3.771 \(\int \frac{1}{\sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}} \, dx\)

Optimal. Leaf size=99 \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}} \]

[Out]

(Sqrt[2]*ArcTan[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b, (-I)*c]])/(Sqrt[2]*Sqrt[Sqrt[b^2 - c^2] + Sqrt[b^2 - c
^2]*Cosh[x + I*ArcTan[b, (-I)*c]]])])/(b^2 - c^2)^(1/4)

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Rubi [A]  time = 0.112632, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3115, 2649, 206} \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]],x]

[Out]

(Sqrt[2]*ArcTan[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b, (-I)*c]])/(Sqrt[2]*Sqrt[Sqrt[b^2 - c^2] + Sqrt[b^2 - c
^2]*Cosh[x + I*ArcTan[b, (-I)*c]]])])/(b^2 - c^2)^(1/4)

Rule 3115

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Int[1/Sqrt[a +
Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 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 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{1}{\sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}} \, dx &=\int \frac{1}{\sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}} \, dx\\ &=2 i \operatorname{Subst}\left (\int \frac{1}{2 \sqrt{b^2-c^2}-x^2} \, dx,x,-\frac{i \sqrt{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )\\ &=\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}}\\ \end{align*}

Mathematica [C]  time = 29.1132, size = 211, normalized size = 2.13 \[ -\frac{\sqrt{2} \left (c \sqrt{b^2-c^2} \sinh (x)+b \sqrt{b^2-c^2} \cosh (x)+b^2-c^2\right ) \sqrt{-\frac{c \sqrt{b^2-c^2} \sinh (x)+b \sqrt{b^2-c^2} \cosh (x)-b^2+c^2}{b^2-c^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-c^2}-b \cosh (x)-c \sinh (x)}{\sqrt{b^2-c^2}}}}{\sqrt{2}}\right ),1\right )}{\sqrt{b^2-c^2} (b \sinh (x)+c \cosh (x)) \sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]],x]

[Out]

-((Sqrt[2]*EllipticF[ArcSin[Sqrt[(Sqrt[b^2 - c^2] - b*Cosh[x] - c*Sinh[x])/Sqrt[b^2 - c^2]]/Sqrt[2]], 1]*(b^2
- c^2 + b*Sqrt[b^2 - c^2]*Cosh[x] + c*Sqrt[b^2 - c^2]*Sinh[x])*Sqrt[-((-b^2 + c^2 + b*Sqrt[b^2 - c^2]*Cosh[x]
+ c*Sqrt[b^2 - c^2]*Sinh[x])/(b^2 - c^2))])/(Sqrt[b^2 - c^2]*(c*Cosh[x] + b*Sinh[x])*Sqrt[Sqrt[b^2 - c^2] + b*
Cosh[x] + c*Sinh[x]]))

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Maple [A]  time = 0.279, size = 129, normalized size = 1.3 \begin{align*}{\frac{1}{\sinh \left ( x \right ) }\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}}\arctan \left ({\cosh \left ( x \right ) \sqrt{\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) }{\frac{1}{\sqrt{-\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}}}}} \right ){\frac{1}{\sqrt{\sqrt{{b}^{2}-{c}^{2}} \left ( \sinh \left ( x \right ) -1 \right ) }}}{\frac{1}{\sqrt{-{(\sinh \left ( x \right ){b}^{2}-\sinh \left ( x \right ){c}^{2}-{b}^{2}+{c}^{2}){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(b^2-c^2)^(1/2)*(sinh(x)-1)*sinh(x)^2)^(1/2)/((b^2-c^2)^(1/2)*(sinh(x)-1))^(1/2)*arctan(((b^2-c^2)^(1/2)*(si
nh(x)-1))^(1/2)*cosh(x)/(-(b^2-c^2)^(1/2)*(sinh(x)-1)*sinh(x)^2)^(1/2))/sinh(x)/(-(sinh(x)*b^2-sinh(x)*c^2-b^2
+c^2)/(b^2-c^2)^(1/2))^(1/2)

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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 ) + \sqrt{b^{2} - c^{2}}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2)), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cosh{\left (x \right )} + c \sinh{\left (x \right )} + \sqrt{b^{2} - c^{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*cosh(x) + c*sinh(x) + sqrt(b**2 - c**2)), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError