Integrand size = 28, antiderivative size = 137 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {b^2 \sin ^3(c+d x)}{3 d}+\frac {3 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 b^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^7(c+d x)}{7 d}+\frac {b^2 \sin ^7(c+d x)}{7 d} \] Output:
-2/7*a*b*cos(d*x+c)^7/d+a^2*sin(d*x+c)/d-a^2*sin(d*x+c)^3/d+1/3*b^2*sin(d* x+c)^3/d+3/5*a^2*sin(d*x+c)^5/d-2/5*b^2*sin(d*x+c)^5/d-1/7*a^2*sin(d*x+c)^ 7/d+1/7*b^2*sin(d*x+c)^7/d
Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {-30 a b \cos ^7(c+d x)+105 a^2 \sin (c+d x)-35 \left (3 a^2-b^2\right ) \sin ^3(c+d x)+21 \left (3 a^2-2 b^2\right ) \sin ^5(c+d x)-15 \left (a^2-b^2\right ) \sin ^7(c+d x)}{105 d} \] Input:
Integrate[Cos[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]
Output:
(-30*a*b*Cos[c + d*x]^7 + 105*a^2*Sin[c + d*x] - 35*(3*a^2 - b^2)*Sin[c + d*x]^3 + 21*(3*a^2 - 2*b^2)*Sin[c + d*x]^5 - 15*(a^2 - b^2)*Sin[c + d*x]^7 )/(105*d)
Time = 0.36 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3042, 3569, 2009}
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.
| \(\displaystyle \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^5 (a \cos (c+d x)+b \sin (c+d x))^2dx\) |
| \(\Big \downarrow \) 3569 |
| \(\displaystyle \int \left (a^2 \cos ^7(c+d x)+2 a b \sin (c+d x) \cos ^6(c+d x)+b^2 \sin ^2(c+d x) \cos ^5(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \sin ^7(c+d x)}{7 d}+\frac {3 a^2 \sin ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {a^2 \sin (c+d x)}{d}-\frac {2 a b \cos ^7(c+d x)}{7 d}+\frac {b^2 \sin ^7(c+d x)}{7 d}-\frac {2 b^2 \sin ^5(c+d x)}{5 d}+\frac {b^2 \sin ^3(c+d x)}{3 d}\) |
Input:
Int[Cos[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]
Output:
(-2*a*b*Cos[c + d*x]^7)/(7*d) + (a^2*Sin[c + d*x])/d - (a^2*Sin[c + d*x]^3 )/d + (b^2*Sin[c + d*x]^3)/(3*d) + (3*a^2*Sin[c + d*x]^5)/(5*d) - (2*b^2*S in[c + d*x]^5)/(5*d) - (a^2*Sin[c + d*x]^7)/(7*d) + (b^2*Sin[c + d*x]^7)/( 7*d)
Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*(a *cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] && Inte gerQ[m] && IGtQ[n, 0]
Time = 3.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.72
| method | result | size |
| parts | \(\frac {a^{2} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7 d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{7}-\frac {2 \sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}-\frac {2 a b \cos \left (d x +c \right )^{7}}{7 d}\) | \(99\) |
| derivativedivides | \(\frac {\frac {a^{2} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}-\frac {2 a b \cos \left (d x +c \right )^{7}}{7}+b^{2} \left (-\frac {\cos \left (d x +c \right )^{6} \sin \left (d x +c \right )}{7}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{35}\right )}{d}\) | \(108\) |
| default | \(\frac {\frac {a^{2} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}-\frac {2 a b \cos \left (d x +c \right )^{7}}{7}+b^{2} \left (-\frac {\cos \left (d x +c \right )^{6} \sin \left (d x +c \right )}{7}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{35}\right )}{d}\) | \(108\) |
| risch | \(-\frac {5 a b \cos \left (d x +c \right )}{32 d}+\frac {35 a^{2} \sin \left (d x +c \right )}{64 d}+\frac {5 b^{2} \sin \left (d x +c \right )}{64 d}-\frac {a b \cos \left (7 d x +7 c \right )}{224 d}+\frac {\sin \left (7 d x +7 c \right ) a^{2}}{448 d}-\frac {\sin \left (7 d x +7 c \right ) b^{2}}{448 d}-\frac {a b \cos \left (5 d x +5 c \right )}{32 d}+\frac {7 \sin \left (5 d x +5 c \right ) a^{2}}{320 d}-\frac {3 \sin \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {3 a b \cos \left (3 d x +3 c \right )}{32 d}+\frac {7 \sin \left (3 d x +3 c \right ) a^{2}}{64 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{192 d}\) | \(193\) |
| norman | \(\frac {-\frac {4 a b}{7 d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {4 \left (3 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 \left (3 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {8 \left (53 a^{2}+38 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{35 d}+\frac {2 \left (129 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {2 \left (129 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d}-\frac {12 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {20 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(250\) |
| parallelrisch | \(\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12} a b +4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} a^{2}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} b^{2}}{3}+\frac {86 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a^{2}}{5}-\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} b^{2}}{15}-20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a b +\frac {424 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a^{2}}{35}+\frac {304 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b^{2}}{35}+\frac {86 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a^{2}}{5}-\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b^{2}}{15}-12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a b +4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{2}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{3}+2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 a b}{7}}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(261\) |
