Integrand size = 12, antiderivative size = 137 \[ \int (a+b \cos (c+d x))^4 \, dx=\frac {1}{8} \left (8 a^4+24 a^2 b^2+3 b^4\right ) x+\frac {a b \left (19 a^2+16 b^2\right ) \sin (c+d x)}{6 d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {7 a b (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {b (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d} \] Output:
1/8*(8*a^4+24*a^2*b^2+3*b^4)*x+1/6*a*b*(19*a^2+16*b^2)*sin(d*x+c)/d+1/24*b ^2*(26*a^2+9*b^2)*cos(d*x+c)*sin(d*x+c)/d+7/12*a*b*(a+b*cos(d*x+c))^2*sin( d*x+c)/d+1/4*b*(a+b*cos(d*x+c))^3*sin(d*x+c)/d
Time = 0.59 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int (a+b \cos (c+d x))^4 \, dx=\frac {12 \left (8 a^4+24 a^2 b^2+3 b^4\right ) (c+d x)+96 a b \left (4 a^2+3 b^2\right ) \sin (c+d x)+24 b^2 \left (6 a^2+b^2\right ) \sin (2 (c+d x))+32 a b^3 \sin (3 (c+d x))+3 b^4 \sin (4 (c+d x))}{96 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^4,x]
Output:
(12*(8*a^4 + 24*a^2*b^2 + 3*b^4)*(c + d*x) + 96*a*b*(4*a^2 + 3*b^2)*Sin[c + d*x] + 24*b^2*(6*a^2 + b^2)*Sin[2*(c + d*x)] + 32*a*b^3*Sin[3*(c + d*x)] + 3*b^4*Sin[4*(c + d*x)])/(96*d)
Time = 0.50 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3135, 3042, 3232, 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.
| \(\displaystyle \int (a+b \cos (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4dx\) |
| \(\Big \downarrow \) 3135 |
| \(\displaystyle \frac {1}{4} \int (a+b \cos (c+d x))^2 \left (4 a^2+7 b \cos (c+d x) a+3 b^2\right )dx+\frac {b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (4 a^2+7 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 b^2\right )dx+\frac {b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int (a+b \cos (c+d x)) \left (a \left (12 a^2+23 b^2\right )+b \left (26 a^2+9 b^2\right ) \cos (c+d x)\right )dx+\frac {7 a b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a \left (12 a^2+23 b^2\right )+b \left (26 a^2+9 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {7 a b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {2 a b \left (19 a^2+16 b^2\right ) \sin (c+d x)}{d}+\frac {b^2 \left (26 a^2+9 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} x \left (8 a^4+24 a^2 b^2+3 b^4\right )\right )+\frac {7 a b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^4,x]
Output:
(b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + ((7*a*b*(a + b*Cos[c + d*x ])^2*Sin[c + d*x])/(3*d) + ((3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*x)/2 + (2*a*b* (19*a^2 + 16*b^2)*Sin[c + d*x])/d + (b^2*(26*a^2 + 9*b^2)*Cos[c + d*x]*Sin [c + d*x])/(2*d))/3)/4
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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[1/(m + 1) Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ [{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]
Time = 15.83 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.74
| method | result | size |
| parallelrisch | \(\frac {24 \left (6 a^{2} b^{2}+b^{4}\right ) \sin \left (2 d x +2 c \right )+32 \sin \left (3 d x +3 c \right ) a \,b^{3}+3 \sin \left (4 d x +4 c \right ) b^{4}+96 \left (4 a^{3} b +3 a \,b^{3}\right ) \sin \left (d x +c \right )+96 \left (a^{4}+3 a^{2} b^{2}+\frac {3}{8} b^{4}\right ) d x}{96 d}\) | \(101\) |
| derivativedivides | \(\frac {b^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a \,b^{3} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+6 a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 \sin \left (d x +c \right ) a^{3} b +a^{4} \left (d x +c \right )}{d}\) | \(116\) |
| default | \(\frac {b^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a \,b^{3} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+6 a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 \sin \left (d x +c \right ) a^{3} b +a^{4} \left (d x +c \right )}{d}\) | \(116\) |
| parts | \(a^{4} x +\frac {b^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {4 \sin \left (d x +c \right ) a^{3} b}{d}+\frac {6 a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a \,b^{3} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}\) | \(120\) |
| risch | \(a^{4} x +3 a^{2} b^{2} x +\frac {3 b^{4} x}{8}+\frac {4 \sin \left (d x +c \right ) a^{3} b}{d}+\frac {3 \sin \left (d x +c \right ) a \,b^{3}}{d}+\frac {\sin \left (4 d x +4 c \right ) b^{4}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) a \,b^{3}}{3 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) b^{4}}{4 d}\) | \(124\) |
| norman | \(\frac {\left (a^{4}+3 a^{2} b^{2}+\frac {3}{8} b^{4}\right ) x +\left (a^{4}+3 a^{2} b^{2}+\frac {3}{8} b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (4 a^{4}+12 a^{2} b^{2}+\frac {3}{2} b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (4 a^{4}+12 a^{2} b^{2}+\frac {3}{2} b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (6 a^{4}+18 a^{2} b^{2}+\frac {9}{4} b^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {b \left (32 a^{3}-24 a^{2} b +32 b^{2} a -5 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {b \left (32 a^{3}+24 a^{2} b +32 b^{2} a +5 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b \left (288 a^{3}-72 a^{2} b +160 b^{2} a +9 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {b \left (288 a^{3}+72 a^{2} b +160 b^{2} a -9 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(321\) |
