\(\int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx\) [388]

Optimal result

Integrand size = 24, antiderivative size = 157 \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=4 a \left (5 a^2+3 b^2\right ) x+\frac {4 a \left (15 a^2+4 b^2\right ) \cos (d+e x)}{3 e}+\frac {4 b \left (15 a^2+4 b^2\right ) \sin (d+e x)}{3 e}+\frac {8 (a+b \cos (d+e x)-a \sin (d+e x))^2 (a \cos (d+e x)+b \sin (d+e x))}{3 e}+\frac {20 (a+b \cos (d+e x)-a \sin (d+e x)) \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{3 e} \]

[Out]

4*a*(5*a^2+3*b^2)*x+4/3*a*(15*a^2+4*b^2)*cos(e*x+d)/e+4/3*b*(15*a^2+4*b^2)*sin(e*x+d)/e+8/3*(a+b*cos(e*x+d)-a*
sin(e*x+d))^2*(a*cos(e*x+d)+b*sin(e*x+d))/e+20/3*(a+b*cos(e*x+d)-a*sin(e*x+d))*(a^2*cos(e*x+d)+a*b*sin(e*x+d))
/e

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3199, 3225, 2717, 2718} \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=\frac {4 b \left (15 a^2+4 b^2\right ) \sin (d+e x)}{3 e}+\frac {4 a \left (15 a^2+4 b^2\right ) \cos (d+e x)}{3 e}+4 a x \left (5 a^2+3 b^2\right )+\frac {20 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right ) (a (-\sin (d+e x))+a+b \cos (d+e x))}{3 e}+\frac {8 (a \cos (d+e x)+b \sin (d+e x)) (a (-\sin (d+e x))+a+b \cos (d+e x))^2}{3 e} \]

[In]

Int[(2*a + 2*b*Cos[d + e*x] - 2*a*Sin[d + e*x])^3,x]

[Out]

4*a*(5*a^2 + 3*b^2)*x + (4*a*(15*a^2 + 4*b^2)*Cos[d + e*x])/(3*e) + (4*b*(15*a^2 + 4*b^2)*Sin[d + e*x])/(3*e)
+ (8*(a + b*Cos[d + e*x] - a*Sin[d + e*x])^2*(a*Cos[d + e*x] + b*Sin[d + e*x]))/(3*e) + (20*(a + b*Cos[d + e*x
] - a*Sin[d + e*x])*(a^2*Cos[d + e*x] + a*b*Sin[d + e*x]))/(3*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]

Rule 3225

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

Rubi steps \begin{align*} \text {integral}& = \frac {8 (a+b \cos (d+e x)-a \sin (d+e x))^2 (a \cos (d+e x)+b \sin (d+e x))}{3 e}+\frac {1}{3} \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x)) \left (4 \left (5 a^2+2 b^2\right )+20 a b \cos (d+e x)-20 a^2 \sin (d+e x)\right ) \, dx \\ & = \frac {8 (a+b \cos (d+e x)-a \sin (d+e x))^2 (a \cos (d+e x)+b \sin (d+e x))}{3 e}+\frac {20 (a+b \cos (d+e x)-a \sin (d+e x)) \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{3 e}+\frac {\int \left (48 a^2 \left (5 a^2+3 b^2\right )+16 a b \left (15 a^2+4 b^2\right ) \cos (d+e x)-16 a^2 \left (15 a^2+4 b^2\right ) \sin (d+e x)\right ) \, dx}{12 a} \\ & = 4 a \left (5 a^2+3 b^2\right ) x+\frac {8 (a+b \cos (d+e x)-a \sin (d+e x))^2 (a \cos (d+e x)+b \sin (d+e x))}{3 e}+\frac {20 (a+b \cos (d+e x)-a \sin (d+e x)) \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{3 e}-\frac {1}{3} \left (4 a \left (15 a^2+4 b^2\right )\right ) \int \sin (d+e x) \, dx+\frac {1}{3} \left (4 b \left (15 a^2+4 b^2\right )\right ) \int \cos (d+e x) \, dx \\ & = 4 a \left (5 a^2+3 b^2\right ) x+\frac {4 a \left (15 a^2+4 b^2\right ) \cos (d+e x)}{3 e}+\frac {4 b \left (15 a^2+4 b^2\right ) \sin (d+e x)}{3 e}+\frac {8 (a+b \cos (d+e x)-a \sin (d+e x))^2 (a \cos (d+e x)+b \sin (d+e x))}{3 e}+\frac {20 (a+b \cos (d+e x)-a \sin (d+e x)) \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right )}{3 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.67 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87 \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=\frac {2 \left (6 a \left (5 a^2+3 b^2\right ) (d+e x)+9 a \left (5 a^2+b^2\right ) \cos (d+e x)+18 a^2 b \cos (2 (d+e x))-a \left (a^2-3 b^2\right ) \cos (3 (d+e x))+9 b \left (5 a^2+b^2\right ) \sin (d+e x)-9 a \left (a^2-b^2\right ) \sin (2 (d+e x))+b \left (-3 a^2+b^2\right ) \sin (3 (d+e x))\right )}{3 e} \]

