3.8.39 \(\int (a+b \cosh (x)+c \sinh (x))^3 \, dx\) [739]

3.8.39.1 Optimal result
3.8.39.2 Mathematica [A] (verified)
3.8.39.3 Rubi [A] (verified)
3.8.39.4 Maple [A] (verified)
3.8.39.5 Fricas [A] (verification not implemented)
3.8.39.6 Sympy [A] (verification not implemented)
3.8.39.7 Maxima [A] (verification not implemented)
3.8.39.8 Giac [B] (verification not implemented)
3.8.39.9 Mupad [B] (verification not implemented)

3.8.39.1 Optimal result

Integrand size = 12, antiderivative size = 119 \[ \int (a+b \cosh (x)+c \sinh (x))^3 \, dx=\frac {1}{2} a \left (2 a^2+3 b^2-3 c^2\right ) x+\frac {1}{6} c \left (11 a^2+4 b^2-4 c^2\right ) \cosh (x)+\frac {1}{6} b \left (11 a^2+4 b^2-4 c^2\right ) \sinh (x)+\frac {5}{6} (a c \cosh (x)+a b \sinh (x)) (a+b \cosh (x)+c \sinh (x))+\frac {1}{3} (c \cosh (x)+b \sinh (x)) (a+b \cosh (x)+c \sinh (x))^2 \]

output
1/2*a*(2*a^2+3*b^2-3*c^2)*x+1/6*c*(11*a^2+4*b^2-4*c^2)*cosh(x)+1/6*b*(11*a 
^2+4*b^2-4*c^2)*sinh(x)+5/6*(a*c*cosh(x)+a*b*sinh(x))*(a+b*cosh(x)+c*sinh( 
x))+1/3*(c*cosh(x)+b*sinh(x))*(a+b*cosh(x)+c*sinh(x))^2
 
3.8.39.2 Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97 \[ \int (a+b \cosh (x)+c \sinh (x))^3 \, dx=\frac {1}{12} \left (6 a \left (2 a^2+3 b^2-3 c^2\right ) x+9 c \left (4 a^2+b^2-c^2\right ) \cosh (x)+18 a b c \cosh (2 x)+c \left (3 b^2+c^2\right ) \cosh (3 x)+9 b \left (4 a^2+b^2-c^2\right ) \sinh (x)+9 a \left (b^2+c^2\right ) \sinh (2 x)+b \left (b^2+3 c^2\right ) \sinh (3 x)\right ) \]

input
Integrate[(a + b*Cosh[x] + c*Sinh[x])^3,x]
 
output
(6*a*(2*a^2 + 3*b^2 - 3*c^2)*x + 9*c*(4*a^2 + b^2 - c^2)*Cosh[x] + 18*a*b* 
c*Cosh[2*x] + c*(3*b^2 + c^2)*Cosh[3*x] + 9*b*(4*a^2 + b^2 - c^2)*Sinh[x] 
+ 9*a*(b^2 + c^2)*Sinh[2*x] + b*(b^2 + 3*c^2)*Sinh[3*x])/12
 
3.8.39.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 3599, 3042, 3625, 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.

\(\displaystyle \int (a+b \cosh (x)+c \sinh (x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \cos (i x)-i c \sin (i x))^3dx\)

\(\Big \downarrow \) 3599

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3625

\(\displaystyle \frac {1}{3} \left (\frac {\int \left (3 \left (2 a^2+3 b^2-3 c^2\right ) a^2+b \left (11 a^2+4 b^2-4 c^2\right ) \cosh (x) a+c \left (11 a^2+4 b^2-4 c^2\right ) \sinh (x) a\right )dx}{2 a}+\frac {5}{2} (a b \sinh (x)+a c \cosh (x)) (a+b \cosh (x)+c \sinh (x))\right )+\frac {1}{3} (b \sinh (x)+c \cosh (x)) (a+b \cosh (x)+c \sinh (x))^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {3 a^2 x \left (2 a^2+3 b^2-3 c^2\right )+a b \sinh (x) \left (11 a^2+4 b^2-4 c^2\right )+a c \cosh (x) \left (11 a^2+4 b^2-4 c^2\right )}{2 a}+\frac {5}{2} (a b \sinh (x)+a c \cosh (x)) (a+b \cosh (x)+c \sinh (x))\right )+\frac {1}{3} (b \sinh (x)+c \cosh (x)) (a+b \cosh (x)+c \sinh (x))^2\)

