\(\int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) [48]

Optimal result

Integrand size = 19, antiderivative size = 55 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {1}{2} \left (a^2+b^2\right ) x-\frac {(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3152, 8} \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {1}{2} x \left (a^2+b^2\right )-\frac {(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d} \]

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

((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)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3152

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] + Dist[(n - 1)*((a^2 + b^2)/n), Int[(
a*Cos[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[
(n - 1)/2] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d}+\frac {1}{2} \left (a^2+b^2\right ) \int 1 \, 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {2 \left (a^2+b^2\right ) (c+d x)-2 a b \cos (2 (c+d x))+\left (a^2-b^2\right ) \sin (2 (c+d x))}{4 d} \]

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(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)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {\sin \left (2 d x +2 c \right ) \left (a^{2}-b^{2}\right )+2 a^{2} x d +2 b^{2} d x -2 a b \cos \left (2 d x +2 c \right )+2 a b}{4 d}\)	\(57\)
risch	\(\frac {a^{2} x}{2}+\frac {x \,b^{2}}{2}-\frac {a b \cos \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 d}-\frac {\sin \left (2 d x +2 c \right ) b^{2}}{4 d}\)	\(64\)
derivativedivides	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-a b \cos \left (d x +c \right )^{2}+b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(70\)
default	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-a b \cos \left (d x +c \right )^{2}+b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(70\)
parts	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a b \sin \left (d x +c \right )^{2}}{d}\)	\(74\)
norman	\(\frac {\left (\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) x +\left (\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {\left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\left (a^{2}+b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\)	\(140\)

[In]

int((cos(d*x+c)*a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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*a*b)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {2 \, a b \cos \left (d x + c\right )^{2} - {\left (a^{2} + b^{2}\right )} d x - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(2*a*b*cos(d*x + c)^2 - (a^2 + b^2)*d*x - (a^2 - b^2)*cos(d*x + c)*sin(d*x + c))/d

Sympy [B] (verification not implemented)

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 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {a b \cos ^{2}{\left (c + d x \right )}}{d} + \frac {b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \]

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

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*co
s(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))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=-\frac {a b \cos \left (d x + c\right )^{2}}{d} + \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{4 \, d} + \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{4 \, d} \]

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {1}{2} \, {\left (a^{2} + b^{2}\right )} x - \frac {a b \cos \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 21.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2\,x}{2}+\frac {b^2\,x}{2}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {a\,b\,\cos \left (2\,c+2\,d\,x\right )}{2\,d} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^2,x)

[Out]

(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)