| orering | \(\text {Expression too large to display}\) | \(3642\) |
Input:
int(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/7*a^2/d*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c) +b^2/d*(1/7*sin(d*x+c)^7-2/5*sin(d*x+c)^5+1/3*sin(d*x+c)^3)-2/7*a*b*cos(d* x+c)^7/d
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {30 \, a b \cos \left (d x + c\right )^{7} - {\left (15 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, a^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \] Input:
integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
-1/105*(30*a*b*cos(d*x + c)^7 - (15*(a^2 - b^2)*cos(d*x + c)^6 + 3*(6*a^2 + b^2)*cos(d*x + c)^4 + 4*(6*a^2 + b^2)*cos(d*x + c)^2 + 48*a^2 + 8*b^2)*s in(d*x + c))/d
Time = 0.60 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.36 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {16 a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {2 a b \cos ^{7}{\left (c + d x \right )}}{7 d} + \frac {8 b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{2} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)
Output:
Piecewise((16*a**2*sin(c + d*x)**7/(35*d) + 8*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*a**2*sin(c + d*x)**3*cos(c + d*x)**4/d + a**2*sin(c + d *x)*cos(c + d*x)**6/d - 2*a*b*cos(c + d*x)**7/(7*d) + 8*b**2*sin(c + d*x)* *7/(105*d) + 4*b**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + b**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))**2*cos( c)**5, True))
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.72 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {30 \, a b \cos \left (d x + c\right )^{7} + 3 \, {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{2} - {\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{105 \, d} \] Input:
integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
-1/105*(30*a*b*cos(d*x + c)^7 + 3*(5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*a^2 - (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*b^2)/d
Time = 0.17 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.13 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {a b \cos \left (7 \, d x + 7 \, c\right )}{224 \, d} - \frac {a b \cos \left (5 \, d x + 5 \, c\right )}{32 \, d} - \frac {3 \, a b \cos \left (3 \, d x + 3 \, c\right )}{32 \, d} - \frac {5 \, a b \cos \left (d x + c\right )}{32 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (7 \, a^{2} - 3 \, b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (21 \, a^{2} - b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (7 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \] Input:
integrate(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
-1/224*a*b*cos(7*d*x + 7*c)/d - 1/32*a*b*cos(5*d*x + 5*c)/d - 3/32*a*b*cos (3*d*x + 3*c)/d - 5/32*a*b*cos(d*x + c)/d + 1/448*(a^2 - b^2)*sin(7*d*x + 7*c)/d + 1/320*(7*a^2 - 3*b^2)*sin(5*d*x + 5*c)/d + 1/192*(21*a^2 - b^2)*s in(3*d*x + 3*c)/d + 5/64*(7*a^2 + b^2)*sin(d*x + c)/d
Time = 16.51 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.28 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {16\,a^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {8\,b^2\,\sin \left (c+d\,x\right )}{105\,d}+\frac {8\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{35\,d}+\frac {6\,a^2\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d}+\frac {4\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{105\,d}+\frac {b^2\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}-\frac {b^2\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d}-\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^7}{7\,d} \] Input:
int(cos(c + d*x)^5*(a*cos(c + d*x) + b*sin(c + d*x))^2,x)
Output:
(16*a^2*sin(c + d*x))/(35*d) + (8*b^2*sin(c + d*x))/(105*d) + (8*a^2*cos(c + d*x)^2*sin(c + d*x))/(35*d) + (6*a^2*cos(c + d*x)^4*sin(c + d*x))/(35*d ) + (a^2*cos(c + d*x)^6*sin(c + d*x))/(7*d) + (4*b^2*cos(c + d*x)^2*sin(c + d*x))/(105*d) + (b^2*cos(c + d*x)^4*sin(c + d*x))/(35*d) - (b^2*cos(c + d*x)^6*sin(c + d*x))/(7*d) - (2*a*b*cos(c + d*x)^7)/(7*d)
Time = 0.17 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.19 \[ \int \cos ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a b -90 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a b +90 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a b -30 \cos \left (d x +c \right ) a b -15 \sin \left (d x +c \right )^{7} a^{2}+15 \sin \left (d x +c \right )^{7} b^{2}+63 \sin \left (d x +c \right )^{5} a^{2}-42 \sin \left (d x +c \right )^{5} b^{2}-105 \sin \left (d x +c \right )^{3} a^{2}+35 \sin \left (d x +c \right )^{3} b^{2}+105 \sin \left (d x +c \right ) a^{2}+30 a b}{105 d} \] Input:
int(cos(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)
Output:
(30*cos(c + d*x)*sin(c + d*x)**6*a*b - 90*cos(c + d*x)*sin(c + d*x)**4*a*b + 90*cos(c + d*x)*sin(c + d*x)**2*a*b - 30*cos(c + d*x)*a*b - 15*sin(c + d*x)**7*a**2 + 15*sin(c + d*x)**7*b**2 + 63*sin(c + d*x)**5*a**2 - 42*sin( c + d*x)**5*b**2 - 105*sin(c + d*x)**3*a**2 + 35*sin(c + d*x)**3*b**2 + 10 5*sin(c + d*x)*a**2 + 30*a*b)/(105*d)