input
Int[(a + b*Cosh[x] + c*Sinh[x])^3,x]
 
output
((c*Cosh[x] + b*Sinh[x])*(a + b*Cosh[x] + c*Sinh[x])^2)/3 + ((5*(a*c*Cosh[ 
x] + a*b*Sinh[x])*(a + b*Cosh[x] + c*Sinh[x]))/2 + (3*a^2*(2*a^2 + 3*b^2 - 
 3*c^2)*x + a*c*(11*a^2 + 4*b^2 - 4*c^2)*Cosh[x] + a*b*(11*a^2 + 4*b^2 - 4 
*c^2)*Sinh[x])/(2*a))/3
 

3.8.39.3.1 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 3599
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[1/n   Int[Simp[n*a^2 + ( 
n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x 
], x]*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, 
 c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]
 

rule 3625
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(n_.)*((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_) 
]), x_Symbol] :> Simp[(B*c - b*C - a*C*Cos[d + e*x] + a*B*Sin[d + e*x])*((a 
 + b*Cos[d + e*x] + c*Sin[d + e*x])^n/(a*e*(n + 1))), x] + Simp[1/(a*(n + 1 
))   Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1)*Simp[a*(b*B + c*C)*n 
 + a^2*A*(n + 1) + (n*(a^2*B - B*c^2 + b*c*C) + a*b*A*(n + 1))*Cos[d + e*x] 
 + (n*(b*B*c + a^2*C - b^2*C) + a*c*A*(n + 1))*Sin[d + e*x], x], x], x] /; 
FreeQ[{a, b, c, d, e, A, B, C}, x] && GtQ[n, 0] && NeQ[a^2 - b^2 - c^2, 0]
 
3.8.39.4 Maple [A] (verified)

Time = 4.82 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82

method result size
parts \(a^{3} x +c^{3} \left (-\frac {2}{3}+\frac {\sinh \left (x \right )^{2}}{3}\right ) \cosh \left (x \right )+c \,b^{2} \cosh \left (x \right )^{3}+3 a \,b^{2} \left (\frac {\cosh \left (x \right ) \sinh \left (x \right )}{2}+\frac {x}{2}\right )+\frac {b \left (c \sinh \left (x \right )+a \right )^{3}}{c}+b^{3} \left (\frac {2}{3}+\frac {\cosh \left (x \right )^{2}}{3}\right ) \sinh \left (x \right )+3 c \,a^{2} \cosh \left (x \right )+3 a \,c^{2} \left (\frac {\cosh \left (x \right ) \sinh \left (x \right )}{2}-\frac {x}{2}\right )\) \(97\)
default \(a^{3} x +3 \sinh \left (x \right ) a^{2} b +3 c \,a^{2} \cosh \left (x \right )+3 a \,b^{2} \left (\frac {\cosh \left (x \right ) \sinh \left (x \right )}{2}+\frac {x}{2}\right )+3 c a b \cosh \left (x \right )^{2}+3 a \,c^{2} \left (\frac {\cosh \left (x \right ) \sinh \left (x \right )}{2}-\frac {x}{2}\right )+b^{3} \left (\frac {2}{3}+\frac {\cosh \left (x \right )^{2}}{3}\right ) \sinh \left (x \right )+c \,b^{2} \cosh \left (x \right )^{3}+c^{2} b \sinh \left (x \right )^{3}+c^{3} \left (-\frac {2}{3}+\frac {\sinh \left (x \right )^{2}}{3}\right ) \cosh \left (x \right )\) \(110\)
risch \(\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{8}+a^{3} x -\frac {3 a \,c^{2} x}{2}+\frac {3 a \,b^{2} x}{2}+\frac {{\mathrm e}^{3 x} b^{3}}{24}+\frac {3 a^{2} b \,{\mathrm e}^{x}}{2}+\frac {3 b^{3} {\mathrm e}^{x}}{8}-\frac {3 \,{\mathrm e}^{-2 x} a \,b^{2}}{8}+\frac {{\mathrm e}^{3 x} c^{2} b}{8}+\frac {3 \,{\mathrm e}^{2 x} a \,c^{2}}{8}+\frac {3 \,{\mathrm e}^{x} c \,a^{2}}{2}+\frac {3 \,{\mathrm e}^{x} c \,b^{2}}{8}-\frac {3 \,{\mathrm e}^{x} c^{2} b}{8}-\frac {3 \,{\mathrm e}^{-x} a^{2} b}{2}+\frac {3 \,{\mathrm e}^{-x} c \,a^{2}}{2}+\frac {3 \,{\mathrm e}^{-x} c \,b^{2}}{8}+\frac {3 \,{\mathrm e}^{-x} c^{2} b}{8}-\frac {3 \,{\mathrm e}^{-2 x} a \,c^{2}}{8}+\frac {{\mathrm e}^{-3 x} c \,b^{2}}{8}-\frac {{\mathrm e}^{-3 x} c^{2} b}{8}+\frac {{\mathrm e}^{3 x} c \,b^{2}}{8}-\frac {3 \,{\mathrm e}^{x} c^{3}}{8}-\frac {3 \,{\mathrm e}^{-x} b^{3}}{8}-\frac {3 \,{\mathrm e}^{-x} c^{3}}{8}-\frac {{\mathrm e}^{-3 x} b^{3}}{24}+\frac {{\mathrm e}^{-3 x} c^{3}}{24}+\frac {{\mathrm e}^{3 x} c^{3}}{24}+\frac {3 \,{\mathrm e}^{2 x} c a b}{4}+\frac {3 \,{\mathrm e}^{-2 x} c a b}{4}\) \(259\)

