b^3)*sinh(x)^2 + 2*(a*b*cosh(x) + a*b*sinh(x))*sqrt(a^2 - b^2)*log(((a + b
)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - 2*sqrt(a^2 - b^2)*(cosh(x) + sinh(x)) + a - b)/(
(a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - a + b)))/((a^4 - 2*a^2*b^2 + b^4)*cosh(x)
+ (a^4 - 2*a^2*b^2 + b^4)*sinh(x)), -1/2*(a^3 + a^2*b - a*b^2 - b^3 - (a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^2 -
2*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)*sinh(x) - (a^3 - a^2*b - a*b^2 + b^3)*sinh(x)^2 + 4*(a*b*cosh(x) + a*b*s
inh(x))*sqrt(-a^2 + b^2)*arctan(sqrt(-a^2 + b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))))/((a^4 - 2*a^2*b^2 + b^4
)*cosh(x) + (a^4 - 2*a^2*b^2 + b^4)*sinh(x))]

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giac [A]  time = 0.12, size = 71, normalized size = 0.99 \[ -\frac {2 \, a b \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {e^{\left (-x\right )}}{2 \, {\left (a - b\right )}} + \frac {e^{x}}{2 \, {\left (a + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b*arctan((a*e^x + b*e^x)/sqrt(-a^2 + b^2))/((a^2 - b^2)*sqrt(-a^2 + b^2)) - 1/2*e^(-x)/(a - b) + 1/2*e^x/
(a + b)

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maple [A]  time = 0.12, size = 92, normalized size = 1.28 \[ -\frac {2 a b \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {-a^{2}+b^{2}}}-\frac {4}{\left (4 a -4 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {4}{\left (4 a +4 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b/(a+b)/(a-b)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-4/(4*a-4*b)/(tanh(1/2*x
)+1)-4/(4*a+4*b)/(tanh(1/2*x)-1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.53, size = 158, normalized size = 2.19 \[ \frac {{\mathrm {e}}^x}{2\,a+2\,b}-\frac {{\mathrm {e}}^{-x}}{2\,a-2\,b}+\frac {2\,\mathrm {atan}\left (\frac {a\,b\,{\mathrm {e}}^x\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{a^3\,\sqrt {a^2\,b^2}+b^3\,\sqrt {a^2\,b^2}-a\,b^2\,\sqrt {a^2\,b^2}-a^2\,b\,\sqrt {a^2\,b^2}}\right )\,\sqrt {a^2\,b^2}}{\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(a + b*coth(x)),x)

[Out]

exp(x)/(2*a + 2*b) - exp(-x)/(2*a - 2*b) + (2*atan((a*b*exp(x)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(a^3
*(a^2*b^2)^(1/2) + b^3*(a^2*b^2)^(1/2) - a*b^2*(a^2*b^2)^(1/2) - a^2*b*(a^2*b^2)^(1/2)))*(a^2*b^2)^(1/2))/(b^6
- a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh {\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(x)/(a + b*coth(x)), x)

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