Integrand size = 19, antiderivative size = 55 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {1}{2} \left (a^2+b^2\right ) x-\frac {(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d} \] Output:
1/2*(a^2+b^2)*x-1/2*(b*cos(d*x+c)-a*sin(d*x+c))*(a*cos(d*x+c)+b*sin(d*x+c) )/d
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {2 \left (a^2+b^2\right ) (c+d x)-2 a b \cos (2 (c+d x))+\left (a^2-b^2\right ) \sin (2 (c+d x))}{4 d} \] Input:
Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]
Output:
(2*(a^2 + b^2)*(c + d*x) - 2*a*b*Cos[2*(c + d*x)] + (a^2 - b^2)*Sin[2*(c + d*x)])/(4*d)
Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3552, 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.
\(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^2dx\) |
\(\Big \downarrow \) 3552 |
\(\displaystyle \frac {1}{2} \left (a^2+b^2\right ) \int 1dx-\frac {(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{2} x \left (a^2+b^2\right )-\frac {(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d}\) |
Input:
Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]
Output:
((a^2 + b^2)*x)/2 - ((b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b *Sin[c + d*x]))/(2*d)
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(-(b*Cos[c + d*x] - a*Sin[c + d*x]))*((a*Cos[c + d*x] + b* Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[(n - 1)*((a^2 + b^2)/n) Int[(a*Co s[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && N eQ[a^2 + b^2, 0] && !IntegerQ[(n - 1)/2] && GtQ[n, 1]
Time = 0.40 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {\sin \left (2 d x +2 c \right ) \left (a^{2}-b^{2}\right )+2 a^{2} x d +2 b^{2} d x -2 a b \cos \left (2 d x +2 c \right )+2 b a}{4 d}\) | \(57\) |
risch | \(\frac {a^{2} x}{2}+\frac {b^{2} x}{2}-\frac {a b \cos \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d}-\frac {\sin \left (2 d x +2 c \right ) b^{2}}{4 d}\) | \(64\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-b a \cos \left (d x +c \right )^{2}+b^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(70\) |
default | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-b a \cos \left (d x +c \right )^{2}+b^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(70\) |
parts | \(\frac {a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b a \sin \left (d x +c \right )^{2}}{d}\) | \(74\) |
norman | \(\frac {\left (\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) x +\left (\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\left (a^{2}+b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {4 b a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(140\) |
orering | \(x \left (\cos \left (d x +c \right ) a +b \sin \left (d x +c \right )\right )^{2}-\frac {\left (\cos \left (d x +c \right ) a +b \sin \left (d x +c \right )\right ) \left (-\sin \left (d x +c \right ) d a +b d \cos \left (d x +c \right )\right )}{2 d^{2}}+\frac {x \left (2 \left (-\sin \left (d x +c \right ) d a +b d \cos \left (d x +c \right )\right )^{2}+2 \left (\cos \left (d x +c \right ) a +b \sin \left (d x +c \right )\right ) \left (-d^{2} \cos \left (d x +c \right ) a -b \,d^{2} \sin \left (d x +c \right )\right )\right )}{4 d^{2}}\) | \(140\) |
Input:
int((cos(d*x+c)*a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/4*(sin(2*d*x+2*c)*(a^2-b^2)+2*a^2*x*d+2*b^2*d*x-2*a*b*cos(2*d*x+2*c)+2*b *a)/d
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {2 \, a b \cos \left (d x + c\right )^{2} - {\left (a^{2} + b^{2}\right )} d x - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \] Input:
integrate((a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-1/2*(2*a*b*cos(d*x + c)^2 - (a^2 + b^2)*d*x - (a^2 - b^2)*cos(d*x + c)*si n(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (49) = 98\).
Time = 0.10 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.33 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {a b \sin ^{2}{\left (c + d x \right )}}{d} + \frac {b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \] Input:
integrate((a*cos(d*x+c)+b*sin(d*x+c))**2,x)
Output:
Piecewise((a**2*x*sin(c + d*x)**2/2 + a**2*x*cos(c + d*x)**2/2 + a**2*sin( c + d*x)*cos(c + d*x)/(2*d) + a*b*sin(c + d*x)**2/d + b**2*x*sin(c + d*x)* *2/2 + b**2*x*cos(c + d*x)**2/2 - b**2*sin(c + d*x)*cos(c + d*x)/(2*d), Ne (d, 0)), (x*(a*cos(c) + b*sin(c))**2, True))
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {a b \cos \left (d x + c\right )^{2}}{d} + \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{4 \, d} + \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{4 \, d} \] Input:
integrate((a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-a*b*cos(d*x + c)^2/d + 1/4*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2/d + 1/4*( 2*d*x + 2*c - sin(2*d*x + 2*c))*b^2/d
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {1}{2} \, {\left (a^{2} + b^{2}\right )} x - \frac {a b \cos \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \] Input:
integrate((a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/2*(a^2 + b^2)*x - 1/2*a*b*cos(2*d*x + 2*c)/d + 1/4*(a^2 - b^2)*sin(2*d*x + 2*c)/d
Time = 15.61 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2\,x}{2}+\frac {b^2\,x}{2}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {a\,b\,\cos \left (2\,c+2\,d\,x\right )}{2\,d} \] Input:
int((a*cos(c + d*x) + b*sin(c + d*x))^2,x)
Output:
(a^2*x)/2 + (b^2*x)/2 + (a^2*sin(2*c + 2*d*x))/(4*d) - (b^2*sin(2*c + 2*d* x))/(4*d) - (a*b*cos(2*c + 2*d*x))/(2*d)
Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.95 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {\cos \left (d x +c \right )^{2} a^{2} d x -2 \cos \left (d x +c \right )^{2} a b +\cos \left (d x +c \right )^{2} b^{2} d x +\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}+\sin \left (d x +c \right )^{2} a^{2} d x +\sin \left (d x +c \right )^{2} b^{2} d x}{2 d} \] Input:
int((a*cos(d*x+c)+b*sin(d*x+c))^2,x)
Output:
(cos(c + d*x)**2*a**2*d*x - 2*cos(c + d*x)**2*a*b + cos(c + d*x)**2*b**2*d *x + cos(c + d*x)*sin(c + d*x)*a**2 - cos(c + d*x)*sin(c + d*x)*b**2 + sin (c + d*x)**2*a**2*d*x + sin(c + d*x)**2*b**2*d*x)/(2*d)