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

Optimal. Leaf size=157 \[ \frac{4 b \left (15 a^2+4 b^2\right ) \sin (d+e x)}{3 e}+\frac{4 a \left (15 a^2+4 b^2\right ) \cos (d+e x)}{3 e}+4 a x \left (5 a^2+3 b^2\right )+\frac{20 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right ) (a (-\sin (d+e x))+a+b \cos (d+e x))}{3 e}+\frac{8 (a \cos (d+e x)+b \sin (d+e x)) (a (-\sin (d+e x))+a+b \cos (d+e x))^2}{3 e} \]

[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)

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Rubi [A]  time = 0.135751, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3120, 3146, 2637, 2638} \[ \frac{4 b \left (15 a^2+4 b^2\right ) \sin (d+e x)}{3 e}+\frac{4 a \left (15 a^2+4 b^2\right ) \cos (d+e x)}{3 e}+4 a x \left (5 a^2+3 b^2\right )+\frac{20 \left (a^2 \cos (d+e x)+a b \sin (d+e x)\right ) (a (-\sin (d+e x))+a+b \cos (d+e x))}{3 e}+\frac{8 (a \cos (d+e x)+b \sin (d+e x)) (a (-\sin (d+e x))+a+b \cos (d+e x))^2}{3 e} \]

Antiderivative was successfully verified.

[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 3120

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*Sin
[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 3146

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]

Rule 2637

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

Rule 2638

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

Rubi steps

\begin{align*} \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^3 \, 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{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]  time = 0.442698, size = 136, normalized size = 0.87 \[ \frac{2 \left (6 a \left (5 a^2+3 b^2\right ) (d+e x)-9 a \left (a^2-b^2\right ) \sin (2 (d+e x))+9 b \left (5 a^2+b^2\right ) \sin (d+e x)+b \left (b^2-3 a^2\right ) \sin (3 (d+e x))+9 a \left (5 a^2+b^2\right ) \cos (d+e x)-a \left (a^2-3 b^2\right ) \cos (3 (d+e x))+18 a^2 b \cos (2 (d+e x))\right )}{3 e} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.07, size = 176, normalized size = 1.1 \begin{align*} 8\,{\frac{{a}^{3} \left ( ex+d \right ) +3\,\sin \left ( ex+d \right ){a}^{2}b+3\,{a}^{3}\cos \left ( ex+d \right ) +3\,a{b}^{2} \left ( 1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) +3\, \left ( \cos \left ( ex+d \right ) \right ) ^{2}{a}^{2}b+3\,{a}^{3} \left ( -1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) +1/3\,{b}^{3} \left ( 2+ \left ( \cos \left ( ex+d \right ) \right ) ^{2} \right ) \sin \left ( ex+d \right ) + \left ( \cos \left ( ex+d \right ) \right ) ^{3}a{b}^{2}+{a}^{2}b \left ( \sin \left ( ex+d \right ) \right ) ^{3}+1/3\,{a}^{3} \left ( 2+ \left ( \sin \left ( ex+d \right ) \right ) ^{2} \right ) \cos \left ( ex+d \right ) }{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/e*(a^3*(e*x+d)+3*sin(e*x+d)*a^2*b+3*a^3*cos(e*x+d)+3*a*b^2*(1/2*sin(e*x+d)*cos(e*x+d)+1/2*e*x+1/2*d)+3*cos(e
*x+d)^2*a^2*b+3*a^3*(-1/2*sin(e*x+d)*cos(e*x+d)+1/2*e*x+1/2*d)+1/3*b^3*(2+cos(e*x+d)^2)*sin(e*x+d)+cos(e*x+d)^
3*a*b^2+a^2*b*sin(e*x+d)^3+1/3*a^3*(2+sin(e*x+d)^2)*cos(e*x+d))

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Maxima [A]  time = 0.991341, size = 254, normalized size = 1.62 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.21198, size = 292, normalized size = 1.86 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.974717, size = 291, normalized size = 1.85 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.11012, size = 204, normalized size = 1.3 \begin{align*} 12 \, a^{2} b \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} - \frac{2}{3} \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (3 \, x e + 3 \, d\right ) e^{\left (-1\right )} + 6 \,{\left (5 \, a^{3} + a b^{2}\right )} \cos \left (x e + d\right ) e^{\left (-1\right )} - \frac{2}{3} \,{\left (3 \, a^{2} b - b^{3}\right )} e^{\left (-1\right )} \sin \left (3 \, x e + 3 \, d\right ) - 6 \,{\left (a^{3} - a b^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + 6 \,{\left (5 \, a^{2} b + b^{3}\right )} e^{\left (-1\right )} \sin \left (x e + d\right ) + 4 \,{\left (5 \, a^{3} + 3 \, a b^{2}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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*x*e + 2*d)*e^(-1) - 2/3*(a^3 - 3*a*b^2)*cos(3*x*e + 3*d)*e^(-1) + 6*(5*a^3 + a*b^2)*cos(x*e + d
)*e^(-1) - 2/3*(3*a^2*b - b^3)*e^(-1)*sin(3*x*e + 3*d) - 6*(a^3 - a*b^2)*e^(-1)*sin(2*x*e + 2*d) + 6*(5*a^2*b
+ b^3)*e^(-1)*sin(x*e + d) + 4*(5*a^3 + 3*a*b^2)*x