3.754 \(\int (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x))^3 \, dx\)

Optimal. Leaf size=136 \[ \frac{5}{2} x \left (b^2-c^2\right )^{3/2}+\frac{5}{2} b \left (b^2-c^2\right ) \sinh (x)+\frac{5}{2} c \left (b^2-c^2\right ) \cosh (x)+\frac{1}{3} (b \sinh (x)+c \cosh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2+\frac{5}{6} \sqrt{b^2-c^2} (b \sinh (x)+c \cosh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right ) \]

[Out]

(5*(b^2 - c^2)^(3/2)*x)/2 + (5*c*(b^2 - c^2)*Cosh[x])/2 + (5*b*(b^2 - c^2)*Sinh[x])/2 + (5*Sqrt[b^2 - c^2]*(c*
Cosh[x] + b*Sinh[x])*(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]))/6 + ((c*Cosh[x] + b*Sinh[x])*(Sqrt[b^2 - c^2]
+ b*Cosh[x] + c*Sinh[x])^2)/3

________________________________________________________________________________________

Rubi [A]  time = 0.0896551, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3113, 2637, 2638} \[ \frac{5}{2} x \left (b^2-c^2\right )^{3/2}+\frac{5}{2} b \left (b^2-c^2\right ) \sinh (x)+\frac{5}{2} c \left (b^2-c^2\right ) \cosh (x)+\frac{1}{3} (b \sinh (x)+c \cosh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2+\frac{5}{6} \sqrt{b^2-c^2} (b \sinh (x)+c \cosh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(b^2 - c^2)^(3/2)*x)/2 + (5*c*(b^2 - c^2)*Cosh[x])/2 + (5*b*(b^2 - c^2)*Sinh[x])/2 + (5*Sqrt[b^2 - c^2]*(c*
Cosh[x] + b*Sinh[x])*(Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]))/6 + ((c*Cosh[x] + b*Sinh[x])*(Sqrt[b^2 - c^2]
+ b*Cosh[x] + c*Sinh[x])^2)/3

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 )^3 \, dx &=\frac{1}{3} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2+\frac{1}{3} \left (5 \sqrt{b^2-c^2}\right ) \int \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2 \, dx\\ &=\frac{5}{6} \sqrt{b^2-c^2} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )+\frac{1}{3} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2+\frac{1}{2} \left (5 \left (b^2-c^2\right )\right ) \int \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right ) \, dx\\ &=\frac{5}{2} \left (b^2-c^2\right )^{3/2} x+\frac{5}{6} \sqrt{b^2-c^2} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )+\frac{1}{3} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2+\frac{1}{2} \left (5 b \left (b^2-c^2\right )\right ) \int \cosh (x) \, dx+\frac{1}{2} \left (5 c \left (b^2-c^2\right )\right ) \int \sinh (x) \, dx\\ &=\frac{5}{2} \left (b^2-c^2\right )^{3/2} x+\frac{5}{2} c \left (b^2-c^2\right ) \cosh (x)+\frac{5}{2} b \left (b^2-c^2\right ) \sinh (x)+\frac{5}{6} \sqrt{b^2-c^2} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )+\frac{1}{3} (c \cosh (x)+b \sinh (x)) \left (\sqrt{b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2\\ \end{align*}

Mathematica [A]  time = 0.257189, size = 134, normalized size = 0.99 \[ \frac{1}{12} \left (30 x (b-c) (b+c) \sqrt{b^2-c^2}+45 b \left (b^2-c^2\right ) \sinh (x)+9 \sqrt{b^2-c^2} \left (b^2+c^2\right ) \sinh (2 x)+b \left (b^2+3 c^2\right ) \sinh (3 x)+45 c \left (b^2-c^2\right ) \cosh (x)+18 b c \sqrt{b^2-c^2} \cosh (2 x)+c \left (3 b^2+c^2\right ) \cosh (3 x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*(b - c)*(b + c)*Sqrt[b^2 - c^2]*x + 45*c*(b^2 - c^2)*Cosh[x] + 18*b*c*Sqrt[b^2 - c^2]*Cosh[2*x] + c*(3*b^2
+ c^2)*Cosh[3*x] + 45*b*(b^2 - c^2)*Sinh[x] + 9*Sqrt[b^2 - c^2]*(b^2 + c^2)*Sinh[2*x] + b*(b^2 + 3*c^2)*Sinh[
3*x])/12

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 202, normalized size = 1.5 \begin{align*}{b}^{3} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{2}}{3}} \right ) \sinh \left ( x \right ) +3\,c{b}^{2} \left ( 1/3\,\cosh \left ( x \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}+1/3\,\cosh \left ( x \right ) \right ) +3\,\sqrt{{b}^{2}-{c}^{2}}{b}^{2} \left ( 1/2\,\cosh \left ( x \right ) \sinh \left ( x \right ) +x/2 \right ) +3\,b{c}^{2} \left ( 1/3\,\sinh \left ( x \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}-1/3\,\sinh \left ( x \right ) \right ) +3\,\sqrt{{b}^{2}-{c}^{2}}bc \left ( \cosh \left ( x \right ) \right ) ^{2}+3\,{b}^{3}\sinh \left ( x \right ) -3\,b{c}^{2}\sinh \left ( x \right ) +{c}^{3} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( x \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( x \right ) +3\,\sqrt{{b}^{2}-{c}^{2}}{c}^{2} \left ( 1/2\,\cosh \left ( x \right ) \sinh \left ( x \right ) -x/2 \right ) +3\,{b}^{2}\cosh \left ( x \right ) c-3\,{c}^{3}\cosh \left ( x \right ) +\sqrt{{b}^{2}-{c}^{2}}{b}^{2}x-\sqrt{{b}^{2}-{c}^{2}}{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))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.03399, size = 217, normalized size = 1.6 \begin{align*} b^{2} c \cosh \left (x\right )^{3} + b c^{2} \sinh \left (x\right )^{3} + \frac{1}{24} \, c^{3}{\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} + e^{\left (-3 \, x\right )} - 9 \, e^{x}\right )} + \frac{1}{24} \, b^{3}{\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} - e^{\left (-3 \, x\right )} + 9 \, e^{x}\right )} +{\left (b^{2} - c^{2}\right )}^{\frac{3}{2}} x + 3 \,{\left (b^{2} - c^{2}\right )}{\left (c \cosh \left (x\right ) + b \sinh \left (x\right )\right )} + \frac{3}{8} \,{\left (8 \, b c \cosh \left (x\right )^{2} + b^{2}{\left (4 \, x + e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}\right )} - c^{2}{\left (4 \, x - e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}\right )} \sqrt{b^{2} - c^{2}} \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))^3,x, algorithm="maxima")

[Out]

b^2*c*cosh(x)^3 + b*c^2*sinh(x)^3 + 1/24*c^3*(e^(3*x) - 9*e^(-x) + e^(-3*x) - 9*e^x) + 1/24*b^3*(e^(3*x) - 9*e
^(-x) - e^(-3*x) + 9*e^x) + (b^2 - c^2)^(3/2)*x + 3*(b^2 - c^2)*(c*cosh(x) + b*sinh(x)) + 3/8*(8*b*c*cosh(x)^2
+ b^2*(4*x + e^(2*x) - e^(-2*x)) - c^2*(4*x - e^(2*x) + e^(-2*x)))*sqrt(b^2 - c^2)

________________________________________________________________________________________

Fricas [B]  time = 2.64613, size = 1665, normalized size = 12.24 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="fricas")

[Out]

1/24*((b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^6 + 6*(b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)*sinh(x)^5 + (b^3 +
3*b^2*c + 3*b*c^2 + c^3)*sinh(x)^6 + 45*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x)^4 + 15*(3*b^3 + 3*b^2*c - 3*b*c^2
- 3*c^3 + (b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^2)*sinh(x)^4 + 20*((b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^
3 + 9*(b^3 + b^2*c - b*c^2 - c^3)*cosh(x))*sinh(x)^3 - b^3 + 3*b^2*c - 3*b*c^2 + c^3 - 45*(b^3 - b^2*c - b*c^2
+ c^3)*cosh(x)^2 + 15*((b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^4 - 3*b^3 + 3*b^2*c + 3*b*c^2 - 3*c^3 + 18*(b^
3 + b^2*c - b*c^2 - c^3)*cosh(x)^2)*sinh(x)^2 + 6*((b^3 + 3*b^2*c + 3*b*c^2 + c^3)*cosh(x)^5 + 30*(b^3 + b^2*c
- b*c^2 - c^3)*cosh(x)^3 - 15*(b^3 - b^2*c - b*c^2 + c^3)*cosh(x))*sinh(x) + 3*(3*(b^2 + 2*b*c + c^2)*cosh(x)
^5 + 15*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x)^4 + 3*(b^2 + 2*b*c + c^2)*sinh(x)^5 + 20*(b^2 - c^2)*x*cosh(x)^3 +
10*(3*(b^2 + 2*b*c + c^2)*cosh(x)^2 + 2*(b^2 - c^2)*x)*sinh(x)^3 + 30*((b^2 + 2*b*c + c^2)*cosh(x)^3 + 2*(b^2
- c^2)*x*cosh(x))*sinh(x)^2 - 3*(b^2 - 2*b*c + c^2)*cosh(x) + 3*(5*(b^2 + 2*b*c + c^2)*cosh(x)^4 + 20*(b^2 -
c^2)*x*cosh(x)^2 - b^2 + 2*b*c - c^2)*sinh(x))*sqrt(b^2 - c^2))/(cosh(x)^3 + 3*cosh(x)^2*sinh(x) + 3*cosh(x)*s
inh(x)^2 + sinh(x)^3)

________________________________________________________________________________________

Sympy [B]  time = 2.25663, size = 298, normalized size = 2.19 \begin{align*} - \frac{2 b^{3} \sinh ^{3}{\left (x \right )}}{3} + b^{3} \sinh{\left (x \right )} \cosh ^{2}{\left (x \right )} + 3 b^{3} \sinh{\left (x \right )} + b^{2} c \cosh ^{3}{\left (x \right )} + 3 b^{2} c \cosh{\left (x \right )} - \frac{3 b^{2} x \sqrt{b^{2} - c^{2}} \sinh ^{2}{\left (x \right )}}{2} + \frac{3 b^{2} x \sqrt{b^{2} - c^{2}} \cosh ^{2}{\left (x \right )}}{2} + b^{2} x \sqrt{b^{2} - c^{2}} + \frac{3 b^{2} \sqrt{b^{2} - c^{2}} \sinh{\left (x \right )} \cosh{\left (x \right )}}{2} + b c^{2} \sinh ^{3}{\left (x \right )} - 3 b c^{2} \sinh{\left (x \right )} + 3 b c \sqrt{b^{2} - c^{2}} \sinh ^{2}{\left (x \right )} + c^{3} \sinh ^{2}{\left (x \right )} \cosh{\left (x \right )} - \frac{2 c^{3} \cosh ^{3}{\left (x \right )}}{3} - 3 c^{3} \cosh{\left (x \right )} + \frac{3 c^{2} x \sqrt{b^{2} - c^{2}} \sinh ^{2}{\left (x \right )}}{2} - \frac{3 c^{2} x \sqrt{b^{2} - c^{2}} \cosh ^{2}{\left (x \right )}}{2} - c^{2} x \sqrt{b^{2} - c^{2}} + \frac{3 c^{2} \sqrt{b^{2} - c^{2}} \sinh{\left (x \right )} \cosh{\left (x \right )}}{2} \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))**3,x)

[Out]

-2*b**3*sinh(x)**3/3 + b**3*sinh(x)*cosh(x)**2 + 3*b**3*sinh(x) + b**2*c*cosh(x)**3 + 3*b**2*c*cosh(x) - 3*b**
2*x*sqrt(b**2 - c**2)*sinh(x)**2/2 + 3*b**2*x*sqrt(b**2 - c**2)*cosh(x)**2/2 + b**2*x*sqrt(b**2 - c**2) + 3*b*
*2*sqrt(b**2 - c**2)*sinh(x)*cosh(x)/2 + b*c**2*sinh(x)**3 - 3*b*c**2*sinh(x) + 3*b*c*sqrt(b**2 - c**2)*sinh(x
)**2 + c**3*sinh(x)**2*cosh(x) - 2*c**3*cosh(x)**3/3 - 3*c**3*cosh(x) + 3*c**2*x*sqrt(b**2 - c**2)*sinh(x)**2/
2 - 3*c**2*x*sqrt(b**2 - c**2)*cosh(x)**2/2 - c**2*x*sqrt(b**2 - c**2) + 3*c**2*sqrt(b**2 - c**2)*sinh(x)*cosh
(x)/2

________________________________________________________________________________________

Giac [A]  time = 1.15932, size = 262, normalized size = 1.93 \begin{align*} \frac{5}{2} \,{\left (b^{2} - c^{2}\right )}^{\frac{3}{2}} x + \frac{3}{8} \,{\left (b^{2} + 2 \, b c + c^{2}\right )} \sqrt{b^{2} - c^{2}} e^{\left (2 \, x\right )} + \frac{1}{24} \,{\left (b^{3} + 3 \, b^{2} c + 3 \, b c^{2} + c^{3}\right )} e^{\left (3 \, x\right )} - \frac{1}{24} \,{\left (b^{3} - 3 \, b^{2} c + 3 \, b c^{2} - c^{3} + 45 \,{\left (b^{3} - b^{2} c - b c^{2} + c^{3}\right )} e^{\left (2 \, x\right )} + 9 \,{\left (\sqrt{b^{2} - c^{2}} b^{2} - 2 \, \sqrt{b^{2} - c^{2}} b c + \sqrt{b^{2} - c^{2}} c^{2}\right )} e^{x}\right )} e^{\left (-3 \, x\right )} + \frac{15}{8} \,{\left (b^{3} + b^{2} c - b c^{2} - c^{3}\right )} e^{x} \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))^3,x, algorithm="giac")

[Out]

5/2*(b^2 - c^2)^(3/2)*x + 3/8*(b^2 + 2*b*c + c^2)*sqrt(b^2 - c^2)*e^(2*x) + 1/24*(b^3 + 3*b^2*c + 3*b*c^2 + c^
3)*e^(3*x) - 1/24*(b^3 - 3*b^2*c + 3*b*c^2 - c^3 + 45*(b^3 - b^2*c - b*c^2 + c^3)*e^(2*x) + 9*(sqrt(b^2 - c^2)
*b^2 - 2*sqrt(b^2 - c^2)*b*c + sqrt(b^2 - c^2)*c^2)*e^x)*e^(-3*x) + 15/8*(b^3 + b^2*c - b*c^2 - c^3)*e^x