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

Optimal. Leaf size=170 \[ \frac {b \left (11 a^2+4 \left (b^2+c^2\right )\right ) \sin (d+e x)}{6 e}-\frac {c \left (11 a^2+4 \left (b^2+c^2\right )\right ) \cos (d+e x)}{6 e}+\frac {1}{2} a x \left (2 a^2+3 \left (b^2+c^2\right )\right )-\frac {(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))^2}{3 e}-\frac {5 (a c \cos (d+e x)-a b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))}{6 e} \]

[Out]

1/2*a*(2*a^2+3*b^2+3*c^2)*x-1/6*c*(11*a^2+4*b^2+4*c^2)*cos(e*x+d)/e+1/6*b*(11*a^2+4*b^2+4*c^2)*sin(e*x+d)/e-5/
6*(a*c*cos(e*x+d)-a*b*sin(e*x+d))*(a+b*cos(e*x+d)+c*sin(e*x+d))/e-1/3*(c*cos(e*x+d)-b*sin(e*x+d))*(a+b*cos(e*x
+d)+c*sin(e*x+d))^2/e

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Rubi [A]  time = 0.19, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3120, 3146, 2637, 2638} \[ \frac {b \left (11 a^2+4 \left (b^2+c^2\right )\right ) \sin (d+e x)}{6 e}-\frac {c \left (11 a^2+4 \left (b^2+c^2\right )\right ) \cos (d+e x)}{6 e}+\frac {1}{2} a x \left (2 a^2+3 \left (b^2+c^2\right )\right )-\frac {(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))^2}{3 e}-\frac {5 (a c \cos (d+e x)-a b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))}{6 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*a^2 + 3*(b^2 + c^2))*x)/2 - (c*(11*a^2 + 4*(b^2 + c^2))*Cos[d + e*x])/(6*e) + (b*(11*a^2 + 4*(b^2 + c^2)
)*Sin[d + e*x])/(6*e) - (5*(a*c*Cos[d + e*x] - a*b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x]))/(6*e)
- ((c*Cos[d + e*x] - b*Sin[d + e*x])*(a + b*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 (a+b \cos (d+e x)+c \sin (d+e x))^3 \, dx &=-\frac {(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))^2}{3 e}+\frac {1}{3} \int (a+b \cos (d+e x)+c \sin (d+e x)) \left (3 a^2+2 \left (b^2+c^2\right )+5 a b \cos (d+e x)+5 a c \sin (d+e x)\right ) \, dx\\ &=-\frac {5 (a c \cos (d+e x)-a b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))}{6 e}-\frac {(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))^2}{3 e}+\frac {\int \left (3 a^2 \left (2 a^2+3 \left (b^2+c^2\right )\right )+a b \left (11 a^2+4 \left (b^2+c^2\right )\right ) \cos (d+e x)+a c \left (11 a^2+4 \left (b^2+c^2\right )\right ) \sin (d+e x)\right ) \, dx}{6 a}\\ &=\frac {1}{2} a \left (2 a^2+3 \left (b^2+c^2\right )\right ) x-\frac {5 (a c \cos (d+e x)-a b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))}{6 e}-\frac {(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))^2}{3 e}+\frac {1}{6} \left (b \left (11 a^2+4 \left (b^2+c^2\right )\right )\right ) \int \cos (d+e x) \, dx+\frac {1}{6} \left (c \left (11 a^2+4 \left (b^2+c^2\right )\right )\right ) \int \sin (d+e x) \, dx\\ &=\frac {1}{2} a \left (2 a^2+3 \left (b^2+c^2\right )\right ) x-\frac {c \left (11 a^2+4 \left (b^2+c^2\right )\right ) \cos (d+e x)}{6 e}+\frac {b \left (11 a^2+4 \left (b^2+c^2\right )\right ) \sin (d+e x)}{6 e}-\frac {5 (a c \cos (d+e x)-a b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))}{6 e}-\frac {(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))^2}{3 e}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 144, normalized size = 0.85 \[ \frac {6 a \left (2 a^2+3 \left (b^2+c^2\right )\right ) (d+e x)+9 b \left (4 a^2+b^2+c^2\right ) \sin (d+e x)-9 c \left (4 a^2+b^2+c^2\right ) \cos (d+e x)+9 a \left (b^2-c^2\right ) \sin (2 (d+e x))-18 a b c \cos (2 (d+e x))+b \left (b^2-3 c^2\right ) \sin (3 (d+e x))+c \left (c^2-3 b^2\right ) \cos (3 (d+e x))}{12 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*(2*a^2 + 3*(b^2 + c^2))*(d + e*x) - 9*c*(4*a^2 + b^2 + c^2)*Cos[d + e*x] - 18*a*b*c*Cos[2*(d + e*x)] + c*
(-3*b^2 + c^2)*Cos[3*(d + e*x)] + 9*b*(4*a^2 + b^2 + c^2)*Sin[d + e*x] + 9*a*(b^2 - c^2)*Sin[2*(d + e*x)] + b*
(b^2 - 3*c^2)*Sin[3*(d + e*x)])/(12*e)

