\(\int (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x))^3 \, dx\) [514]

Optimal result

Integrand size = 24, antiderivative size = 136 \[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx=\frac {5}{2} \left (b^2-c^2\right )^{3/2} x+\frac {5}{2} c \left (b^2-c^2\right ) \cosh (x)+\frac {5}{2} b \left (b^2-c^2\right ) \sinh (x)+\frac {5}{6} \sqrt {b^2-c^2} (c \cosh (x)+b \sinh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )+\frac {1}{3} (c \cosh (x)+b \sinh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2 \] Output:

5/2*(b^2-c^2)^(3/2)*x+5/2*c*(b^2-c^2)*cosh(x)+5/2*b*(b^2-c^2)*sinh(x)+5/6* 
(b^2-c^2)^(1/2)*(c*cosh(x)+b*sinh(x))*((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x) 
)+1/3*(c*cosh(x)+b*sinh(x))*((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^2
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.99 \[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx=\frac {1}{12} \left (30 (b-c) (b+c) \sqrt {b^2-c^2} x+45 c \left (b^2-c^2\right ) \cosh (x)+18 b c \sqrt {b^2-c^2} \cosh (2 x)+c \left (3 b^2+c^2\right ) \cosh (3 x)+45 b \left (b^2-c^2\right ) \sinh (x)+9 \sqrt {b^2-c^2} \left (b^2+c^2\right ) \sinh (2 x)+b \left (b^2+3 c^2\right ) \sinh (3 x)\right ) \] Input:

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

Output:

(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)*S 
inh[x] + 9*Sqrt[b^2 - c^2]*(b^2 + c^2)*Sinh[2*x] + b*(b^2 + 3*c^2)*Sinh[3* 
x])/12
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 3592, 3042, 3592, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (\sqrt {b^2-c^2}+b \cos (i x)-i c \sin (i x)\right )^3dx\)

\(\Big \downarrow \) 3592

\(\displaystyle \frac {5}{3} \sqrt {b^2-c^2} \int \left (b \cosh (x)+c \sinh (x)+\sqrt {b^2-c^2}\right )^2dx+\frac {1}{3} (b \sinh (x)+c \cosh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} (b \sinh (x)+c \cosh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2+\frac {5}{3} \sqrt {b^2-c^2} \int \left (b \cos (i x)-i c \sin (i x)+\sqrt {b^2-c^2}\right )^2dx\)

\(\Big \downarrow \) 3592

\(\displaystyle \frac {5}{3} \sqrt {b^2-c^2} \left (\frac {3}{2} \sqrt {b^2-c^2} \int \left (b \cosh (x)+c \sinh (x)+\sqrt {b^2-c^2}\right )dx+\frac {1}{2} (b \sinh (x)+c \cosh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )\right )+\frac {1}{3} (b \sinh (x)+c \cosh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} (b \sinh (x)+c \cosh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^2+\frac {5}{3} \sqrt {b^2-c^2} \left (\frac {3}{2} \sqrt {b^2-c^2} \left (x \sqrt {b^2-c^2}+b \sinh (x)+c \cosh (x)\right )+\frac {1}{2} (b \sinh (x)+c \cosh (x)) \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3592

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] + Simp[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]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.34

\[\sqrt {b^{2}-c^{2}}\, b^{2} x -\sqrt {b^{2}-c^{2}}\, c^{2} x +3 \sinh \left (x \right ) b^{3}-3 \sinh \left (x \right ) b \,c^{2}+3 \cosh \left (x \right ) b^{2} c -3 \cosh \left (x \right ) c^{3}+3 \sqrt {b^{2}-c^{2}}\, b^{2} \left (\frac {\cosh \left (x \right ) \sinh \left (x \right )}{2}+\frac {x}{2}\right )+3 \sqrt {b^{2}-c^{2}}\, b c \cosh \left (x \right )^{2}+3 \sqrt {b^{2}-c^{2}}\, c^{2} \left (\frac {\cosh \left (x \right ) \sinh \left (x \right )}{2}-\frac {x}{2}\right )+b^{3} \left (\frac {2}{3}+\frac {\cosh \left (x \right )^{2}}{3}\right ) \sinh \left (x \right )+b^{2} c \cosh \left (x \right )^{3}+b \,c^{2} \sinh \left (x \right )^{3}+c^{3} \left (-\frac {2}{3}+\frac {\sinh \left (x \right )^{2}}{3}\right ) \cosh \left (x \right )\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 664 vs. \(2 (118) = 236\).

Time = 0.12 (sec) , antiderivative size = 664, normalized size of antiderivative = 4.88 \[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx =\text {Too large to display} \] Input:

integrate(((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^3,x, algorithm="fricas")
 

Output:

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*cos 
h(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)*sinh(x)^2 + sinh(x)^3)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (124) = 248\).

Time = 0.18 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.19 \[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx=- \frac {2 b^{3} \sinh ^{3}{\left (x \right )}}{3} + b^{3} \sinh {\left (x \right )} \cosh ^{2}{\left (x \right )} + 3 b^{3} \sinh {\left (x \right )} + b^{2} c \cosh ^{3}{\left (x \right )} + 3 b^{2} c \cosh {\left (x \right )} - \frac {3 b^{2} x \sqrt {b^{2} - c^{2}} \sinh ^{2}{\left (x \right )}}{2} + \frac {3 b^{2} x \sqrt {b^{2} - c^{2}} \cosh ^{2}{\left (x \right )}}{2} + b^{2} x \sqrt {b^{2} - c^{2}} + \frac {3 b^{2} \sqrt {b^{2} - c^{2}} \sinh {\left (x \right )} \cosh {\left (x \right )}}{2} + b c^{2} \sinh ^{3}{\left (x \right )} - 3 b c^{2} \sinh {\left (x \right )} + 3 b c \sqrt {b^{2} - c^{2}} \sinh ^{2}{\left (x \right )} + c^{3} \sinh ^{2}{\left (x \right )} \cosh {\left (x \right )} - \frac {2 c^{3} \cosh ^{3}{\left (x \right )}}{3} - 3 c^{3} \cosh {\left (x \right )} + \frac {3 c^{2} x \sqrt {b^{2} - c^{2}} \sinh ^{2}{\left (x \right )}}{2} - \frac {3 c^{2} x \sqrt {b^{2} - c^{2}} \cosh ^{2}{\left (x \right )}}{2} - c^{2} x \sqrt {b^{2} - c^{2}} + \frac {3 c^{2} \sqrt {b^{2} - c^{2}} \sinh {\left (x \right )} \cosh {\left (x \right )}}{2} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.18 \[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx=b^{2} c \cosh \left (x\right )^{3} + b c^{2} \sinh \left (x\right )^{3} + \frac {1}{24} \, c^{3} {\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} + e^{\left (-3 \, x\right )} - 9 \, e^{x}\right )} + \frac {1}{24} \, b^{3} {\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} - e^{\left (-3 \, x\right )} + 9 \, e^{x}\right )} + {\left (b^{2} - c^{2}\right )}^{\frac {3}{2}} x + 3 \, {\left (b^{2} - c^{2}\right )} {\left (c \cosh \left (x\right ) + b \sinh \left (x\right )\right )} + \frac {3}{8} \, {\left (8 \, b c \cosh \left (x\right )^{2} + b^{2} {\left (4 \, x + e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}\right )} - c^{2} {\left (4 \, x - e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}\right )} \sqrt {b^{2} - c^{2}} \] Input:

integrate(((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^3,x, algorithm="maxima")
 

Output:

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)
 

Giac [F]

\[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx=\int { {\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + \sqrt {b^{2} - c^{2}}\right )}^{3} \,d x } \] Input:

integrate(((b^2-c^2)^(1/2)+b*cosh(x)+c*sinh(x))^3,x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.06 \[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx=\frac {11\,b^3\,\mathrm {sinh}\left (x\right )}{3}+\frac {c^3\,{\mathrm {cosh}\left (x\right )}^3}{3}+\frac {5\,x\,{\left (b^2-c^2\right )}^{3/2}}{2}-4\,c^3\,\mathrm {cosh}\left (x\right )+\frac {b^3\,{\mathrm {cosh}\left (x\right )}^2\,\mathrm {sinh}\left (x\right )}{3}+3\,b^2\,c\,\mathrm {cosh}\left (x\right )-4\,b\,c^2\,\mathrm {sinh}\left (x\right )+b^2\,c\,{\mathrm {cosh}\left (x\right )}^3+3\,b\,c\,{\mathrm {cosh}\left (x\right )}^2\,\sqrt {b^2-c^2}+\frac {3\,b^2\,\mathrm {cosh}\left (x\right )\,\mathrm {sinh}\left (x\right )\,\sqrt {b^2-c^2}}{2}+\frac {3\,c^2\,\mathrm {cosh}\left (x\right )\,\mathrm {sinh}\left (x\right )\,\sqrt {b^2-c^2}}{2}+b\,c^2\,{\mathrm {cosh}\left (x\right )}^2\,\mathrm {sinh}\left (x\right ) \] Input:

