3.363 \(\int (2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 157, normalized size of antiderivative = 1.00, 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} \[ \frac {4 a \left (15 a^2+4 c^2\right ) \sin (d+e x)}{3 e}-\frac {4 c \left (15 a^2+4 c^2\right ) \cos (d+e x)}{3 e}+4 a x \left (5 a^2+3 c^2\right )-\frac {20 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right ) (a \cos (d+e x)+a+c \sin (d+e x))}{3 e}-\frac {8 (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+a+c \sin (d+e x))^2}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.41, size = 135, normalized size = 0.86 \[ \frac {2 \left (6 a \left (5 a^2+3 c^2\right ) (d+e x)+9 a \left (5 a^2+c^2\right ) \sin (d+e x)+9 a \left (a^2-c^2\right ) \sin (2 (d+e x))+a \left (a^2-3 c^2\right ) \sin (3 (d+e x))-9 c \left (5 a^2+c^2\right ) \cos (d+e x)+c \left (c^2-3 a^2\right ) \cos (3 (d+e x))-18 a^2 c \cos (2 (d+e x))\right )}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.99, size = 134, normalized size = 0.85 \[ -\frac {4 \, {\left (18 \, a^{2} c \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, a^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} - 3 \, {\left (5 \, a^{3} + 3 \, a c^{2}\right )} e x + 6 \, {\left (3 \, a^{2} c + c^{3}\right )} \cos \left (e x + d\right ) - {\left (22 \, a^{3} + 6 \, a c^{2} + 2 \, {\left (a^{3} - 3 \, a c^{2}\right )} \cos \left (e x + d\right )^{2} + 9 \, {\left (a^{3} - a c^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )}}{3 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(18*a^2*c*cos(e*x + d)^2 + 2*(3*a^2*c - c^3)*cos(e*x + d)^3 - 3*(5*a^3 + 3*a*c^2)*e*x + 6*(3*a^2*c + c^3)
*cos(e*x + d) - (22*a^3 + 6*a*c^2 + 2*(a^3 - 3*a*c^2)*cos(e*x + d)^2 + 9*(a^3 - a*c^2)*cos(e*x + d))*sin(e*x +
d))/e

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giac [A]  time = 0.18, size = 151, normalized size = 0.96 \[ -12 \, a^{2} c \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} - \frac {2}{3} \, {\left (3 \, a^{2} c - c^{3}\right )} \cos \left (3 \, x e + 3 \, d\right ) e^{\left (-1\right )} - 6 \, {\left (5 \, a^{2} c + c^{3}\right )} \cos \left (x e + d\right ) e^{\left (-1\right )} + \frac {2}{3} \, {\left (a^{3} - 3 \, a c^{2}\right )} e^{\left (-1\right )} \sin \left (3 \, x e + 3 \, d\right ) + 6 \, {\left (a^{3} - a c^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + 6 \, {\left (5 \, a^{3} + a c^{2}\right )} e^{\left (-1\right )} \sin \left (x e + d\right ) + 4 \, {\left (5 \, a^{3} + 3 \, a c^{2}\right )} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12*a^2*c*cos(2*x*e + 2*d)*e^(-1) - 2/3*(3*a^2*c - c^3)*cos(3*x*e + 3*d)*e^(-1) - 6*(5*a^2*c + c^3)*cos(x*e +
d)*e^(-1) + 2/3*(a^3 - 3*a*c^2)*e^(-1)*sin(3*x*e + 3*d) + 6*(a^3 - a*c^2)*e^(-1)*sin(2*x*e + 2*d) + 6*(5*a^3 +
a*c^2)*e^(-1)*sin(x*e + d) + 4*(5*a^3 + 3*a*c^2)*x

