3.41 \(\int \frac {\cosh (x)}{\sqrt {a-a \cosh (x)}} \, dx\)

Optimal. Leaf size=53 \[ \frac {2 \sinh (x)}{\sqrt {a-a \cosh (x)}}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a-a \cosh (x)}}\right )}{\sqrt {a}} \]

[Out]

-arctan(1/2*sinh(x)*a^(1/2)*2^(1/2)/(a-a*cosh(x))^(1/2))*2^(1/2)/a^(1/2)+2*sinh(x)/(a-a*cosh(x))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2751, 2649, 206} \[ \frac {2 \sinh (x)}{\sqrt {a-a \cosh (x)}}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {2} \sqrt {a-a \cosh (x)}}\right )}{\sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/Sqrt[a - a*Cosh[x]],x]

[Out]

-((Sqrt[2]*ArcTan[(Sqrt[a]*Sinh[x])/(Sqrt[2]*Sqrt[a - a*Cosh[x]])])/Sqrt[a]) + (2*Sinh[x])/Sqrt[a - a*Cosh[x]]

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 2649

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

Rule 2751

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 35, normalized size = 0.66 \[ \frac {2 \sinh \left (\frac {x}{2}\right ) \left (2 \cosh \left (\frac {x}{2}\right )+\log \left (\tanh \left (\frac {x}{4}\right )\right )\right )}{\sqrt {a-a \cosh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/Sqrt[a - a*Cosh[x]],x]

[Out]

(2*(2*Cosh[x/2] + Log[Tanh[x/4]])*Sinh[x/2])/Sqrt[a - a*Cosh[x]]

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fricas [B]  time = 0.55, size = 92, normalized size = 1.74 \[ \frac {\sqrt {2} a \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a}{\cosh \relax (x) + \sinh \relax (x)}} \sqrt {-\frac {1}{a}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - \cosh \relax (x) - \sinh \relax (x) - 1}{\cosh \relax (x) + \sinh \relax (x) - 1}\right ) - 2 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a}{\cosh \relax (x) + \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x) + 1\right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)/(a-a*cosh(x))^(1/2),x, algorithm="fricas")

[Out]

(sqrt(2)*a*sqrt(-1/a)*log((2*sqrt(2)*sqrt(1/2)*sqrt(-a/(cosh(x) + sinh(x)))*sqrt(-1/a)*(cosh(x) + sinh(x)) - c
osh(x) - sinh(x) - 1)/(cosh(x) + sinh(x) - 1)) - 2*sqrt(1/2)*sqrt(-a/(cosh(x) + sinh(x)))*(cosh(x) + sinh(x) +
1))/a

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giac [C]  time = 0.16, size = 90, normalized size = 1.70 \[ -\sqrt {2} {\left (\frac {2 \, {\left (i \, \arctan \left (-i\right ) - 1\right )} \mathrm {sgn}\left (-e^{x} + 1\right )}{\sqrt {-a}} + \frac {2 \, \arctan \left (\frac {\sqrt {-a e^{x}}}{\sqrt {a}}\right )}{\sqrt {a} \mathrm {sgn}\left (-e^{x} + 1\right )} + \frac {1}{\sqrt {-a e^{x}} \mathrm {sgn}\left (-e^{x} + 1\right )} - \frac {\sqrt {-a e^{x}}}{a \mathrm {sgn}\left (-e^{x} + 1\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)/(a-a*cosh(x))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.34, size = 40, normalized size = 0.75 \[ \frac {\sinh \left (\frac {x}{2}\right ) \left (4 \cosh \left (\frac {x}{2}\right )+\ln \left (-1+\cosh \left (\frac {x}{2}\right )\right )-\ln \left (\cosh \left (\frac {x}{2}\right )+1\right )\right )}{\sqrt {-2 a \left (\sinh ^{2}\left (\frac {x}{2}\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(a-a*cosh(x))^(1/2),x)

[Out]

sinh(1/2*x)*(4*cosh(1/2*x)+ln(-1+cosh(1/2*x))-ln(cosh(1/2*x)+1))/(-2*a*sinh(1/2*x)^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \relax (x)}{\sqrt {-a \cosh \relax (x) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)/(a-a*cosh(x))^(1/2),x, algorithm="maxima")

[Out]

integrate(cosh(x)/sqrt(-a*cosh(x) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {cosh}\relax (x)}{\sqrt {a-a\,\mathrm {cosh}\relax (x)}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(a - a*cosh(x))^(1/2),x)

[Out]

int(cosh(x)/(a - a*cosh(x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh {\relax (x )}}{\sqrt {- a \left (\cosh {\relax (x )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)/(a-a*cosh(x))**(1/2),x)

[Out]

Integral(cosh(x)/sqrt(-a*(cosh(x) - 1)), x)

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