input
int((a+b*cosh(x)+c*sinh(x))^3,x,method=_RETURNVERBOSE)
 
output
a^3*x+c^3*(-2/3+1/3*sinh(x)^2)*cosh(x)+c*b^2*cosh(x)^3+3*a*b^2*(1/2*cosh(x 
)*sinh(x)+1/2*x)+b*(c*sinh(x)+a)^3/c+b^3*(2/3+1/3*cosh(x)^2)*sinh(x)+3*c*a 
^2*cosh(x)+3*a*c^2*(1/2*cosh(x)*sinh(x)-1/2*x)
 
3.8.39.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.34 \[ \int (a+b \cosh (x)+c \sinh (x))^3 \, dx=\frac {3}{2} \, a b c \cosh \left (x\right )^{2} + \frac {1}{12} \, {\left (3 \, b^{2} c + c^{3}\right )} \cosh \left (x\right )^{3} + \frac {1}{12} \, {\left (b^{3} + 3 \, b c^{2}\right )} \sinh \left (x\right )^{3} + \frac {1}{4} \, {\left (6 \, a b c + {\left (3 \, b^{2} c + c^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2} - 3 \, a c^{2}\right )} x - \frac {3}{4} \, {\left (c^{3} - {\left (4 \, a^{2} + b^{2}\right )} c\right )} \cosh \left (x\right ) + \frac {1}{4} \, {\left (12 \, a^{2} b + 3 \, b^{3} - 3 \, b c^{2} + {\left (b^{3} + 3 \, b c^{2}\right )} \cosh \left (x\right )^{2} + 6 \, {\left (a b^{2} + a c^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) \]

input
integrate((a+b*cosh(x)+c*sinh(x))^3,x, algorithm="fricas")
 
output
3/2*a*b*c*cosh(x)^2 + 1/12*(3*b^2*c + c^3)*cosh(x)^3 + 1/12*(b^3 + 3*b*c^2 
)*sinh(x)^3 + 1/4*(6*a*b*c + (3*b^2*c + c^3)*cosh(x))*sinh(x)^2 + 1/2*(2*a 
^3 + 3*a*b^2 - 3*a*c^2)*x - 3/4*(c^3 - (4*a^2 + b^2)*c)*cosh(x) + 1/4*(12* 
a^2*b + 3*b^3 - 3*b*c^2 + (b^3 + 3*b*c^2)*cosh(x)^2 + 6*(a*b^2 + a*c^2)*co 
sh(x))*sinh(x)
 
