### 3.777 $$\int \frac{1}{\sqrt{-\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}} \, dx$$

Optimal. Leaf size=102 $-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{-\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}}$

[Out]

-((Sqrt[2]*ArcTanh[((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 = 0.0938029, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.107, Rules used = {3115, 2649, 204} $-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt{2} \sqrt{-\sqrt{b^2-c^2}+\sqrt{b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((Sqrt[2]*ArcTanh[((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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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} \tanh ^{-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 = 30.6454, size = 52609, normalized size = 515.77 $\text{Result too large to show}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [A]  time = 0.323, 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