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

Optimal. Leaf size=108 \[ -\frac{2 i (A b-a B) \sqrt{\frac{a+b \cosh (x)}{a+b}} \text{EllipticF}\left (\frac{i x}{2},\frac{2 b}{a+b}\right )}{b \sqrt{a+b \cosh (x)}}-\frac{2 i B \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{b \sqrt{\frac{a+b \cosh (x)}{a+b}}} \]

[Out]

((-2*I)*B*Sqrt[a + b*Cosh[x]]*EllipticE[(I/2)*x, (2*b)/(a + b)])/(b*Sqrt[(a + b*Cosh[x])/(a + b)]) - ((2*I)*(A
*b - a*B)*Sqrt[(a + b*Cosh[x])/(a + b)]*EllipticF[(I/2)*x, (2*b)/(a + b)])/(b*Sqrt[a + b*Cosh[x]])

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Rubi [A]  time = 0.111749, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2752, 2663, 2661, 2655, 2653} \[ -\frac{2 i (A b-a B) \sqrt{\frac{a+b \cosh (x)}{a+b}} F\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{b \sqrt{a+b \cosh (x)}}-\frac{2 i B \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{b \sqrt{\frac{a+b \cosh (x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*B*Sqrt[a + b*Cosh[x]]*EllipticE[(I/2)*x, (2*b)/(a + b)])/(b*Sqrt[(a + b*Cosh[x])/(a + b)]) - ((2*I)*(A
*b - a*B)*Sqrt[(a + b*Cosh[x])/(a + b)]*EllipticF[(I/2)*x, (2*b)/(a + b)])/(b*Sqrt[a + b*Cosh[x]])

Rule 2752

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

Rule 2663

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

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{A+B \cosh (x)}{\sqrt{a+b \cosh (x)}} \, dx &=\frac{B \int \sqrt{a+b \cosh (x)} \, dx}{b}+\frac{(A b-a B) \int \frac{1}{\sqrt{a+b \cosh (x)}} \, dx}{b}\\ &=\frac{\left (B \sqrt{a+b \cosh (x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cosh (x)}{a+b}} \, dx}{b \sqrt{\frac{a+b \cosh (x)}{a+b}}}+\frac{\left ((A b-a B) \sqrt{\frac{a+b \cosh (x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cosh (x)}{a+b}}} \, dx}{b \sqrt{a+b \cosh (x)}}\\ &=-\frac{2 i B \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{b \sqrt{\frac{a+b \cosh (x)}{a+b}}}-\frac{2 i (A b-a B) \sqrt{\frac{a+b \cosh (x)}{a+b}} F\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{b \sqrt{a+b \cosh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.443158, size = 80, normalized size = 0.74 \[ -\frac{2 i \sqrt{\frac{a+b \cosh (x)}{a+b}} \left ((A b-a B) \text{EllipticF}\left (\frac{i x}{2},\frac{2 b}{a+b}\right )+B (a+b) E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )\right )}{b \sqrt{a+b \cosh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[(a + b*Cosh[x])/(a + b)]*((a + b)*B*EllipticE[(I/2)*x, (2*b)/(a + b)] + (A*b - a*B)*EllipticF[(I/
2)*x, (2*b)/(a + b)]))/(b*Sqrt[a + b*Cosh[x]])

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Maple [A]  time = 0.096, size = 218, normalized size = 2. \begin{align*} 2\,{\frac{\sqrt{- \left ( \sinh \left ( x/2 \right ) \right ) ^{2}}\sqrt{ \left ( 2\, \left ( \cosh \left ( x/2 \right ) \right ) ^{2}b+a-b \right ) \left ( \sinh \left ( x/2 \right ) \right ) ^{2}}}{\sqrt{2\,b \left ( \sinh \left ( x/2 \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \sinh \left ( x/2 \right ) \right ) ^{2}}\sinh \left ( x/2 \right ) \sqrt{2\, \left ( \sinh \left ( x/2 \right ) \right ) ^{2}b+a+b}} \left ( A{\it EllipticF} \left ( \cosh \left ( x/2 \right ) \sqrt{-2\,{\frac{b}{a-b}}},1/2\,\sqrt{-2\,{\frac{a-b}{b}}} \right ) +B{\it EllipticF} \left ( \cosh \left ( x/2 \right ) \sqrt{-2\,{\frac{b}{a-b}}},1/2\,\sqrt{-2\,{\frac{a-b}{b}}} \right ) -2\,B{\it EllipticE} \left ( \cosh \left ( x/2 \right ) \sqrt{-2\,{\frac{b}{a-b}}},1/2\,\sqrt{-2\,{\frac{a-b}{b}}} \right ) \right ) \sqrt{{\frac{2\, \left ( \cosh \left ( x/2 \right ) \right ) ^{2}b+a-b}{a-b}}}{\frac{1}{\sqrt{-2\,{\frac{b}{a-b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(A*EllipticF(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*(-2*(a-b)/b)^(1/2))+B*EllipticF(cosh(1/2*x)*(-2*b/(a-b))^(1/
2),1/2*(-2*(a-b)/b)^(1/2))-2*B*EllipticE(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*(-2*(a-b)/b)^(1/2)))*(-sinh(1/2*x)
^2)^(1/2)*((2*cosh(1/2*x)^2*b+a-b)/(a-b))^(1/2)*((2*cosh(1/2*x)^2*b+a-b)*sinh(1/2*x)^2)^(1/2)/(-2*b/(a-b))^(1/
2)/(2*b*sinh(1/2*x)^4+(a+b)*sinh(1/2*x)^2)^(1/2)/sinh(1/2*x)/(2*sinh(1/2*x)^2*b+a+b)^(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{b \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+b*cosh(x))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \cosh \left (x\right ) + A}{\sqrt{b \cosh \left (x\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cosh(x) + A)/sqrt(b*cosh(x) + a), x)

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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 + b \cosh{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cosh \left (x\right ) + A}{\sqrt{b \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+b*cosh(x))^(1/2),x, algorithm="giac")

[Out]

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