[In]

Integrate[(2*a + 2*b*Cos[d + e*x] - 2*a*Sin[d + e*x])^3,x]

[Out]

(2*(6*a*(5*a^2 + 3*b^2)*(d + e*x) + 9*a*(5*a^2 + b^2)*Cos[d + e*x] + 18*a^2*b*Cos[2*(d + e*x)] - a*(a^2 - 3*b^
2)*Cos[3*(d + e*x)] + 9*b*(5*a^2 + b^2)*Sin[d + e*x] - 9*a*(a^2 - b^2)*Sin[2*(d + e*x)] + b*(-3*a^2 + b^2)*Sin
[3*(d + e*x)]))/(3*e)

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.97

method	result	size
parallelrisch	\(\frac {\frac {2 \left (-a^{3}+3 a \,b^{2}\right ) \cos \left (3 e x +3 d \right )}{3}+6 \left (-a^{3}+a \,b^{2}\right ) \sin \left (2 e x +2 d \right )+\frac {2 \left (-3 a^{2} b +b^{3}\right ) \sin \left (3 e x +3 d \right )}{3}+12 a^{2} b \cos \left (2 e x +2 d \right )+6 \left (5 a^{3}+a \,b^{2}\right ) \cos \left (e x +d \right )+6 \left (5 a^{2} b +b^{3}\right ) \sin \left (e x +d \right )+\frac {4 \left (15 e x +22\right ) a^{3}}{3}-12 a^{2} b +4 \left (3 e x +2\right ) b^{2} a}{e}\)	\(152\)
parts	\(\frac {8 a \,b^{2} \cos \left (e x +d \right )^{3}+24 a \,b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {8 a^{2} b \left (\sin \left (e x +d \right )-1\right )^{3}}{e}+8 a^{3} x +\frac {24 a^{3} \cos \left (e x +d \right )}{e}+\frac {24 a^{3} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {8 b^{3} \left (2+\cos \left (e x +d \right )^{2}\right ) \sin \left (e x +d \right )}{3 e}+\frac {8 a^{3} \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3 e}\)	\(166\)
derivativedivides	\(\frac {\frac {8 b^{3} \left (2+\cos \left (e x +d \right )^{2}\right ) \sin \left (e x +d \right )}{3}+8 a \,b^{2} \cos \left (e x +d \right )^{3}+24 a \,b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+8 a^{2} b \sin \left (e x +d \right )^{3}+24 a^{2} b \cos \left (e x +d \right )^{2}+24 \sin \left (e x +d \right ) a^{2} b +\frac {8 a^{3} \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3}+24 a^{3} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+24 a^{3} \cos \left (e x +d \right )+8 a^{3} \left (e x +d \right )}{e}\)	\(176\)
default	\(\frac {\frac {8 b^{3} \left (2+\cos \left (e x +d \right )^{2}\right ) \sin \left (e x +d \right )}{3}+8 a \,b^{2} \cos \left (e x +d \right )^{3}+24 a \,b^{2} \left (\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+8 a^{2} b \sin \left (e x +d \right )^{3}+24 a^{2} b \cos \left (e x +d \right )^{2}+24 \sin \left (e x +d \right ) a^{2} b +\frac {8 a^{3} \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3}+24 a^{3} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+24 a^{3} \cos \left (e x +d \right )+8 a^{3} \left (e x +d \right )}{e}\)	\(176\)
