### 3.755 $$\int (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x))^2 \, dx$$

Optimal. Leaf size=90 $\frac{3}{2} x \left (b^2-c^2\right )+\frac{3}{2} b \sqrt{b^2-c^2} \sinh (x)+\frac{3}{2} c \sqrt{b^2-c^2} \cosh (x)+\frac{1}{2} (b \sinh (x)+c \cosh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )$

[Out]

(3*(b^2 - c^2)*x)/2 + (3*c*Sqrt[b^2 - c^2]*Cosh[x])/2 + (3*b*Sqrt[b^2 - c^2]*Sinh[x])/2 + ((c*Cosh[x] + b*Sinh
[x])*(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]))/2

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Rubi [A]  time = 0.048117, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {3113, 2637, 2638} $\frac{3}{2} x \left (b^2-c^2\right )+\frac{3}{2} b \sqrt{b^2-c^2} \sinh (x)+\frac{3}{2} c \sqrt{b^2-c^2} \cosh (x)+\frac{1}{2} (b \sinh (x)+c \cosh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*(b^2 - c^2)*x)/2 + (3*c*Sqrt[b^2 - c^2]*Cosh[x])/2 + (3*b*Sqrt[b^2 - c^2]*Sinh[x])/2 + ((c*Cosh[x] + b*Sinh
[x])*(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]))/2

Rule 3113

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> -Simp[((c*Cos[d
+ e*x] - b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1))/(e*n), x] + Dist[(a*(2*n - 1))/n, Int[
(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]
&& GtQ[n, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2 \, dx &=\frac{1}{2} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )+\frac{1}{2} \left (3 \sqrt{b^2-c^2}\right ) \int \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right ) \, dx\\ &=\frac{3}{2} \left (b^2-c^2\right ) x+\frac{1}{2} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )+\frac{1}{2} \left (3 b \sqrt{b^2-c^2}\right ) \int \cosh (x) \, dx+\frac{1}{2} \left (3 c \sqrt{b^2-c^2}\right ) \int \sinh (x) \, dx\\ &=\frac{3}{2} \left (b^2-c^2\right ) x+\frac{3}{2} c \sqrt{b^2-c^2} \cosh (x)+\frac{3}{2} b \sqrt{b^2-c^2} \sinh (x)+\frac{1}{2} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )\\ \end{align*}

Mathematica [A]  time = 0.113133, size = 72, normalized size = 0.8 $\frac{1}{4} \left (8 b \sqrt{b^2-c^2} \sinh (x)+\left (b^2+c^2\right ) \sinh (2 x)+8 c \sqrt{b^2-c^2} \cosh (x)+6 x (b-c) (b+c)+2 b c \cosh (2 x)\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(6*(b - c)*(b + c)*x + 8*c*Sqrt[b^2 - c^2]*Cosh[x] + 2*b*c*Cosh[2*x] + 8*b*Sqrt[b^2 - c^2]*Sinh[x] + (b^2 + c^
2)*Sinh[2*x])/4

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Maple [A]  time = 0.033, size = 80, normalized size = 0.9 \begin{align*}{b}^{2} \left ({\frac{\cosh \left ( x \right ) \sinh \left ( x \right ) }{2}}+{\frac{x}{2}} \right ) +b \left ( \cosh \left ( x \right ) \right ) ^{2}c+{c}^{2} \left ({\frac{\cosh \left ( x \right ) \sinh \left ( x \right ) }{2}}-{\frac{x}{2}} \right ) +2\,b\sinh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}+2\,c\cosh \left ( x \right ) \sqrt{{b}^{2}-{c}^{2}}+{b}^{2}x-{c}^{2}x \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.17802, size = 107, normalized size = 1.19 \begin{align*} b c \cosh \left (x\right )^{2} + \frac{1}{8} \, b^{2}{\left (4 \, x + e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}\right )} - \frac{1}{8} \, c^{2}{\left (4 \, x - e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )} + b^{2} x - c^{2} x + 2 \, \sqrt{b^{2} - c^{2}}{\left (c \cosh \left (x\right ) + b \sinh \left (x\right )\right )} \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))^2,x, algorithm="maxima")

[Out]

b*c*cosh(x)^2 + 1/8*b^2*(4*x + e^(2*x) - e^(-2*x)) - 1/8*c^2*(4*x - e^(2*x) + e^(-2*x)) + b^2*x - c^2*x + 2*sq
rt(b^2 - c^2)*(c*cosh(x) + b*sinh(x))

