### 3.770 $$\int \sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)} \, dx$$

Optimal. Leaf size=37 $\frac{2 (b \sinh (x)+c \cosh (x))}{\sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}}$

[Out]

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

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Rubi [A]  time = 0.0386524, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.038, Rules used = {3112} $\frac{2 (b \sinh (x)+c \cosh (x))}{\sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3112

Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Simp[(-2*(c*Cos[d
+ e*x] - b*Sin[d + e*x]))/(e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]), x] /; FreeQ[{a, b, c, d, e}, x] && E
qQ[a^2 - b^2 - c^2, 0]

Rubi steps

\begin{align*} \int \sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)} \, dx &=\frac{2 (c \cosh (x)+b \sinh (x))}{\sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}}\\ \end{align*}

Mathematica [C]  time = 68.2719, size = 455, normalized size = 12.3 $\frac{4 (b-c) (b+c)^2 \left (2 b c \sqrt{b^2-c^2} \sinh (x)+b \left (2 b \sqrt{b^2-c^2}-2 b^2+c^2\right ) \cosh (x)+\left (\sinh \left (\frac{x}{2}\right )-\cosh \left (\frac{x}{2}\right )\right ) \left (\left (2 b^2 \left (2 \sqrt{b^2-c^2}+c\right )+b c \left (3 c-2 \sqrt{b^2-c^2}\right )-c^2 \left (\sqrt{b^2-c^2}+c\right )-4 b^3\right ) \sinh \left (\frac{x}{2}\right )+c \left (b \left (2 \sqrt{b^2-c^2}+c\right )+c \left (c-\sqrt{b^2-c^2}\right )-2 b^2\right ) \cosh \left (\frac{x}{2}\right )\right ) \sqrt{\frac{\left (\sqrt{b^2-c^2}-b-c\right ) (\sinh (x)+\cosh (x))}{\sqrt{b^2-c^2}-b+c}} E\left (\left .\sin ^{-1}\left (\sqrt{\frac{\left (-b-c+\sqrt{b^2-c^2}\right ) (\cosh (x)+\sinh (x))}{-b+c+\sqrt{b^2-c^2}}}\right )\right |1\right )-2 b^2 \sqrt{b^2-c^2}+c^2 \sqrt{b^2-c^2}-2 b^2 c \sinh (x)+2 b^3-2 b c^2+c^3 \sinh (x)\right )}{\sqrt{b^2-c^2} \left (-\sqrt{b^2-c^2}+b+c\right )^2 \left (b \sqrt{b^2-c^2}-b^2+c^2\right ) \sqrt{\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*(b - c)*(b + c)^2*(2*b^3 - 2*b*c^2 - 2*b^2*Sqrt[b^2 - c^2] + c^2*Sqrt[b^2 - c^2] + b*(-2*b^2 + c^2 + 2*b*Sq
rt[b^2 - c^2])*Cosh[x] - 2*b^2*c*Sinh[x] + c^3*Sinh[x] + 2*b*c*Sqrt[b^2 - c^2]*Sinh[x] + EllipticE[ArcSin[Sqrt
[((-b - c + Sqrt[b^2 - c^2])*(Cosh[x] + Sinh[x]))/(-b + c + Sqrt[b^2 - c^2])]], 1]*(-Cosh[x/2] + Sinh[x/2])*(c
*(-2*b^2 + c*(c - Sqrt[b^2 - c^2]) + b*(c + 2*Sqrt[b^2 - c^2]))*Cosh[x/2] + (-4*b^3 + b*c*(3*c - 2*Sqrt[b^2 -
c^2]) - c^2*(c + Sqrt[b^2 - c^2]) + 2*b^2*(c + 2*Sqrt[b^2 - c^2]))*Sinh[x/2])*Sqrt[((-b - c + Sqrt[b^2 - c^2])
*(Cosh[x] + Sinh[x]))/(-b + c + Sqrt[b^2 - c^2])]))/(Sqrt[b^2 - c^2]*(b + c - Sqrt[b^2 - c^2])^2*(-b^2 + c^2 +
b*Sqrt[b^2 - c^2])*Sqrt[Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]])

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Maple [B]  time = 0.442, size = 201, normalized size = 5.4 \begin{align*}{ \left ( -{b}^{2}+{c}^{2} \right ) \cosh \left ( x \right ){\frac{1}{\sqrt{{b}^{2}-{c}^{2}}}}{\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}}}}}}}}}+{\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}}\sqrt{{b}^{2}-{c}^{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((b*cosh(x)+c*sinh(x)+(b^2-c^2)^(1/2))^(1/2),x)