risch	\(20 a^{3} x +12 a \,b^{2} x +\frac {30 a^{3} \cos \left (e x +d \right )}{e}+\frac {6 a \cos \left (e x +d \right ) b^{2}}{e}+\frac {30 b \sin \left (e x +d \right ) a^{2}}{e}+\frac {6 b^{3} \sin \left (e x +d \right )}{e}-\frac {2 a^{3} \cos \left (3 e x +3 d \right )}{3 e}+\frac {2 a \cos \left (3 e x +3 d \right ) b^{2}}{e}-\frac {2 b \sin \left (3 e x +3 d \right ) a^{2}}{e}+\frac {2 b^{3} \sin \left (3 e x +3 d \right )}{3 e}+\frac {12 a^{2} b \cos \left (2 e x +2 d \right )}{e}-\frac {6 a^{3} \sin \left (2 e x +2 d \right )}{e}+\frac {6 a \sin \left (2 e x +2 d \right ) b^{2}}{e}\)	\(196\)
norman	\(\frac {\left (20 a^{3}+12 a \,b^{2}\right ) x +\left (20 a^{3}+12 a \,b^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6}+\left (60 a^{3}+36 a \,b^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\left (60 a^{3}+36 a \,b^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\frac {\left (128 a^{3}-96 a^{2} b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e}+\frac {176 a^{3}+48 a \,b^{2}}{3 e}-\frac {8 \left (3 a^{3}-6 a^{2} b -3 a \,b^{2}-2 b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e}+\frac {8 \left (3 a^{3}+6 a^{2} b -3 a \,b^{2}+2 b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}}{e}+\frac {3 \left (16 a^{3}-32 a^{2} b +16 a \,b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{e}+\frac {32 b \left (15 a^{2}+b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{3 e}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\)	\(287\)

[In]

int((2*a+2*b*cos(e*x+d)-2*a*sin(e*x+d))^3,x,method=_RETURNVERBOSE)

[Out]

2/3*((-a^3+3*a*b^2)*cos(3*e*x+3*d)+9*(-a^3+a*b^2)*sin(2*e*x+2*d)+(-3*a^2*b+b^3)*sin(3*e*x+3*d)+18*a^2*b*cos(2*
e*x+2*d)+9*(5*a^3+a*b^2)*cos(e*x+d)+9*(5*a^2*b+b^3)*sin(e*x+d)+2*(15*e*x+22)*a^3-18*a^2*b+6*(3*e*x+2)*b^2*a)/e

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.80 \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=\frac {4 \, {\left (18 \, a^{2} b \cos \left (e x + d\right )^{2} + 24 \, a^{3} \cos \left (e x + d\right ) - 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (e x + d\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} e x + {\left (24 \, a^{2} b + 4 \, b^{3} - 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (e x + d\right )^{2} - 9 \, {\left (a^{3} - a b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )}}{3 \, e} \]

[In]

integrate((2*a+2*b*cos(e*x+d)-2*a*sin(e*x+d))^3,x, algorithm="fricas")

[Out]

