Integrand size = 12, antiderivative size = 90 \[ \int (a+b \cosh (c+d x))^3 \, dx=\frac {1}{2} a \left (2 a^2+3 b^2\right ) x+\frac {2 b \left (4 a^2+b^2\right ) \sinh (c+d x)}{3 d}+\frac {5 a b^2 \cosh (c+d x) \sinh (c+d x)}{6 d}+\frac {b (a+b \cosh (c+d x))^2 \sinh (c+d x)}{3 d} \] Output:
1/2*a*(2*a^2+3*b^2)*x+2/3*b*(4*a^2+b^2)*sinh(d*x+c)/d+5/6*a*b^2*cosh(d*x+c )*sinh(d*x+c)/d+1/3*b*(a+b*cosh(d*x+c))^2*sinh(d*x+c)/d
Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int (a+b \cosh (c+d x))^3 \, dx=\frac {12 a^3 c+18 a b^2 c+12 a^3 d x+18 a b^2 d x+9 b \left (4 a^2+b^2\right ) \sinh (c+d x)+9 a b^2 \sinh (2 (c+d x))+b^3 \sinh (3 (c+d x))}{12 d} \] Input:
Integrate[(a + b*Cosh[c + d*x])^3,x]
Output:
(12*a^3*c + 18*a*b^2*c + 12*a^3*d*x + 18*a*b^2*d*x + 9*b*(4*a^2 + b^2)*Sin h[c + d*x] + 9*a*b^2*Sinh[2*(c + d*x)] + b^3*Sinh[3*(c + d*x)])/(12*d)
Time = 0.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3135, 3042, 3213}
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.
\(\displaystyle \int (a+b \cosh (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^3dx\) |
\(\Big \downarrow \) 3135 |
\(\displaystyle \frac {1}{3} \int (a+b \cosh (c+d x)) \left (3 a^2+5 b \cosh (c+d x) a+2 b^2\right )dx+\frac {b \sinh (c+d x) (a+b \cosh (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sinh (c+d x) (a+b \cosh (c+d x))^2}{3 d}+\frac {1}{3} \int \left (a+b \sin \left (i c+i d x+\frac {\pi }{2}\right )\right ) \left (3 a^2+5 b \sin \left (i c+i d x+\frac {\pi }{2}\right ) a+2 b^2\right )dx\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {1}{3} \left (\frac {2 b \left (4 a^2+b^2\right ) \sinh (c+d x)}{d}+\frac {3}{2} a x \left (2 a^2+3 b^2\right )+\frac {5 a b^2 \sinh (c+d x) \cosh (c+d x)}{2 d}\right )+\frac {b \sinh (c+d x) (a+b \cosh (c+d x))^2}{3 d}\) |
Input:
Int[(a + b*Cosh[c + d*x])^3,x]
Output:
(b*(a + b*Cosh[c + d*x])^2*Sinh[c + d*x])/(3*d) + ((3*a*(2*a^2 + 3*b^2)*x) /2 + (2*b*(4*a^2 + b^2)*Sinh[c + d*x])/d + (5*a*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(2*d))/3
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[1/n Int[(a + b* Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*x] , x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 229.86 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74
method | result | size |
parallelrisch | \(\frac {9 b^{2} a \sinh \left (2 d x +2 c \right )+b^{3} \sinh \left (3 d x +3 c \right )+9 \left (4 a^{2} b +b^{3}\right ) \sinh \left (d x +c \right )+12 \left (a^{2}+\frac {3 b^{2}}{2}\right ) a d x}{12 d}\) | \(67\) |
derivativedivides | \(\frac {b^{3} \left (\frac {2}{3}+\frac {\cosh \left (d x +c \right )^{2}}{3}\right ) \sinh \left (d x +c \right )+3 b^{2} a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \sinh \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\) | \(77\) |
default | \(\frac {b^{3} \left (\frac {2}{3}+\frac {\cosh \left (d x +c \right )^{2}}{3}\right ) \sinh \left (d x +c \right )+3 b^{2} a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \sinh \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\) | \(77\) |
parts | \(a^{3} x +\frac {b^{3} \left (\frac {2}{3}+\frac {\cosh \left (d x +c \right )^{2}}{3}\right ) \sinh \left (d x +c \right )}{d}+\frac {3 a^{2} b \sinh \left (d x +c \right )}{d}+\frac {3 b^{2} a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(78\) |
risch | \(a^{3} x +\frac {3 a \,b^{2} x}{2}+\frac {b^{3} {\mathrm e}^{3 d x +3 c}}{24 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} b^{2} a}{8 d}+\frac {3 b \,{\mathrm e}^{d x +c} a^{2}}{2 d}+\frac {3 \,{\mathrm e}^{d x +c} b^{3}}{8 d}-\frac {3 b \,{\mathrm e}^{-d x -c} a^{2}}{2 d}-\frac {3 b^{3} {\mathrm e}^{-d x -c}}{8 d}-\frac {3 \,{\mathrm e}^{-2 d x -2 c} b^{2} a}{8 d}-\frac {b^{3} {\mathrm e}^{-3 d x -3 c}}{24 d}\) | \(148\) |
Input:
int((a+b*cosh(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/12*(9*b^2*a*sinh(2*d*x+2*c)+b^3*sinh(3*d*x+3*c)+9*(4*a^2*b+b^3)*sinh(d*x +c)+12*(a^2+3/2*b^2)*a*d*x)/d