3.8.39.6 Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.65 \[ \int (a+b \cosh (x)+c \sinh (x))^3 \, dx=a^{3} x + 3 a^{2} b \sinh {\left (x \right )} + 3 a^{2} c \cosh {\left (x \right )} - \frac {3 a b^{2} x \sinh ^{2}{\left (x \right )}}{2} + \frac {3 a b^{2} x \cosh ^{2}{\left (x \right )}}{2} + \frac {3 a b^{2} \sinh {\left (x \right )} \cosh {\left (x \right )}}{2} + 3 a b c \cosh ^{2}{\left (x \right )} + \frac {3 a c^{2} x \sinh ^{2}{\left (x \right )}}{2} - \frac {3 a c^{2} x \cosh ^{2}{\left (x \right )}}{2} + \frac {3 a c^{2} \sinh {\left (x \right )} \cosh {\left (x \right )}}{2} - \frac {2 b^{3} \sinh ^{3}{\left (x \right )}}{3} + b^{3} \sinh {\left (x \right )} \cosh ^{2}{\left (x \right )} + b^{2} c \cosh ^{3}{\left (x \right )} + b c^{2} \sinh ^{3}{\left (x \right )} + c^{3} \sinh ^{2}{\left (x \right )} \cosh {\left (x \right )} - \frac {2 c^{3} \cosh ^{3}{\left (x \right )}}{3} \]

input
integrate((a+b*cosh(x)+c*sinh(x))**3,x)
 
output
a**3*x + 3*a**2*b*sinh(x) + 3*a**2*c*cosh(x) - 3*a*b**2*x*sinh(x)**2/2 + 3 
*a*b**2*x*cosh(x)**2/2 + 3*a*b**2*sinh(x)*cosh(x)/2 + 3*a*b*c*cosh(x)**2 + 
 3*a*c**2*x*sinh(x)**2/2 - 3*a*c**2*x*cosh(x)**2/2 + 3*a*c**2*sinh(x)*cosh 
(x)/2 - 2*b**3*sinh(x)**3/3 + b**3*sinh(x)*cosh(x)**2 + b**2*c*cosh(x)**3 
+ b*c**2*sinh(x)**3 + c**3*sinh(x)**2*cosh(x) - 2*c**3*cosh(x)**3/3
 
3.8.39.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15 \[ \int (a+b \cosh (x)+c \sinh (x))^3 \, dx=b^{2} c \cosh \left (x\right )^{3} + b c^{2} \sinh \left (x\right )^{3} + a^{3} x + \frac {1}{24} \, c^{3} {\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} + e^{\left (-3 \, x\right )} - 9 \, e^{x}\right )} + \frac {1}{24} \, b^{3} {\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} - e^{\left (-3 \, x\right )} + 9 \, e^{x}\right )} + 3 \, {\left (c \cosh \left (x\right ) + b \sinh \left (x\right )\right )} a^{2} + \frac {3}{8} \, {\left (8 \, b c \cosh \left (x\right )^{2} + b^{2} {\left (4 \, x + e^{\left (2 \, x\right )} - e^{\left (-2 \, x\right )}\right )} - c^{2} {\left (4 \, x - e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}\right )}\right )} a \]

input
integrate((a+b*cosh(x)+c*sinh(x))^3,x, algorithm="maxima")
 
output
b^2*c*cosh(x)^3 + b*c^2*sinh(x)^3 + a^3*x + 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) + 3* 
(c*cosh(x) + b*sinh(x))*a^2 + 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)))*a
 
3.8.39.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (109) = 218\).

