Integrand size = 19, antiderivative size = 71 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=a b x+\frac {2 \left (a^2+b^2\right ) \sin (c+d x)}{3 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {(a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \] Output:
a*b*x+2/3*(a^2+b^2)*sin(d*x+c)/d+1/3*a*b*cos(d*x+c)*sin(d*x+c)/d+1/3*(a+b* cos(d*x+c))^2*sin(d*x+c)/d
Time = 0.44 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3 \left (4 a^2+3 b^2\right ) \sin (c+d x)+b (12 a (c+d x)+6 a \sin (2 (c+d x))+b \sin (3 (c+d x)))}{12 d} \] Input:
Integrate[Cos[c + d*x]*(a + b*Cos[c + d*x])^2,x]
Output:
(3*(4*a^2 + 3*b^2)*Sin[c + d*x] + b*(12*a*(c + d*x) + 6*a*Sin[2*(c + d*x)] + b*Sin[3*(c + d*x)]))/(12*d)
Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3232, 27, 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.
\(\displaystyle \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 3232 |
\(\displaystyle \frac {1}{3} \int 2 (b+a \cos (c+d x)) (a+b \cos (c+d x))dx+\frac {\sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \int (b+a \cos (c+d x)) (a+b \cos (c+d x))dx+\frac {\sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \int \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {\sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {2}{3} \left (\frac {\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac {a b \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a b x}{2}\right )+\frac {\sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\) |
Input:
Int[Cos[c + d*x]*(a + b*Cos[c + d*x])^2,x]
Output:
((a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + (2*((3*a*b*x)/2 + ((a^2 + b^ 2)*Sin[c + d*x])/d + (a*b*Cos[c + d*x]*Sin[c + d*x])/(2*d)))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 4.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86
method | result | size |
parallelrisch | \(\frac {12 a b x d +12 a^{2} \sin \left (d x +c \right )+9 \sin \left (d x +c \right ) b^{2}+\sin \left (3 d x +3 c \right ) b^{2}+6 a b \sin \left (2 d x +2 c \right )}{12 d}\) | \(61\) |
derivativedivides | \(\frac {a^{2} \sin \left (d x +c \right )+2 a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {b^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(63\) |
default | \(\frac {a^{2} \sin \left (d x +c \right )+2 a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {b^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(63\) |
risch | \(a b x +\frac {a^{2} \sin \left (d x +c \right )}{d}+\frac {3 b^{2} \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) b^{2}}{12 d}+\frac {a b \sin \left (2 d x +2 c \right )}{2 d}\) | \(66\) |
parts | \(\frac {a^{2} \sin \left (d x +c \right )}{d}+\frac {b^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}+\frac {2 a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(68\) |
norman | \(\frac {a b x +a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {4 \left (3 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 \left (a^{2}-a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 \left (a^{2}+a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+3 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 a b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(145\) |
orering | \(x \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}-\frac {49 \left (-d \sin \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}-2 \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right ) d \sin \left (d x +c \right ) b \right )}{36 d^{2}}+\frac {49 x \left (-d^{2} \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}+4 d^{2} \sin \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right ) b +2 d^{2} \sin \left (d x +c \right )^{2} b^{2} \cos \left (d x +c \right )-2 \cos \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right ) d^{2} b \right )}{36 d^{2}}-\frac {7 \left (d^{3} \sin \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}+14 d^{3} \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right ) \sin \left (d x +c \right ) b -6 \sin \left (d x +c \right )^{3} b^{2} d^{3}+6 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} b^{2} d^{3}\right )}{18 d^{4}}+\frac {7 x \left (d^{4} \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}-16 d^{4} \sin \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right ) b -44 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) b^{2} d^{4}+14 d^{4} \cos \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right ) b +6 d^{4} \cos \left (d x +c \right )^{3} b^{2}\right )}{18 d^{4}}-\frac {-d^{5} \sin \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}-62 d^{5} \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right ) \sin \left (d x +c \right ) b +60 d^{5} \sin \left (d x +c \right )^{3} b^{2}-120 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} b^{2} d^{5}}{36 d^{6}}+\frac {x \left (-d^{6} \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right )^{2}+64 d^{6} \sin \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right ) b +482 d^{6} \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-62 d^{6} \cos \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right ) b -120 d^{6} \cos \left (d x +c \right )^{3} b^{2}\right )}{36 d^{6}}\) | \(595\) |
Input:
int(cos(d*x+c)*(a+cos(d*x+c)*b)^2,x,method=_RETURNVERBOSE)
Output:
1/12*(12*a*b*x*d+12*a^2*sin(d*x+c)+9*sin(d*x+c)*b^2+sin(3*d*x+3*c)*b^2+6*a *b*sin(2*d*x+2*c))/d
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3 \, a b d x + {\left (b^{2} \cos \left (d x + c\right )^{2} + 3 \, a b \cos \left (d x + c\right ) + 3 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))^2,x, algorithm="fricas")
Output:
1/3*(3*a*b*d*x + (b^2*cos(d*x + c)^2 + 3*a*b*cos(d*x + c) + 3*a^2 + 2*b^2) *sin(d*x + c))/d
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.51 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} \sin {\left (c + d x \right )}}{d} + a b x \sin ^{2}{\left (c + d x \right )} + a b x \cos ^{2}{\left (c + d x \right )} + \frac {a b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {2 b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))**2,x)
Output:
Piecewise((a**2*sin(c + d*x)/d + a*b*x*sin(c + d*x)**2 + a*b*x*cos(c + d*x )**2 + a*b*sin(c + d*x)*cos(c + d*x)/d + 2*b**2*sin(c + d*x)**3/(3*d) + b* *2*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(a + b*cos(c))**2*cos(c), True))
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.85 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))^2,x, algorithm="maxima")
Output:
1/6*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b - 2*(sin(d*x + c)^3 - 3*sin(d* x + c))*b^2 + 6*a^2*sin(d*x + c))/d
Time = 0.51 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.85 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=a b x + \frac {b^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {a b \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (4 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*cos(d*x+c))^2,x, algorithm="giac")
Output:
a*b*x + 1/12*b^2*sin(3*d*x + 3*c)/d + 1/2*a*b*sin(2*d*x + 2*c)/d + 1/4*(4* a^2 + 3*b^2)*sin(d*x + c)/d
Time = 39.72 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,b^2\,\sin \left (c+d\,x\right )}{3\,d}+a\,b\,x+\frac {b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)*(a + b*cos(c + d*x))^2,x)
Output:
(a^2*sin(c + d*x))/d + (2*b^2*sin(c + d*x))/(3*d) + a*b*x + (b^2*cos(c + d *x)^2*sin(c + d*x))/(3*d) + (a*b*cos(c + d*x)*sin(c + d*x))/d
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -\sin \left (d x +c \right )^{3} b^{2}+3 \sin \left (d x +c \right ) a^{2}+3 \sin \left (d x +c \right ) b^{2}+3 a b d x}{3 d} \] Input:
int(cos(d*x+c)*(a+b*cos(d*x+c))^2,x)
Output:
(3*cos(c + d*x)*sin(c + d*x)*a*b - sin(c + d*x)**3*b**2 + 3*sin(c + d*x)*a **2 + 3*sin(c + d*x)*b**2 + 3*a*b*d*x)/(3*d)