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

Output:

(11*b^3*sinh(x))/3 + (c^3*cosh(x)^3)/3 + (5*x*(b^2 - c^2)^(3/2))/2 - 4*c^3 
*cosh(x) + (b^3*cosh(x)^2*sinh(x))/3 + 3*b^2*c*cosh(x) - 4*b*c^2*sinh(x) + 
 b^2*c*cosh(x)^3 + 3*b*c*cosh(x)^2*(b^2 - c^2)^(1/2) + (3*b^2*cosh(x)*sinh 
(x)*(b^2 - c^2)^(1/2))/2 + (3*c^2*cosh(x)*sinh(x)*(b^2 - c^2)^(1/2))/2 + b 
*c^2*cosh(x)^2*sinh(x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.87 \[ \int \left (\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)\right )^3 \, dx=\cosh \left (x \right )^{3} b^{2} c -\frac {2 \cosh \left (x \right )^{3} c^{3}}{3}+\frac {3 \sqrt {b^{2}-c^{2}}\, \cosh \left (x \right )^{2} b^{2} x}{2}+3 \sqrt {b^{2}-c^{2}}\, \cosh \left (x \right )^{2} b c -\frac {3 \sqrt {b^{2}-c^{2}}\, \cosh \left (x \right )^{2} c^{2} x}{2}+\cosh \left (x \right )^{2} \sinh \left (x \right ) b^{3}+\frac {3 \sqrt {b^{2}-c^{2}}\, \cosh \left (x \right ) \sinh \left (x \right ) b^{2}}{2}+\frac {3 \sqrt {b^{2}-c^{2}}\, \cosh \left (x \right ) \sinh \left (x \right ) c^{2}}{2}+\cosh \left (x \right ) \sinh \left (x \right )^{2} c^{3}+3 \cosh \left (x \right ) b^{2} c -3 \cosh \left (x \right ) c^{3}-\frac {3 \sqrt {b^{2}-c^{2}}\, \sinh \left (x \right )^{2} b^{2} x}{2}+\frac {3 \sqrt {b^{2}-c^{2}}\, \sinh \left (x \right )^{2} c^{2} x}{2}+\sqrt {b^{2}-c^{2}}\, b^{2} x -\sqrt {b^{2}-c^{2}}\, c^{2} x -\frac {2 \sinh \left (x \right )^{3} b^{3}}{3}+\sinh \left (x \right )^{3} b \,c^{2}+3 \sinh \left (x \right ) b^{3}-3 \sinh \left (x \right ) b \,c^{2} \] Input:

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

Output:

(6*cosh(x)**3*b**2*c - 4*cosh(x)**3*c**3 + 9*sqrt(b**2 - c**2)*cosh(x)**2* 
b**2*x + 18*sqrt(b**2 - c**2)*cosh(x)**2*b*c - 9*sqrt(b**2 - c**2)*cosh(x) 
**2*c**2*x + 6*cosh(x)**2*sinh(x)*b**3 + 9*sqrt(b**2 - c**2)*cosh(x)*sinh( 
x)*b**2 + 9*sqrt(b**2 - c**2)*cosh(x)*sinh(x)*c**2 + 6*cosh(x)*sinh(x)**2* 
c**3 + 18*cosh(x)*b**2*c - 18*cosh(x)*c**3 - 9*sqrt(b**2 - c**2)*sinh(x)** 
2*b**2*x + 9*sqrt(b**2 - c**2)*sinh(x)**2*c**2*x + 6*sqrt(b**2 - c**2)*b** 
2*x - 6*sqrt(b**2 - c**2)*c**2*x - 4*sinh(x)**3*b**3 + 6*sinh(x)**3*b*c**2 
 + 18*sinh(x)*b**3 - 18*sinh(x)*b*c**2)/6