Integrand size = 19, antiderivative size = 54 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {a x}{2}+\frac {b \sin (c+d x)}{d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d}-\frac {b \sin ^3(c+d x)}{3 d} \] Output:
1/2*a*x+b*sin(d*x+c)/d+1/2*a*cos(d*x+c)*sin(d*x+c)/d-1/3*b*sin(d*x+c)^3/d
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {a (c+d x)}{2 d}+\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}+\frac {a \sin (2 (c+d x))}{4 d} \] Input:
Integrate[Cos[c + d*x]^2*(a + b*Cos[c + d*x]),x]
Output:
(a*(c + d*x))/(2*d) + (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d) + (a*S in[2*(c + d*x)])/(4*d)
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3227, 3042, 3113, 2009, 3115, 24}
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.
\(\displaystyle \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle a \int \cos ^2(c+d x)dx+b \int \cos ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+b \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {b \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {b \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\) |
Input:
Int[Cos[c + d*x]^2*(a + b*Cos[c + d*x]),x]
Output:
a*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (b*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Time = 3.92 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81
method | result | size |
parallelrisch | \(\frac {6 a x d +9 b \sin \left (d x +c \right )+b \sin \left (3 d x +3 c \right )+3 a \sin \left (2 d x +2 c \right )}{12 d}\) | \(44\) |
risch | \(\frac {a x}{2}+\frac {3 b \sin \left (d x +c \right )}{4 d}+\frac {b \sin \left (3 d x +3 c \right )}{12 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(48\) |
derivativedivides | \(\frac {\frac {b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(49\) |
default | \(\frac {\frac {b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(49\) |
parts | \(\frac {a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}\) | \(51\) |
norman | \(\frac {\frac {\left (a +2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a x}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {4 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {\left (a -2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(123\) |
orering | \(x \cos \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right )-\frac {49 \left (-2 \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right ) d \sin \left (d x +c \right )-b \cos \left (d x +c \right )^{2} d \sin \left (d x +c \right )\right )}{36 d^{2}}+\frac {49 x \left (2 d^{2} \sin \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right )+4 d^{2} \sin \left (d x +c \right )^{2} b \cos \left (d x +c \right )-2 \cos \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right ) d^{2}-b \cos \left (d x +c \right )^{3} d^{2}\right )}{36 d^{2}}-\frac {7 \left (8 d^{3} \sin \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right ) \cos \left (d x +c \right )-6 d^{3} \sin \left (d x +c \right )^{3} b +13 d^{3} \sin \left (d x +c \right ) b \cos \left (d x +c \right )^{2}\right )}{18 d^{4}}+\frac {7 x \left (8 d^{4} \cos \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right )-52 d^{4} \sin \left (d x +c \right )^{2} b \cos \left (d x +c \right )-8 d^{4} \sin \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right )+13 d^{4} \cos \left (d x +c \right )^{3} b \right )}{18 d^{4}}-\frac {-32 d^{5} \cos \left (d x +c \right ) \left (a +\cos \left (d x +c \right ) b \right ) \sin \left (d x +c \right )-151 d^{5} \cos \left (d x +c \right )^{2} b \sin \left (d x +c \right )+60 d^{5} \sin \left (d x +c \right )^{3} b}{36 d^{6}}+\frac {x \left (32 d^{6} \sin \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right )+514 d^{6} \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b -32 d^{6} \cos \left (d x +c \right )^{2} \left (a +\cos \left (d x +c \right ) b \right )-151 d^{6} \cos \left (d x +c \right )^{3} b \right )}{36 d^{6}}\) | \(466\) |
Input:
int(cos(d*x+c)^2*(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
1/12*(6*a*x*d+9*b*sin(d*x+c)+b*sin(3*d*x+3*c)+3*a*sin(2*d*x+2*c))/d
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {3 \, a d x + {\left (2 \, b \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + 4 \, b\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
1/6*(3*a*d*x + (2*b*cos(d*x + c)^2 + 3*a*cos(d*x + c) + 4*b)*sin(d*x + c)) /d
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.70 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**2*(a+b*cos(d*x+c)),x)
Output:
Piecewise((a*x*sin(c + d*x)**2/2 + a*x*cos(c + d*x)**2/2 + a*sin(c + d*x)* cos(c + d*x)/(2*d) + 2*b*sin(c + d*x)**3/(3*d) + b*sin(c + d*x)*cos(c + d* x)**2/d, Ne(d, 0)), (x*(a + b*cos(c))*cos(c)**2, True))
Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b}{12 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+b*cos(d*x+c)),x, algorithm="maxima")
Output:
1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*b)/d
Time = 0.49 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {1}{2} \, a x + \frac {b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, b \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
1/2*a*x + 1/12*b*sin(3*d*x + 3*c)/d + 1/4*a*sin(2*d*x + 2*c)/d + 3/4*b*sin (d*x + c)/d
Time = 39.75 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {a\,x}{2}+\frac {2\,b\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \] Input:
int(cos(c + d*x)^2*(a + b*cos(c + d*x)),x)
Output:
(a*x)/2 + (2*b*sin(c + d*x))/(3*d) + (a*cos(c + d*x)*sin(c + d*x))/(2*d) + (b*cos(c + d*x)^2*sin(c + d*x))/(3*d)
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -2 \sin \left (d x +c \right )^{3} b +6 \sin \left (d x +c \right ) b +3 a d x}{6 d} \] Input:
int(cos(d*x+c)^2*(a+b*cos(d*x+c)),x)
Output:
(3*cos(c + d*x)*sin(c + d*x)*a - 2*sin(c + d*x)**3*b + 6*sin(c + d*x)*b + 3*a*d*x)/(6*d)