3.764 \(\int \frac{1}{\sqrt{a+b \cosh (x)+c \sinh (x)}} \, dx\)

Optimal. Leaf size=102 \[ -\frac{2 i \sqrt{\frac{a+b \cosh (x)+c \sinh (x)}{a+\sqrt{b^2-c^2}}} \text{EllipticF}\left (\frac{1}{2} \left (i x-\tan ^{-1}(b,-i c)\right ),\frac{2 \sqrt{b^2-c^2}}{a+\sqrt{b^2-c^2}}\right )}{\sqrt{a+b \cosh (x)+c \sinh (x)}} \]

[Out]

((-2*I)*EllipticF[(I*x - ArcTan[b, (-I)*c])/2, (2*Sqrt[b^2 - c^2])/(a + Sqrt[b^2 - c^2])]*Sqrt[(a + b*Cosh[x]
+ c*Sinh[x])/(a + Sqrt[b^2 - c^2])])/Sqrt[a + b*Cosh[x] + c*Sinh[x]]

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Rubi [A]  time = 0.0713375, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3127, 2661} \[ -\frac{2 i \sqrt{\frac{a+b \cosh (x)+c \sinh (x)}{a+\sqrt{b^2-c^2}}} F\left (\frac{1}{2} \left (i x-\tan ^{-1}(b,-i c)\right )|\frac{2 \sqrt{b^2-c^2}}{a+\sqrt{b^2-c^2}}\right )}{\sqrt{a+b \cosh (x)+c \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Cosh[x] + c*Sinh[x]],x]

[Out]

((-2*I)*EllipticF[(I*x - ArcTan[b, (-I)*c])/2, (2*Sqrt[b^2 - c^2])/(a + Sqrt[b^2 - c^2])]*Sqrt[(a + b*Cosh[x]
+ c*Sinh[x])/(a + Sqrt[b^2 - c^2])])/Sqrt[a + b*Cosh[x] + c*Sinh[x]]

Rule 3127

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

Rubi steps

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

Mathematica [C]  time = 0.482625, size = 237, normalized size = 2.32 \[ \frac{2 \text{sech}\left (\tanh ^{-1}\left (\frac{b}{c}\right )+x\right ) \sqrt{a+b \cosh (x)+c \sinh (x)} \sqrt{-\frac{-i c \sqrt{1-\frac{b^2}{c^2}}+b \cosh (x)+c \sinh (x)}{a+i c \sqrt{1-\frac{b^2}{c^2}}}} \sqrt{-\frac{i c \sqrt{1-\frac{b^2}{c^2}}+b \cosh (x)+c \sinh (x)}{a-i c \sqrt{1-\frac{b^2}{c^2}}}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\frac{a+b \cosh (x)+c \sinh (x)}{a+i \sqrt{1-\frac{b^2}{c^2}} c},\frac{a+b \cosh (x)+c \sinh (x)}{a-i \sqrt{1-\frac{b^2}{c^2}} c}\right )}{c \sqrt{1-\frac{b^2}{c^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[a + b*Cosh[x] + c*Sinh[x]],x]

[Out]

(2*AppellF1[1/2, 1/2, 1/2, 3/2, (a + b*Cosh[x] + c*Sinh[x])/(a + I*Sqrt[1 - b^2/c^2]*c), (a + b*Cosh[x] + c*Si
nh[x])/(a - I*Sqrt[1 - b^2/c^2]*c)]*Sech[x + ArcTanh[b/c]]*Sqrt[a + b*Cosh[x] + c*Sinh[x]]*Sqrt[-(((-I)*Sqrt[1
- b^2/c^2]*c + b*Cosh[x] + c*Sinh[x])/(a + I*Sqrt[1 - b^2/c^2]*c))]*Sqrt[-((I*Sqrt[1 - b^2/c^2]*c + b*Cosh[x]
+ c*Sinh[x])/(a - I*Sqrt[1 - b^2/c^2]*c))])/(Sqrt[1 - b^2/c^2]*c)

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Maple [A]  time = 0.244, size = 248, normalized size = 2.4 \begin{align*}{\frac{1}{\sinh \left ( x \right ) }\sqrt{{ \left ( \sinh \left ( x \right ) \right ) ^{2} \left ( -\sinh \left ( x \right ){b}^{2}+\sinh \left ( x \right ){c}^{2}+a\sqrt{{b}^{2}-{c}^{2}} \right ){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}}\ln \left ({ \left ( \cosh \left ( x \right ) \sinh \left ( x \right ) \left ( -{b}^{2}+{c}^{2} \right ) +\cosh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}a+\sqrt{{ \left ( -{b}^{2}+{c}^{2} \right ) \left ( \sinh \left ( x \right ) \right ) ^{3}{\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}+a \left ( \sinh \left ( x \right ) \right ) ^{2}}\sqrt{{b}^{2}-{c}^{2}}\sqrt{{ \left ( -{b}^{2}+{c}^{2} \right ) \sinh \left ( x \right ){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}+a} \right ){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}{\frac{1}{\sqrt{{ \left ( -\sinh \left ( x \right ){b}^{2}+\sinh \left ( x \right ){c}^{2}+a\sqrt{{b}^{2}-{c}^{2}} \right ){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}}}}} \right ) \sqrt{{b}^{2}-{c}^{2}} \left ( -\sinh \left ( x \right ){b}^{2}+\sinh \left ( x \right ){c}^{2}+a\sqrt{{b}^{2}-{c}^{2}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cosh(x)+c*sinh(x))^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cosh \left (x\right ) + c \sinh \left (x\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)+c*sinh(x))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*cosh(x) + c*sinh(x) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{b \cosh \left (x\right ) + c \sinh \left (x\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/sqrt(b*cosh(x) + c*sinh(x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)+c*sinh(x))**(1/2),x)

[Out]

Integral(1/sqrt(a + b*cosh(x) + c*sinh(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cosh \left (x\right ) + c \sinh \left (x\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cosh(x)+c*sinh(x))^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(b*cosh(x) + c*sinh(x) + a), x)