Time = 0.25 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.84 \[ \int (a+b \cosh (x)+c \sinh (x))^3 \, dx=\frac {1}{24} \, b^{3} e^{\left (3 \, x\right )} + \frac {1}{8} \, b^{2} c e^{\left (3 \, x\right )} + \frac {1}{8} \, b c^{2} e^{\left (3 \, x\right )} + \frac {1}{24} \, c^{3} e^{\left (3 \, x\right )} + \frac {3}{8} \, a b^{2} e^{\left (2 \, x\right )} + \frac {3}{4} \, a b c e^{\left (2 \, x\right )} + \frac {3}{8} \, a c^{2} e^{\left (2 \, x\right )} + \frac {3}{2} \, a^{2} b e^{x} + \frac {3}{8} \, b^{3} e^{x} + \frac {3}{2} \, a^{2} c e^{x} + \frac {3}{8} \, b^{2} c e^{x} - \frac {3}{8} \, b c^{2} e^{x} - \frac {3}{8} \, c^{3} e^{x} + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2} - 3 \, a c^{2}\right )} x - \frac {1}{24} \, {\left (b^{3} - 3 \, b^{2} c + 3 \, b c^{2} - c^{3} + 9 \, {\left (4 \, a^{2} b + b^{3} - 4 \, a^{2} c - b^{2} c - b c^{2} + c^{3}\right )} e^{\left (2 \, x\right )} + 9 \, {\left (a b^{2} - 2 \, a b c + a c^{2}\right )} e^{x}\right )} e^{\left (-3 \, x\right )} \]

input
integrate((a+b*cosh(x)+c*sinh(x))^3,x, algorithm="giac")
 
output
1/24*b^3*e^(3*x) + 1/8*b^2*c*e^(3*x) + 1/8*b*c^2*e^(3*x) + 1/24*c^3*e^(3*x 
) + 3/8*a*b^2*e^(2*x) + 3/4*a*b*c*e^(2*x) + 3/8*a*c^2*e^(2*x) + 3/2*a^2*b* 
e^x + 3/8*b^3*e^x + 3/2*a^2*c*e^x + 3/8*b^2*c*e^x - 3/8*b*c^2*e^x - 3/8*c^ 
3*e^x + 1/2*(2*a^3 + 3*a*b^2 - 3*a*c^2)*x - 1/24*(b^3 - 3*b^2*c + 3*b*c^2 
- c^3 + 9*(4*a^2*b + b^3 - 4*a^2*c - b^2*c - b*c^2 + c^3)*e^(2*x) + 9*(a*b 
^2 - 2*a*b*c + a*c^2)*e^x)*e^(-3*x)
 
3.8.39.9 Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.10 \[ \int (a+b \cosh (x)+c \sinh (x))^3 \, dx=a^3\,x+{\mathrm {cosh}\left (x\right )}^3\,\left (b^2\,c-\frac {2\,c^3}{3}\right )+{\mathrm {sinh}\left (x\right )}^3\,\left (b\,c^2-\frac {2\,b^3}{3}\right )+b^3\,{\mathrm {cosh}\left (x\right )}^2\,\mathrm {sinh}\left (x\right )+c^3\,\mathrm {cosh}\left (x\right )\,{\mathrm {sinh}\left (x\right )}^2+3\,a^2\,c\,\mathrm {cosh}\left (x\right )+3\,a^2\,b\,\mathrm {sinh}\left (x\right )+\frac {3\,a\,\mathrm {cosh}\left (x\right )\,\mathrm {sinh}\left (x\right )\,\left (b^2+c^2\right )}{2}+\frac {3\,a\,x\,{\mathrm {cosh}\left (x\right )}^2\,\left (b^2-c^2\right )}{2}+3\,a\,b\,c\,{\mathrm {cosh}\left (x\right )}^2-\frac {3\,a\,x\,{\mathrm {sinh}\left (x\right )}^2\,\left (b^2-c^2\right )}{2} \]

input
int((a + b*cosh(x) + c*sinh(x))^3,x)
 
output
a^3*x + cosh(x)^3*(b^2*c - (2*c^3)/3) + sinh(x)^3*(b*c^2 - (2*b^3)/3) + b^ 
3*cosh(x)^2*sinh(x) + c^3*cosh(x)*sinh(x)^2 + 3*a^2*c*cosh(x) + 3*a^2*b*si 
nh(x) + (3*a*cosh(x)*sinh(x)*(b^2 + c^2))/2 + (3*a*x*cosh(x)^2*(b^2 - c^2) 
)/2 + 3*a*b*c*cosh(x)^2 - (3*a*x*sinh(x)^2*(b^2 - c^2))/2