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87 \[ \int (a+b \cosh (c+d x))^3 \, dx=\frac {b^{3} \sinh \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} d x + 3 \, {\left (b^{3} \cosh \left (d x + c\right )^{2} + 6 \, a b^{2} \cosh \left (d x + c\right ) + 12 \, a^{2} b + 3 \, b^{3}\right )} \sinh \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+b*cosh(d*x+c))^3,x, algorithm="fricas")
Output:
1/12*(b^3*sinh(d*x + c)^3 + 6*(2*a^3 + 3*a*b^2)*d*x + 3*(b^3*cosh(d*x + c) ^2 + 6*a*b^2*cosh(d*x + c) + 12*a^2*b + 3*b^3)*sinh(d*x + c))/d
Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.42 \[ \int (a+b \cosh (c+d x))^3 \, dx=\begin {cases} a^{3} x + \frac {3 a^{2} b \sinh {\left (c + d x \right )}}{d} - \frac {3 a b^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} - \frac {2 b^{3} \sinh ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{3} \sinh {\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cosh {\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*cosh(d*x+c))**3,x)
Output:
Piecewise((a**3*x + 3*a**2*b*sinh(c + d*x)/d - 3*a*b**2*x*sinh(c + d*x)**2 /2 + 3*a*b**2*x*cosh(c + d*x)**2/2 + 3*a*b**2*sinh(c + d*x)*cosh(c + d*x)/ (2*d) - 2*b**3*sinh(c + d*x)**3/(3*d) + b**3*sinh(c + d*x)*cosh(c + d*x)** 2/d, Ne(d, 0)), (x*(a + b*cosh(c))**3, True))
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.29 \[ \int (a+b \cosh (c+d x))^3 \, dx=\frac {3}{8} \, a b^{2} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + a^{3} x + \frac {1}{24} \, b^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} - \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac {3 \, a^{2} b \sinh \left (d x + c\right )}{d} \] Input:
integrate((a+b*cosh(d*x+c))^3,x, algorithm="maxima")
Output:
3/8*a*b^2*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) + a^3*x + 1/24*b^ 3*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d - e^(-3*d*x - 3* c)/d) + 3*a^2*b*sinh(d*x + c)/d
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.46 \[ \int (a+b \cosh (c+d x))^3 \, dx=\frac {b^{3} e^{\left (3 \, d x + 3 \, c\right )}}{24 \, d} + \frac {3 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} - \frac {3 \, a b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac {b^{3} e^{\left (-3 \, d x - 3 \, c\right )}}{24 \, d} + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} x + \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} e^{\left (d x + c\right )}}{8 \, d} - \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} e^{\left (-d x - c\right )}}{8 \, d} \] Input:
integrate((a+b*cosh(d*x+c))^3,x, algorithm="giac")
Output:
1/24*b^3*e^(3*d*x + 3*c)/d + 3/8*a*b^2*e^(2*d*x + 2*c)/d - 3/8*a*b^2*e^(-2 *d*x - 2*c)/d - 1/24*b^3*e^(-3*d*x - 3*c)/d + 1/2*(2*a^3 + 3*a*b^2)*x + 3/ 8*(4*a^2*b + b^3)*e^(d*x + c)/d - 3/8*(4*a^2*b + b^3)*e^(-d*x - c)/d
Time = 2.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.81 \[ \int (a+b \cosh (c+d x))^3 \, dx=\frac {\frac {9\,b^3\,\mathrm {sinh}\left (c+d\,x\right )}{2}+\frac {b^3\,\mathrm {sinh}\left (3\,c+3\,d\,x\right )}{2}+\frac {9\,a\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{2}+18\,a^2\,b\,\mathrm {sinh}\left (c+d\,x\right )+6\,a^3\,d\,x+9\,a\,b^2\,d\,x}{6\,d} \] Input:
int((a + b*cosh(c + d*x))^3,x)
Output:
((9*b^3*sinh(c + d*x))/2 + (b^3*sinh(3*c + 3*d*x))/2 + (9*a*b^2*sinh(2*c + 2*d*x))/2 + 18*a^2*b*sinh(c + d*x) + 6*a^3*d*x + 9*a*b^2*d*x)/(6*d)
Time = 0.21 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.81 \[ \int (a+b \cosh (c+d x))^3 \, dx=\frac {e^{6 d x +6 c} b^{3}+9 e^{5 d x +5 c} a \,b^{2}+36 e^{4 d x +4 c} a^{2} b +9 e^{4 d x +4 c} b^{3}+24 e^{3 d x +3 c} a^{3} d x +36 e^{3 d x +3 c} a \,b^{2} d x -36 e^{2 d x +2 c} a^{2} b -9 e^{2 d x +2 c} b^{3}-9 e^{d x +c} a \,b^{2}-b^{3}}{24 e^{3 d x +3 c} d} \] Input:
int((a+b*cosh(d*x+c))^3,x)
Output:
(e**(6*c + 6*d*x)*b**3 + 9*e**(5*c + 5*d*x)*a*b**2 + 36*e**(4*c + 4*d*x)*a **2*b + 9*e**(4*c + 4*d*x)*b**3 + 24*e**(3*c + 3*d*x)*a**3*d*x + 36*e**(3* c + 3*d*x)*a*b**2*d*x - 36*e**(2*c + 2*d*x)*a**2*b - 9*e**(2*c + 2*d*x)*b* *3 - 9*e**(c + d*x)*a*b**2 - b**3)/(24*e**(3*c + 3*d*x)*d)