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Fricas [B]  time = 2.59504, size = 656, normalized size = 7.29 \begin{align*} \frac{{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )^{4} + 4 \,{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} +{\left (b^{2} + 2 \, b c + c^{2}\right )} \sinh \left (x\right )^{4} + 12 \,{\left (b^{2} - c^{2}\right )} x \cosh \left (x\right )^{2} + 6 \,{\left ({\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (b^{2} - c^{2}\right )} x\right )} \sinh \left (x\right )^{2} - b^{2} + 2 \, b c - c^{2} + 4 \,{\left ({\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )^{3} + 6 \,{\left (b^{2} - c^{2}\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right ) + 8 \,{\left ({\left (b + c\right )} \cosh \left (x\right )^{3} + 3 \,{\left (b + c\right )} \cosh \left (x\right ) \sinh \left (x\right )^{2} +{\left (b + c\right )} \sinh \left (x\right )^{3} -{\left (b - c\right )} \cosh \left (x\right ) +{\left (3 \,{\left (b + c\right )} \cosh \left (x\right )^{2} - b + c\right )} \sinh \left (x\right )\right )} \sqrt{b^{2} - c^{2}}}{8 \,{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}} \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))^2,x, algorithm="fricas")

[Out]

1/8*((b^2 + 2*b*c + c^2)*cosh(x)^4 + 4*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x)^3 + (b^2 + 2*b*c + c^2)*sinh(x)^4 +
12*(b^2 - c^2)*x*cosh(x)^2 + 6*((b^2 + 2*b*c + c^2)*cosh(x)^2 + 2*(b^2 - c^2)*x)*sinh(x)^2 - b^2 + 2*b*c - c^
2 + 4*((b^2 + 2*b*c + c^2)*cosh(x)^3 + 6*(b^2 - c^2)*x*cosh(x))*sinh(x) + 8*((b + c)*cosh(x)^3 + 3*(b + c)*cos
h(x)*sinh(x)^2 + (b + c)*sinh(x)^3 - (b - c)*cosh(x) + (3*(b + c)*cosh(x)^2 - b + c)*sinh(x))*sqrt(b^2 - c^2))
/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)

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Sympy [A]  time = 0.497405, size = 122, normalized size = 1.36 \begin{align*} - \frac{b^{2} x \sinh ^{2}{\left (x \right )}}{2} + \frac{b^{2} x \cosh ^{2}{\left (x \right )}}{2} + b^{2} x + \frac{b^{2} \sinh{\left (x \right )} \cosh{\left (x \right )}}{2} + b c \sinh ^{2}{\left (x \right )} + 2 b \sqrt{b^{2} - c^{2}} \sinh{\left (x \right )} + \frac{c^{2} x \sinh ^{2}{\left (x \right )}}{2} - \frac{c^{2} x \cosh ^{2}{\left (x \right )}}{2} - c^{2} x + \frac{c^{2} \sinh{\left (x \right )} \cosh{\left (x \right )}}{2} + 2 c \sqrt{b^{2} - c^{2}} \cosh{\left (x \right )} \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))**2,x)

[Out]

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

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Giac [A]  time = 1.16837, size = 130, normalized size = 1.44 \begin{align*} \sqrt{b^{2} - c^{2}}{\left (b + c\right )} e^{x} + \frac{3}{2} \,{\left (b^{2} - c^{2}\right )} x + \frac{1}{8} \,{\left (b^{2} + 2 \, b c + c^{2}\right )} e^{\left (2 \, x\right )} - \frac{1}{8} \,{\left (b^{2} - 2 \, b c + c^{2} + 8 \,{\left (\sqrt{b^{2} - c^{2}} b - \sqrt{b^{2} - c^{2}} c\right )} e^{x}\right )} e^{\left (-2 \, x\right )} \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))^2,x, algorithm="giac")

[Out]

sqrt(b^2 - c^2)*(b + c)*e^x + 3/2*(b^2 - c^2)*x + 1/8*(b^2 + 2*b*c + c^2)*e^(2*x) - 1/8*(b^2 - 2*b*c + c^2 + 8
*(sqrt(b^2 - c^2)*b - sqrt(b^2 - c^2)*c)*e^x)*e^(-2*x)