3.104 \(\int \frac{A+B \cosh (x)}{\sqrt{a-a \cosh (x)}} \, dx\)

Optimal. Leaf size=57 \[ \frac{2 B \sinh (x)}{\sqrt{a-a \cosh (x)}}-\frac{\sqrt{2} (A+B) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{2} \sqrt{a-a \cosh (x)}}\right )}{\sqrt{a}} \]

[Out]

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

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Rubi [A]  time = 0.0653115, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2751, 2649, 206} \[ \frac{2 B \sinh (x)}{\sqrt{a-a \cosh (x)}}-\frac{\sqrt{2} (A+B) \tan ^{-1}\left (\frac{\sqrt{a} \sinh (x)}{\sqrt{2} \sqrt{a-a \cosh (x)}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x])/Sqrt[a - a*Cosh[x]],x]

[Out]

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

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)]

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

Mathematica [A]  time = 0.0541786, size = 40, normalized size = 0.7 \[ \frac{2 \sinh \left (\frac{x}{2}\right ) \left ((A+B) \log \left (\tanh \left (\frac{x}{4}\right )\right )+2 B \cosh \left (\frac{x}{2}\right )\right )}{\sqrt{a-a \cosh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[x])/Sqrt[a - a*Cosh[x]],x]

[Out]

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

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Maple [A]  time = 0.066, size = 63, normalized size = 1.1 \begin{align*}{\sinh \left ({\frac{x}{2}} \right ) \left ( \ln \left ( -1+\cosh \left ({\frac{x}{2}} \right ) \right ) A-\ln \left ( 1+\cosh \left ({\frac{x}{2}} \right ) \right ) A+B\ln \left ( -1+\cosh \left ({\frac{x}{2}} \right ) \right ) -B\ln \left ( 1+\cosh \left ({\frac{x}{2}} \right ) \right ) +4\,B\cosh \left ( x/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sinh \left ( x/2 \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cosh \left (x\right ) + A}{\sqrt{-a \cosh \left (x\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.11703, size = 324, normalized size = 5.68 \begin{align*} \frac{\sqrt{2}{\left (A + B\right )} a \sqrt{-\frac{1}{a}} \log \left (\frac{2 \, \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{-\frac{a}{\cosh \left (x\right ) + \sinh \left (x\right )}} \sqrt{-\frac{1}{a}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - \cosh \left (x\right ) - \sinh \left (x\right ) - 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - 2 \, \sqrt{\frac{1}{2}}{\left (B \cosh \left (x\right ) + B \sinh \left (x\right ) + B\right )} \sqrt{-\frac{a}{\cosh \left (x\right ) + \sinh \left (x\right )}}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(2)*(A + B)*a*sqrt(-1/a)*log((2*sqrt(2)*sqrt(1/2)*sqrt(-a/(cosh(x) + sinh(x)))*sqrt(-1/a)*(cosh(x) + sinh
(x)) - cosh(x) - sinh(x) - 1)/(cosh(x) + sinh(x) - 1)) - 2*sqrt(1/2)*(B*cosh(x) + B*sinh(x) + B)*sqrt(-a/(cosh
(x) + sinh(x))))/a

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \cosh{\left (x \right )}}{\sqrt{- a \left (\cosh{\left (x \right )} - 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.19808, size = 157, normalized size = 2.75 \begin{align*} \frac{1}{4} \, \sqrt{2}{\left (\frac{{\left (8 i \, A \sqrt{-a} \arctan \left (-i\right ) + 8 i \, B \sqrt{-a} \arctan \left (-i\right ) - 8 \, B \sqrt{-a}\right )} \mathrm{sgn}\left (-e^{x} + 1\right )}{a} - \frac{8 \,{\left (A + B\right )} \arctan \left (\frac{\sqrt{-a e^{x}}}{\sqrt{a}}\right )}{\sqrt{a} \mathrm{sgn}\left (-e^{x} + 1\right )} - \frac{4 \, B}{\sqrt{-a e^{x}} \mathrm{sgn}\left (-e^{x} + 1\right )} + \frac{4 \, \sqrt{-a e^{x}} B}{a \mathrm{sgn}\left (-e^{x} + 1\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*((8*I*A*sqrt(-a)*arctan(-I) + 8*I*B*sqrt(-a)*arctan(-I) - 8*B*sqrt(-a))*sgn(-e^x + 1)/a - 8*(A + B
)*arctan(sqrt(-a*e^x)/sqrt(a))/(sqrt(a)*sgn(-e^x + 1)) - 4*B/(sqrt(-a*e^x)*sgn(-e^x + 1)) + 4*sqrt(-a*e^x)*B/(
a*sgn(-e^x + 1)))