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maple [A]  time = 0.26, size = 177, normalized size = 1.13 \[ \frac {8 a^{3} \left (e x +d \right )+24 a^{3} \sin \left (e x +d \right )-24 a^{2} c \cos \left (e x +d \right )+24 a^{3} \left (\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-24 a^{2} c \left (\cos ^{2}\left (e x +d \right )\right )+24 a \,c^{2} \left (-\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+\frac {8 a^{3} \left (2+\cos ^{2}\left (e x +d \right )\right ) \sin \left (e x +d \right )}{3}-8 a^{2} c \left (\cos ^{3}\left (e x +d \right )\right )+8 a \,c^{2} \left (\sin ^{3}\left (e x +d \right )\right )-\frac {8 c^{3} \left (2+\sin ^{2}\left (e x +d \right )\right ) \cos \left (e x +d \right )}{3}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 191, normalized size = 1.22 \[ -\frac {8 \, a^{2} c \cos \left (e x + d\right )^{3}}{e} + \frac {8 \, a c^{2} \sin \left (e x + d\right )^{3}}{e} + 8 \, a^{3} x - \frac {8 \, {\left (\sin \left (e x + d\right )^{3} - 3 \, \sin \left (e x + d\right )\right )} a^{3}}{3 \, e} + \frac {8 \, {\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} c^{3}}{3 \, e} - 24 \, a^{2} {\left (\frac {c \cos \left (e x + d\right )}{e} - \frac {a \sin \left (e x + d\right )}{e}\right )} - 6 \, {\left (\frac {4 \, a c \cos \left (e x + d\right )^{2}}{e} - \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} a^{2}}{e} - \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{e}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.57, size = 239, normalized size = 1.52 \[ 20\,a^3\,x-\frac {32\,c^3\,{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4}{e}+\frac {64\,c^3\,{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6}{3\,e}+12\,a\,c^2\,x-\frac {64\,a^2\,c\,{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6}{e}+\frac {40\,a^3\,\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{e}+\frac {80\,a^3\,{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{3\,e}+\frac {64\,a^3\,{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{3\,e}+\frac {16\,a\,c^2\,{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{e}-\frac {64\,a\,c^2\,{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{e}+\frac {24\,a\,c^2\,\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20*a^3*x - (32*c^3*cos(d/2 + (e*x)/2)^4)/e + (64*c^3*cos(d/2 + (e*x)/2)^6)/(3*e) + 12*a*c^2*x - (64*a^2*c*cos(
d/2 + (e*x)/2)^6)/e + (40*a^3*cos(d/2 + (e*x)/2)*sin(d/2 + (e*x)/2))/e + (80*a^3*cos(d/2 + (e*x)/2)^3*sin(d/2
+ (e*x)/2))/(3*e) + (64*a^3*cos(d/2 + (e*x)/2)^5*sin(d/2 + (e*x)/2))/(3*e) + (16*a*c^2*cos(d/2 + (e*x)/2)^3*si
n(d/2 + (e*x)/2))/e - (64*a*c^2*cos(d/2 + (e*x)/2)^5*sin(d/2 + (e*x)/2))/e + (24*a*c^2*cos(d/2 + (e*x)/2)*sin(
d/2 + (e*x)/2))/e

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sympy [A]  time = 0.76, size = 291, normalized size = 1.85 \[ \begin {cases} 12 a^{3} x \sin ^{2}{\left (d + e x \right )} + 12 a^{3} x \cos ^{2}{\left (d + e x \right )} + 8 a^{3} x + \frac {16 a^{3} \sin ^{3}{\left (d + e x \right )}}{3 e} + \frac {8 a^{3} \sin {\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )}}{e} + \frac {12 a^{3} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} + \frac {24 a^{3} \sin {\left (d + e x \right )}}{e} - \frac {8 a^{2} c \cos ^{3}{\left (d + e x \right )}}{e} - \frac {24 a^{2} c \cos ^{2}{\left (d + e x \right )}}{e} - \frac {24 a^{2} c \cos {\left (d + e x \right )}}{e} + 12 a c^{2} x \sin ^{2}{\left (d + e x \right )} + 12 a c^{2} x \cos ^{2}{\left (d + e x \right )} + \frac {8 a c^{2} \sin ^{3}{\left (d + e x \right )}}{e} - \frac {12 a c^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {8 c^{3} \sin ^{2}{\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {16 c^{3} \cos ^{3}{\left (d + e x \right )}}{3 e} & \text {for}\: e \neq 0 \\x \left (2 a \cos {\relax (d )} + 2 a + 2 c \sin {\relax (d )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a+2*a*cos(e*x+d)+2*c*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 + 16*a**3*sin(d + e*x)**3/(3*e) +
8*a**3*sin(d + e*x)*cos(d + e*x)**2/e + 12*a**3*sin(d + e*x)*cos(d + e*x)/e + 24*a**3*sin(d + e*x)/e - 8*a**2*
c*cos(d + e*x)**3/e - 24*a**2*c*cos(d + e*x)**2/e - 24*a**2*c*cos(d + e*x)/e + 12*a*c**2*x*sin(d + e*x)**2 + 1
2*a*c**2*x*cos(d + e*x)**2 + 8*a*c**2*sin(d + e*x)**3/e - 12*a*c**2*sin(d + e*x)*cos(d + e*x)/e - 8*c**3*sin(d
+ e*x)**2*cos(d + e*x)/e - 16*c**3*cos(d + e*x)**3/(3*e), Ne(e, 0)), (x*(2*a*cos(d) + 2*a + 2*c*sin(d))**3, T
rue))

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