Integrand size = 12, antiderivative size = 76 \[ \int (a+b \cos (c+d x))^3 \, dx=a^3 x+\frac {3}{2} a b^2 x+\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)}{d}+\frac {3 a b^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {b^3 \sin ^3(c+d x)}{3 d} \] Output:
a^3*x+3/2*a*b^2*x+b*(3*a^2+b^2)*sin(d*x+c)/d+3/2*a*b^2*cos(d*x+c)*sin(d*x+ c)/d-1/3*b^3*sin(d*x+c)^3/d
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^3 \, dx=\frac {12 a^3 c+18 a b^2 c+12 a^3 d x+18 a b^2 d x+9 b \left (4 a^2+b^2\right ) \sin (c+d x)+9 a b^2 \sin (2 (c+d x))+b^3 \sin (3 (c+d x))}{12 d} \] Input:
Integrate[(a + b*Cos[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 [c + d*x] + 9*a*b^2*Sin[2*(c + d*x)] + b^3*Sin[3*(c + d*x)])/(12*d)
Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.22, 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.
\(\displaystyle \int (a+b \cos (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3dx\) |
\(\Big \downarrow \) 3135 |
\(\displaystyle \frac {1}{3} \int (a+b \cos (c+d x)) \left (3 a^2+5 b \cos (c+d x) a+2 b^2\right )dx+\frac {b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 a^2+5 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 b^2\right )dx+\frac {b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {1}{3} \left (\frac {2 b \left (4 a^2+b^2\right ) \sin (c+d x)}{d}+\frac {3}{2} a x \left (2 a^2+3 b^2\right )+\frac {5 a b^2 \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^3,x]
Output:
(b*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + ((3*a*(2*a^2 + 3*b^2)*x)/2 + (2*b*(4*a^2 + b^2)*Sin[c + d*x])/d + (5*a*b^2*Cos[c + d*x]*Sin[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 = 139.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {9 \sin \left (2 d x +2 c \right ) a \,b^{2}+\sin \left (3 d x +3 c \right ) b^{3}+9 \left (4 a^{2} b +b^{3}\right ) \sin \left (d x +c \right )+12 a d x \left (a^{2}+\frac {3 b^{2}}{2}\right )}{12 d}\) | \(67\) |
derivativedivides | \(\frac {\frac {b^{3} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+3 b^{2} a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 \sin \left (d x +c \right ) a^{2} b +a^{3} \left (d x +c \right )}{d}\) | \(76\) |
default | \(\frac {\frac {b^{3} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+3 b^{2} a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 \sin \left (d x +c \right ) a^{2} b +a^{3} \left (d x +c \right )}{d}\) | \(76\) |
parts | \(a^{3} x +\frac {b^{3} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}+\frac {3 b^{2} a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {3 \sin \left (d x +c \right ) a^{2} b}{d}\) | \(77\) |
risch | \(a^{3} x +\frac {3 a \,b^{2} x}{2}+\frac {3 \sin \left (d x +c \right ) a^{2} b}{d}+\frac {3 \sin \left (d x +c \right ) b^{3}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) b^{3}}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) b^{2} a}{4 d}\) | \(78\) |
norman | \(\frac {\left (a^{3}+\frac {3}{2} b^{2} a \right ) x +\left (a^{3}+\frac {3}{2} b^{2} a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 a^{3}+\frac {9}{2} b^{2} a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (3 a^{3}+\frac {9}{2} b^{2} a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {b \left (6 a^{2}-3 a b +2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {b \left (6 a^{2}+3 a b +2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 b \left (9 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(189\) |
Input:
int((a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
Output:
1/12*(9*sin(2*d*x+2*c)*a*b^2+sin(3*d*x+3*c)*b^3+9*(4*a^2*b+b^3)*sin(d*x+c) +12*a*d*x*(a^2+3/2*b^2))/d
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int (a+b \cos (c+d x))^3 \, dx=\frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} d x + {\left (2 \, b^{3} \cos \left (d x + c\right )^{2} + 9 \, a b^{2} \cos \left (d x + c\right ) + 18 \, a^{2} b + 4 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3,x, algorithm="fricas")
Output:
1/6*(3*(2*a^3 + 3*a*b^2)*d*x + (2*b^3*cos(d*x + c)^2 + 9*a*b^2*cos(d*x + c ) + 18*a^2*b + 4*b^3)*sin(d*x + c))/d
Time = 0.11 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.68 \[ \int (a+b \cos (c+d x))^3 \, dx=\begin {cases} a^{3} x + \frac {3 a^{2} b \sin {\left (c + d x \right )}}{d} + \frac {3 a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*cos(d*x+c))**3,x)
Output:
Piecewise((a**3*x + 3*a**2*b*sin(c + d*x)/d + 3*a*b**2*x*sin(c + d*x)**2/2 + 3*a*b**2*x*cos(c + d*x)**2/2 + 3*a*b**2*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*b**3*sin(c + d*x)**3/(3*d) + b**3*sin(c + d*x)*cos(c + d*x)**2/d, Ne( d, 0)), (x*(a + b*cos(c))**3, True))
Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95 \[ \int (a+b \cos (c+d x))^3 \, dx=a^{3} x + \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2}}{4 \, d} - \frac {{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{3}}{3 \, d} + \frac {3 \, a^{2} b \sin \left (d x + c\right )}{d} \] Input:
integrate((a+b*cos(d*x+c))^3,x, algorithm="maxima")
Output:
a^3*x + 3/4*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b^2/d - 1/3*(sin(d*x + c)^3 - 3*sin(d*x + c))*b^3/d + 3*a^2*b*sin(d*x + c)/d
Time = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95 \[ \int (a+b \cos (c+d x))^3 \, dx=\frac {b^{3} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {3 \, a b^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} x + \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3,x, algorithm="giac")
Output:
1/12*b^3*sin(3*d*x + 3*c)/d + 3/4*a*b^2*sin(2*d*x + 2*c)/d + 1/2*(2*a^3 + 3*a*b^2)*x + 3/4*(4*a^2*b + b^3)*sin(d*x + c)/d
Time = 40.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^3 \, dx=a^3\,x+\frac {3\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {3\,a\,b^2\,x}{2}+\frac {3\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,a^2\,b\,\sin \left (c+d\,x\right )}{d} \] Input:
int((a + b*cos(c + d*x))^3,x)
Output:
a^3*x + (3*b^3*sin(c + d*x))/(4*d) + (b^3*sin(3*c + 3*d*x))/(12*d) + (3*a* b^2*x)/2 + (3*a*b^2*sin(2*c + 2*d*x))/(4*d) + (3*a^2*b*sin(c + d*x))/d
Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int (a+b \cos (c+d x))^3 \, dx=\frac {9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}-2 \sin \left (d x +c \right )^{3} b^{3}+18 \sin \left (d x +c \right ) a^{2} b +6 \sin \left (d x +c \right ) b^{3}+6 a^{3} d x +9 a \,b^{2} d x}{6 d} \] Input:
int((a+b*cos(d*x+c))^3,x)
Output:
(9*cos(c + d*x)*sin(c + d*x)*a*b**2 - 2*sin(c + d*x)**3*b**3 + 18*sin(c + d*x)*a**2*b + 6*sin(c + d*x)*b**3 + 6*a**3*d*x + 9*a*b**2*d*x)/(6*d)