Input:
int((a+cos(d*x+c)*b)^4,x,method=_RETURNVERBOSE)
Output:
1/96*(24*(6*a^2*b^2+b^4)*sin(2*d*x+2*c)+32*sin(3*d*x+3*c)*a*b^3+3*sin(4*d* x+4*c)*b^4+96*(4*a^3*b+3*a*b^3)*sin(d*x+c)+96*(a^4+3*a^2*b^2+3/8*b^4)*d*x) /d
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.70 \[ \int (a+b \cos (c+d x))^4 \, dx=\frac {3 \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} d x + {\left (6 \, b^{4} \cos \left (d x + c\right )^{3} + 32 \, a b^{3} \cos \left (d x + c\right )^{2} + 96 \, a^{3} b + 64 \, a b^{3} + 9 \, {\left (8 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4,x, algorithm="fricas")
Output:
1/24*(3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*d*x + (6*b^4*cos(d*x + c)^3 + 32*a*b^ 3*cos(d*x + c)^2 + 96*a^3*b + 64*a*b^3 + 9*(8*a^2*b^2 + b^4)*cos(d*x + c)) *sin(d*x + c))/d
Time = 0.18 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.75 \[ \int (a+b \cos (c+d x))^4 \, dx=\begin {cases} a^{4} x + \frac {4 a^{3} b \sin {\left (c + d x \right )}}{d} + 3 a^{2} b^{2} x \sin ^{2}{\left (c + d x \right )} + 3 a^{2} b^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {3 a^{2} b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {8 a b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {4 a b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 b^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 b^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{4} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*cos(d*x+c))**4,x)
Output:
Piecewise((a**4*x + 4*a**3*b*sin(c + d*x)/d + 3*a**2*b**2*x*sin(c + d*x)** 2 + 3*a**2*b**2*x*cos(c + d*x)**2 + 3*a**2*b**2*sin(c + d*x)*cos(c + d*x)/ d + 8*a*b**3*sin(c + d*x)**3/(3*d) + 4*a*b**3*sin(c + d*x)*cos(c + d*x)**2 /d + 3*b**4*x*sin(c + d*x)**4/8 + 3*b**4*x*sin(c + d*x)**2*cos(c + d*x)**2 /4 + 3*b**4*x*cos(c + d*x)**4/8 + 3*b**4*sin(c + d*x)**3*cos(c + d*x)/(8*d ) + 5*b**4*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a + b*cos(c) )**4, True))
Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.81 \[ \int (a+b \cos (c+d x))^4 \, dx=a^{4} x + \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b^{2}}{2 \, d} - \frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3}}{3 \, d} + \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4}}{32 \, d} + \frac {4 \, a^{3} b \sin \left (d x + c\right )}{d} \] Input:
integrate((a+b*cos(d*x+c))^4,x, algorithm="maxima")
Output:
a^4*x + 3/2*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2*b^2/d - 4/3*(sin(d*x + c) ^3 - 3*sin(d*x + c))*a*b^3/d + 1/32*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8* sin(2*d*x + 2*c))*b^4/d + 4*a^3*b*sin(d*x + c)/d
Time = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.78 \[ \int (a+b \cos (c+d x))^4 \, dx=\frac {b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {a b^{3} \sin \left (3 \, d x + 3 \, c\right )}{3 \, d} + \frac {1}{8} \, {\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} x + \frac {{\left (6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} \sin \left (d x + c\right )}{d} \] Input:
integrate((a+b*cos(d*x+c))^4,x, algorithm="giac")
Output:
1/32*b^4*sin(4*d*x + 4*c)/d + 1/3*a*b^3*sin(3*d*x + 3*c)/d + 1/8*(8*a^4 + 24*a^2*b^2 + 3*b^4)*x + 1/4*(6*a^2*b^2 + b^4)*sin(2*d*x + 2*c)/d + (4*a^3* b + 3*a*b^3)*sin(d*x + c)/d
Time = 40.64 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.90 \[ \int (a+b \cos (c+d x))^4 \, dx=a^4\,x+\frac {3\,b^4\,x}{8}+3\,a^2\,b^2\,x+\frac {b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {b^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {3\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \] Input:
int((a + b*cos(c + d*x))^4,x)
Output:
a^4*x + (3*b^4*x)/8 + 3*a^2*b^2*x + (b^4*sin(2*c + 2*d*x))/(4*d) + (b^4*si n(4*c + 4*d*x))/(32*d) + (a*b^3*sin(3*c + 3*d*x))/(3*d) + (3*a^2*b^2*sin(2 *c + 2*d*x))/(2*d) + (3*a*b^3*sin(c + d*x))/d + (4*a^3*b*sin(c + d*x))/d
Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int (a+b \cos (c+d x))^4 \, dx=\frac {-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{4}+72 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}-32 \sin \left (d x +c \right )^{3} a \,b^{3}+96 \sin \left (d x +c \right ) a^{3} b +96 \sin \left (d x +c \right ) a \,b^{3}+24 a^{4} d x +72 a^{2} b^{2} d x +9 b^{4} d x}{24 d} \] Input:
int((a+b*cos(d*x+c))^4,x)
Output:
( - 6*cos(c + d*x)*sin(c + d*x)**3*b**4 + 72*cos(c + d*x)*sin(c + d*x)*a** 2*b**2 + 15*cos(c + d*x)*sin(c + d*x)*b**4 - 32*sin(c + d*x)**3*a*b**3 + 9 6*sin(c + d*x)*a**3*b + 96*sin(c + d*x)*a*b**3 + 24*a**4*d*x + 72*a**2*b** 2*d*x + 9*b**4*d*x)/(24*d)