\(\int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx\) [375]

Optimal result

Integrand size = 24, antiderivative size = 81 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=2 \left (3 a^2+c^2\right ) x-\frac {6 a c \cos (d+e x)}{e}-\frac {6 a^2 \sin (d+e x)}{e}-\frac {2 (c \cos (d+e x)+a \sin (d+e x)) (a-a \cos (d+e x)+c \sin (d+e x))}{e} \]

[Out]

2*(3*a^2+c^2)*x-6*a*c*cos(e*x+d)/e-6*a^2*sin(e*x+d)/e-2*(c*cos(e*x+d)+a*sin(e*x+d))*(a-a*cos(e*x+d)+c*sin(e*x+
d))/e

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3199, 2717, 2718} \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=2 x \left (3 a^2+c^2\right )-\frac {6 a^2 \sin (d+e x)}{e}-\frac {6 a c \cos (d+e x)}{e}-\frac {2 (a \sin (d+e x)+c \cos (d+e x)) (a (-\cos (d+e x))+a+c \sin (d+e x))}{e} \]

[In]

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

[Out]

2*(3*a^2 + c^2)*x - (6*a*c*Cos[d + e*x])/e - (6*a^2*Sin[d + e*x])/e - (2*(c*Cos[d + e*x] + a*Sin[d + e*x])*(a
- a*Cos[d + e*x] + c*Sin[d + e*x]))/e

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3199

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(c*Cos[d
+ e*x] - b*Sin[d + e*x]))*((a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1)/(e*n)), x] + Dist[1/n, Int[Simp[n*a^2
+ (n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x], x]*(a + b*Cos[d + e*x] + c*S
in[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (c \cos (d+e x)+a \sin (d+e x)) (a-a \cos (d+e x)+c \sin (d+e x))}{e}+\frac {1}{2} \int \left (4 \left (3 a^2+c^2\right )-12 a^2 \cos (d+e x)+12 a c \sin (d+e x)\right ) \, dx \\ & = 2 \left (3 a^2+c^2\right ) x-\frac {2 (c \cos (d+e x)+a \sin (d+e x)) (a-a \cos (d+e x)+c \sin (d+e x))}{e}-\left (6 a^2\right ) \int \cos (d+e x) \, dx+(6 a c) \int \sin (d+e x) \, dx \\ & = 2 \left (3 a^2+c^2\right ) x-\frac {6 a c \cos (d+e x)}{e}-\frac {6 a^2 \sin (d+e x)}{e}-\frac {2 (c \cos (d+e x)+a \sin (d+e x)) (a-a \cos (d+e x)+c \sin (d+e x))}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=4 \left (\frac {\left (3 a^2+c^2\right ) (d+e x)}{2 e}-\frac {2 a c \cos (d+e x)}{e}+\frac {a c \cos (2 (d+e x))}{2 e}-\frac {2 a^2 \sin (d+e x)}{e}+\frac {\left (a^2-c^2\right ) \sin (2 (d+e x))}{4 e}\right ) \]

[In]

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

[Out]

4*(((3*a^2 + c^2)*(d + e*x))/(2*e) - (2*a*c*Cos[d + e*x])/e + (a*c*Cos[2*(d + e*x)])/(2*e) - (2*a^2*Sin[d + e*
x])/e + ((a^2 - c^2)*Sin[2*(d + e*x)])/(4*e))

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {\left (a^{2}-c^{2}\right ) \sin \left (2 e x +2 d \right )+6 a^{2} e x +2 c^{2} e x -8 a^{2} \sin \left (e x +d \right )-8 a c \cos \left (e x +d \right )+2 a c \cos \left (2 e x +2 d \right )-10 a c}{e}\)	\(77\)
risch	\(6 a^{2} x +2 x \,c^{2}-\frac {8 a c \cos \left (e x +d \right )}{e}-\frac {8 a^{2} \sin \left (e x +d \right )}{e}+\frac {2 a c \cos \left (2 e x +2 d \right )}{e}+\frac {\sin \left (2 e x +2 d \right ) a^{2}}{e}-\frac {\sin \left (2 e x +2 d \right ) c^{2}}{e}\)	\(90\)
derivativedivides	\(\frac {4 a^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+4 a c \cos \left (e x +d \right )^{2}-8 a^{2} \sin \left (e x +d \right )+4 c^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-8 a c \cos \left (e x +d \right )+4 a^{2} \left (e x +d \right )}{e}\)	\(100\)
default	\(\frac {4 a^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+4 a c \cos \left (e x +d \right )^{2}-8 a^{2} \sin \left (e x +d \right )+4 c^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-8 a c \cos \left (e x +d \right )+4 a^{2} \left (e x +d \right )}{e}\)	\(100\)
parts	\(-\frac {8 a \left (\frac {\sin \left (e x +d \right )^{2} c}{2}+a \sin \left (e x +d \right )\right )}{e}+4 a^{2} x +\frac {4 a^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {4 c^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}-\frac {8 a c \cos \left (e x +d \right )}{e}\)	\(107\)
norman	\(\frac {\left (6 a^{2}+2 c^{2}\right ) x +\left (6 a^{2}+2 c^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\left (12 a^{2}+4 c^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\frac {16 a c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{e}-\frac {4 \left (3 a^{2}+c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e}-\frac {4 \left (5 a^{2}-c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{e}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\)	\(147\)