4/3*(18*a^2*b*cos(e*x + d)^2 + 24*a^3*cos(e*x + d) - 2*(a^3 - 3*a*b^2)*cos(e*x + d)^3 + 3*(5*a^3 + 3*a*b^2)*e*
x + (24*a^2*b + 4*b^3 - 2*(3*a^2*b - b^3)*cos(e*x + d)^2 - 9*(a^3 - a*b^2)*cos(e*x + d))*sin(e*x + d))/e

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.85 \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=\begin {cases} 12 a^{3} x \sin ^{2}{\left (d + e x \right )} + 12 a^{3} x \cos ^{2}{\left (d + e x \right )} + 8 a^{3} x + \frac {8 a^{3} \sin ^{2}{\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {12 a^{3} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} + \frac {16 a^{3} \cos ^{3}{\left (d + e x \right )}}{3 e} + \frac {24 a^{3} \cos {\left (d + e x \right )}}{e} + \frac {8 a^{2} b \sin ^{3}{\left (d + e x \right )}}{e} - \frac {24 a^{2} b \sin ^{2}{\left (d + e x \right )}}{e} + \frac {24 a^{2} b \sin {\left (d + e x \right )}}{e} + 12 a b^{2} x \sin ^{2}{\left (d + e x \right )} + 12 a b^{2} x \cos ^{2}{\left (d + e x \right )} + \frac {12 a b^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} + \frac {8 a b^{2} \cos ^{3}{\left (d + e x \right )}}{e} + \frac {16 b^{3} \sin ^{3}{\left (d + e x \right )}}{3 e} + \frac {8 b^{3} \sin {\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \left (- 2 a \sin {\left (d \right )} + 2 a + 2 b \cos {\left (d \right )}\right )^{3} & \text {otherwise} \end {cases} \]

[In]

integrate((2*a+2*b*cos(e*x+d)-2*a*sin(e*x+d))**3,x)

[Out]

Piecewise((12*a**3*x*sin(d + e*x)**2 + 12*a**3*x*cos(d + e*x)**2 + 8*a**3*x + 8*a**3*sin(d + e*x)**2*cos(d + e
*x)/e - 12*a**3*sin(d + e*x)*cos(d + e*x)/e + 16*a**3*cos(d + e*x)**3/(3*e) + 24*a**3*cos(d + e*x)/e + 8*a**2*
b*sin(d + e*x)**3/e - 24*a**2*b*sin(d + e*x)**2/e + 24*a**2*b*sin(d + e*x)/e + 12*a*b**2*x*sin(d + e*x)**2 + 1
2*a*b**2*x*cos(d + e*x)**2 + 12*a*b**2*sin(d + e*x)*cos(d + e*x)/e + 8*a*b**2*cos(d + e*x)**3/e + 16*b**3*sin(
d + e*x)**3/(3*e) + 8*b**3*sin(d + e*x)*cos(d + e*x)**2/e, Ne(e, 0)), (x*(-2*a*sin(d) + 2*a + 2*b*cos(d))**3,
True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.20 \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=\frac {8 \, a b^{2} \cos \left (e x + d\right )^{3}}{e} + \frac {8 \, a^{2} b \sin \left (e x + d\right )^{3}}{e} + 8 \, a^{3} x - \frac {8 \, {\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} a^{3}}{3 \, e} - \frac {8 \, {\left (\sin \left (e x + d\right )^{3} - 3 \, \sin \left (e x + d\right )\right )} b^{3}}{3 \, e} + 24 \, a^{2} {\left (\frac {a \cos \left (e x + d\right )}{e} + \frac {b \sin \left (e x + d\right )}{e}\right )} + 6 \, {\left (\frac {4 \, a b \cos \left (e x + d\right )^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{e}\right )} a \]

[In]

integrate((2*a+2*b*cos(e*x+d)-2*a*sin(e*x+d))^3,x, algorithm="maxima")

[Out]