[Out]

(-b^2+c^2)/(b^2-c^2)^(1/2)/(-(sinh(x)*b^2-sinh(x)*c^2-b^2+c^2)/(b^2-c^2)^(1/2))^(1/2)*cosh(x)+(-(b^2-c^2)^(1/2
)*(sinh(x)-1)*sinh(x)^2)^(1/2)*(b^2-c^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 [B]  time = 1.88173, size = 207, normalized size = 5.59 \begin{align*} \frac{\sqrt{2} \sqrt{2 \, \sqrt{b + c} \sqrt{b - c} e^{\left (-x\right )} +{\left (b - c\right )} e^{\left (-2 \, x\right )} + b + c} \sqrt{b + c} \sqrt{b - c} e^{\left (\frac{1}{2} \, x\right )}}{{\left (b - c\right )} e^{\left (-x\right )} + \sqrt{b + c} \sqrt{b - c}} - \frac{\sqrt{2} \sqrt{2 \, \sqrt{b + c} \sqrt{b - c} e^{\left (-x\right )} +{\left (b - c\right )} e^{\left (-2 \, x\right )} + b + c}{\left (b - c\right )} e^{\left (-\frac{1}{2} \, x\right )}}{{\left (b - c\right )} e^{\left (-x\right )} + \sqrt{b + c} \sqrt{b - c}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*sqrt(2*sqrt(b + c)*sqrt(b - c)*e^(-x) + (b - c)*e^(-2*x) + b + c)*sqrt(b + c)*sqrt(b - c)*e^(1/2*x)/((
b - c)*e^(-x) + sqrt(b + c)*sqrt(b - c)) - sqrt(2)*sqrt(2*sqrt(b + c)*sqrt(b - c)*e^(-x) + (b - c)*e^(-2*x) +
b + c)*(b - c)*e^(-1/2*x)/((b - c)*e^(-x) + sqrt(b + c)*sqrt(b - c))

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Fricas [B]  time = 2.39423, size = 468, normalized size = 12.65 \begin{align*} \frac{2 \, \sqrt{\frac{1}{2}}{\left ({\left (b + c\right )} \cosh \left (x\right )^{2} + 2 \,{\left (b + c\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (b + c\right )} \sinh \left (x\right )^{2} - 2 \, \sqrt{b^{2} - c^{2}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} + b - c\right )} \sqrt{\frac{{\left (b + c\right )} \cosh \left (x\right )^{2} + 2 \,{\left (b + c\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (b + c\right )} \sinh \left (x\right )^{2} + 2 \, \sqrt{b^{2} - c^{2}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} + b - c}{\cosh \left (x\right ) + \sinh \left (x\right )}}}{{\left (b + c\right )} \cosh \left (x\right )^{2} + 2 \,{\left (b + c\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (b + c\right )} \sinh \left (x\right )^{2} - b + c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(1/2)*((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^2 - 2*sqrt(b^2 - c^2)*(cosh(x) +
sinh(x)) + b - c)*sqrt(((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^2 + 2*sqrt(b^2 - c^2)*
(cosh(x) + sinh(x)) + b - c)/(cosh(x) + sinh(x)))/((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sin
h(x)^2 - b + c)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((b*cosh(x)+c*sinh(x)+(b**2-c**2)**(1/2))**(1/2),x)

[Out]

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

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Giac [B]  time = 1.195, size = 140, normalized size = 3.78 \begin{align*} -\frac{\sqrt{2}{\left (\sqrt{b^{2} - c^{2}} e^{\left (\frac{1}{2} \, x\right )} \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} - b + c\right ) -{\left (b \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} - b + c\right ) - c \mathrm{sgn}\left (-\sqrt{b^{2} - c^{2}} e^{x} - b + c\right )\right )} e^{\left (-\frac{1}{2} \, x\right )}\right )}}{\sqrt{b - c}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*(sqrt(b^2 - c^2)*e^(1/2*x)*sgn(-sqrt(b^2 - c^2)*e^x - b + c) - (b*sgn(-sqrt(b^2 - c^2)*e^x - b + c) -
c*sgn(-sqrt(b^2 - c^2)*e^x - b + c))*e^(-1/2*x))/sqrt(b - c)