[In]

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

[Out]

((a^2-c^2)*sin(2*e*x+2*d)+6*a^2*e*x+2*c^2*e*x-8*a^2*sin(e*x+d)-8*a*c*cos(e*x+d)+2*a*c*cos(2*e*x+2*d)-10*a*c)/e

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=\frac {2 \, {\left (2 \, a c \cos \left (e x + d\right )^{2} + {\left (3 \, a^{2} + c^{2}\right )} e x - 4 \, a c \cos \left (e x + d\right ) - {\left (4 \, a^{2} - {\left (a^{2} - c^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )}}{e} \]

[In]

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

[Out]

2*(2*a*c*cos(e*x + d)^2 + (3*a^2 + c^2)*e*x - 4*a*c*cos(e*x + d) - (4*a^2 - (a^2 - c^2)*cos(e*x + d))*sin(e*x
+ d))/e

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (78) = 156\).

Time = 0.10 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.10 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=\begin {cases} 2 a^{2} x \sin ^{2}{\left (d + e x \right )} + 2 a^{2} x \cos ^{2}{\left (d + e x \right )} + 4 a^{2} x + \frac {2 a^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {8 a^{2} \sin {\left (d + e x \right )}}{e} - \frac {4 a c \sin ^{2}{\left (d + e x \right )}}{e} - \frac {8 a c \cos {\left (d + e x \right )}}{e} + 2 c^{2} x \sin ^{2}{\left (d + e x \right )} + 2 c^{2} x \cos ^{2}{\left (d + e x \right )} - \frac {2 c^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \left (- 2 a \cos {\left (d \right )} + 2 a + 2 c \sin {\left (d \right )}\right )^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a**2*x*sin(d + e*x)**2 + 2*a**2*x*cos(d + e*x)**2 + 4*a**2*x + 2*a**2*sin(d + e*x)*cos(d + e*x)/e
- 8*a**2*sin(d + e*x)/e - 4*a*c*sin(d + e*x)**2/e - 8*a*c*cos(d + e*x)/e + 2*c**2*x*sin(d + e*x)**2 + 2*c**2*
x*cos(d + e*x)**2 - 2*c**2*sin(d + e*x)*cos(d + e*x)/e, Ne(e, 0)), (x*(-2*a*cos(d) + 2*a + 2*c*sin(d))**2, Tru
e))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.21 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=4 \, a^{2} x + \frac {4 \, a c \cos \left (e x + d\right )^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} a^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{e} - 8 \, a {\left (\frac {c \cos \left (e x + d\right )}{e} + \frac {a \sin \left (e x + d\right )}{e}\right )} \]

[In]

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

[Out]

4*a^2*x + 4*a*c*cos(e*x + d)^2/e + (2*e*x + 2*d + sin(2*e*x + 2*d))*a^2/e + (2*e*x + 2*d - sin(2*e*x + 2*d))*c
^2/e - 8*a*(c*cos(e*x + d)/e + a*sin(e*x + d)/e)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=2 \, {\left (3 \, a^{2} + c^{2}\right )} x + \frac {2 \, a c \cos \left (2 \, e x + 2 \, d\right )}{e} - \frac {8 \, a c \cos \left (e x + d\right )}{e} - \frac {8 \, a^{2} \sin \left (e x + d\right )}{e} + \frac {{\left (a^{2} - c^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{e} \]

[In]

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

[Out]

2*(3*a^2 + c^2)*x + 2*a*c*cos(2*e*x + 2*d)/e - 8*a*c*cos(e*x + d)/e - 8*a^2*sin(e*x + d)/e + (a^2 - c^2)*sin(2
*e*x + 2*d)/e

Mupad [B] (verification not implemented)

Time = 26.00 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^2 \, dx=\frac {a^2\,\sin \left (2\,d+2\,e\,x\right )-8\,a^2\,\sin \left (d+e\,x\right )-c^2\,\sin \left (2\,d+2\,e\,x\right )+16\,a\,c\,{\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2-4\,a\,c\,{\sin \left (d+e\,x\right )}^2+6\,a^2\,e\,x+2\,c^2\,e\,x}{e} \]

[In]

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

[Out]

(a^2*sin(2*d + 2*e*x) - 8*a^2*sin(d + e*x) - c^2*sin(2*d + 2*e*x) + 16*a*c*sin(d/2 + (e*x)/2)^2 - 4*a*c*sin(d
+ e*x)^2 + 6*a^2*e*x + 2*c^2*e*x)/e