8*a*b^2*cos(e*x + d)^3/e + 8*a^2*b*sin(e*x + d)^3/e + 8*a^3*x - 8/3*(cos(e*x + d)^3 - 3*cos(e*x + d))*a^3/e -
8/3*(sin(e*x + d)^3 - 3*sin(e*x + d))*b^3/e + 24*a^2*(a*cos(e*x + d)/e + b*sin(e*x + d)/e) + 6*(4*a*b*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))*b^2/e)*a

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=\frac {12 \, a^{2} b \cos \left (2 \, e x + 2 \, d\right )}{e} + 4 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} x - \frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (3 \, e x + 3 \, d\right )}{3 \, e} + \frac {6 \, {\left (5 \, a^{3} + a b^{2}\right )} \cos \left (e x + d\right )}{e} - \frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} \sin \left (3 \, e x + 3 \, d\right )}{3 \, e} - \frac {6 \, {\left (a^{3} - a b^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{e} + \frac {6 \, {\left (5 \, a^{2} b + b^{3}\right )} \sin \left (e x + d\right )}{e} \]

[In]

integrate((2*a+2*b*cos(e*x+d)-2*a*sin(e*x+d))^3,x, algorithm="giac")

[Out]

12*a^2*b*cos(2*e*x + 2*d)/e + 4*(5*a^3 + 3*a*b^2)*x - 2/3*(a^3 - 3*a*b^2)*cos(3*e*x + 3*d)/e + 6*(5*a^3 + a*b^
2)*cos(e*x + d)/e - 2/3*(3*a^2*b - b^3)*sin(3*e*x + 3*d)/e - 6*(a^3 - a*b^2)*sin(2*e*x + 2*d)/e + 6*(5*a^2*b +
b^3)*sin(e*x + d)/e

Mupad [B] (verification not implemented)

Time = 27.16 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.86 \[ \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, dx=\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (48\,a^3-96\,a^2\,b+48\,a\,b^2\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\left (24\,a^3+48\,a^2\,b-24\,a\,b^2+16\,b^3\right )+16\,a\,b^2-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (96\,a^2\,b-128\,a^3\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (160\,a^2\,b+\frac {32\,b^3}{3}\right )+\frac {176\,a^3}{3}+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-24\,a^3+48\,a^2\,b+24\,a\,b^2+16\,b^3\right )}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+1\right )}+\frac {8\,a\,\mathrm {atan}\left (\frac {8\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (5\,a^2+3\,b^2\right )}{40\,a^3+24\,a\,b^2}\right )\,\left (5\,a^2+3\,b^2\right )}{e}-\frac {8\,a\,\left (5\,a^2+3\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )-\frac {e\,x}{2}\right )}{e} \]

[In]

int((2*a + 2*b*cos(d + e*x) - 2*a*sin(d + e*x))^3,x)

[Out]

(tan(d/2 + (e*x)/2)^4*(48*a*b^2 - 96*a^2*b + 48*a^3) + tan(d/2 + (e*x)/2)^5*(48*a^2*b - 24*a*b^2 + 24*a^3 + 16
*b^3) + 16*a*b^2 - tan(d/2 + (e*x)/2)^2*(96*a^2*b - 128*a^3) + tan(d/2 + (e*x)/2)^3*(160*a^2*b + (32*b^3)/3) +
(176*a^3)/3 + tan(d/2 + (e*x)/2)*(24*a*b^2 + 48*a^2*b - 24*a^3 + 16*b^3))/(e*(3*tan(d/2 + (e*x)/2)^2 + 3*tan(
d/2 + (e*x)/2)^4 + tan(d/2 + (e*x)/2)^6 + 1)) + (8*a*atan((8*a*tan(d/2 + (e*x)/2)*(5*a^2 + 3*b^2))/(24*a*b^2 +
40*a^3))*(5*a^2 + 3*b^2))/e - (8*a*(5*a^2 + 3*b^2)*(atan(tan(d/2 + (e*x)/2)) - (e*x)/2))/e