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fricas [A]  time = 2.94, size = 147, normalized size = 0.86 \[ -\frac {18 \, a b c \cos \left (e x + d\right )^{2} + 2 \, {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} - 3 \, {\left (2 \, a^{3} + 3 \, a b^{2} + 3 \, a c^{2}\right )} e x + 6 \, {\left (3 \, a^{2} c + c^{3}\right )} \cos \left (e x + d\right ) - {\left (18 \, a^{2} b + 4 \, b^{3} + 6 \, b c^{2} + 2 \, {\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{2} + 9 \, {\left (a b^{2} - a c^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{6 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(18*a*b*c*cos(e*x + d)^2 + 2*(3*b^2*c - c^3)*cos(e*x + d)^3 - 3*(2*a^3 + 3*a*b^2 + 3*a*c^2)*e*x + 6*(3*a^
2*c + c^3)*cos(e*x + d) - (18*a^2*b + 4*b^3 + 6*b*c^2 + 2*(b^3 - 3*b*c^2)*cos(e*x + d)^2 + 9*(a*b^2 - a*c^2)*c
os(e*x + d))*sin(e*x + d))/e

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giac [A]  time = 0.20, size = 167, normalized size = 0.98 \[ -\frac {3}{2} \, a b c \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} - \frac {1}{12} \, {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (3 \, x e + 3 \, d\right ) e^{\left (-1\right )} - \frac {3}{4} \, {\left (4 \, a^{2} c + b^{2} c + c^{3}\right )} \cos \left (x e + d\right ) e^{\left (-1\right )} + \frac {1}{12} \, {\left (b^{3} - 3 \, b c^{2}\right )} e^{\left (-1\right )} \sin \left (3 \, x e + 3 \, d\right ) + \frac {3}{4} \, {\left (a b^{2} - a c^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + \frac {3}{4} \, {\left (4 \, a^{2} b + b^{3} + b c^{2}\right )} e^{\left (-1\right )} \sin \left (x e + d\right ) + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2} + 3 \, a c^{2}\right )} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 177, normalized size = 1.04 \[ \frac {a^{3} \left (e x +d \right )+3 \sin \left (e x +d \right ) a^{2} b -3 a^{2} c \cos \left (e x +d \right )+3 a \,b^{2} \left (\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-3 a b c \left (\cos ^{2}\left (e x +d \right )\right )+3 a \,c^{2} \left (-\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+\frac {b^{3} \left (2+\cos ^{2}\left (e x +d \right )\right ) \sin \left (e x +d \right )}{3}-\left (\cos ^{3}\left (e x +d \right )\right ) b^{2} c +c^{2} b \left (\sin ^{3}\left (e x +d \right )\right )-\frac {c^{3} \left (2+\sin ^{2}\left (e x +d \right )\right ) \cos \left (e x +d \right )}{3}}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(a^3*(e*x+d)+3*sin(e*x+d)*a^2*b-3*a^2*c*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*a*b
*c*cos(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*b^3*(2+cos(e*x+d)^2)*sin(e*x+d)-cos(e*x
+d)^3*b^2*c+c^2*b*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 = 0.33, size = 189, normalized size = 1.11 \[ -\frac {b^{2} c \cos \left (e x + d\right )^{3}}{e} + \frac {b c^{2} \sin \left (e x + d\right )^{3}}{e} + a^{3} x - \frac {{\left (\sin \left (e x + d\right )^{3} - 3 \, \sin \left (e x + d\right )\right )} b^{3}}{3 \, e} + \frac {{\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} c^{3}}{3 \, e} - 3 \, a^{2} {\left (\frac {c \cos \left (e x + d\right )}{e} - \frac {b \sin \left (e x + d\right )}{e}\right )} - \frac {3}{4} \, {\left (\frac {4 \, b c \cos \left (e x + d\right )^{2}}{e} - \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{e} - \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{e}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.70, size = 333, normalized size = 1.96 \[ \frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a^2+3\,b^2+3\,c^2\right )}{2\,a^3+3\,a\,b^2+3\,a\,c^2}\right )\,\left (2\,a^2+3\,b^2+3\,c^2\right )}{e}-\frac {a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )-\frac {e\,x}{2}\right )\,\left (2\,a^2+3\,b^2+3\,c^2\right )}{e}-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (12\,a^2\,c-12\,b\,a\,c+4\,c^3\right )-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (12\,a^2\,b+\frac {4\,b^3}{3}+8\,b\,c^2\right )-\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (6\,a^2\,b+3\,a\,b^2-3\,a\,c^2+2\,b^3\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (6\,c\,a^2-12\,c\,a\,b+6\,c\,b^2\right )+6\,a^2\,c+2\,b^2\,c-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\left (6\,a^2\,b-3\,a\,b^2+3\,a\,c^2+2\,b^3\right )+\frac {4\,c^3}{3}}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*atan((a*tan(d/2 + (e*x)/2)*(2*a^2 + 3*b^2 + 3*c^2))/(3*a*b^2 + 3*a*c^2 + 2*a^3))*(2*a^2 + 3*b^2 + 3*c^2))/e
- (a*(atan(tan(d/2 + (e*x)/2)) - (e*x)/2)*(2*a^2 + 3*b^2 + 3*c^2))/e - (tan(d/2 + (e*x)/2)^2*(12*a^2*c + 4*c^
3 - 12*a*b*c) - tan(d/2 + (e*x)/2)^3*(12*a^2*b + 8*b*c^2 + (4*b^3)/3) - tan(d/2 + (e*x)/2)*(3*a*b^2 + 6*a^2*b
- 3*a*c^2 + 2*b^3) + tan(d/2 + (e*x)/2)^4*(6*a^2*c + 6*b^2*c - 12*a*b*c) + 6*a^2*c + 2*b^2*c - tan(d/2 + (e*x)
/2)^5*(6*a^2*b - 3*a*b^2 + 3*a*c^2 + 2*b^3) + (4*c^3)/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))

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sympy [A]  time = 0.76, size = 294, normalized size = 1.73 \[ \begin {cases} a^{3} x + \frac {3 a^{2} b \sin {\left (d + e x \right )}}{e} - \frac {3 a^{2} c \cos {\left (d + e x \right )}}{e} + \frac {3 a b^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {3 a b^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + \frac {3 a b^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} - \frac {3 a b c \cos ^{2}{\left (d + e x \right )}}{e} + \frac {3 a c^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {3 a c^{2} x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {3 a c^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} + \frac {2 b^{3} \sin ^{3}{\left (d + e x \right )}}{3 e} + \frac {b^{3} \sin {\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )}}{e} - \frac {b^{2} c \cos ^{3}{\left (d + e x \right )}}{e} + \frac {b c^{2} \sin ^{3}{\left (d + e x \right )}}{e} - \frac {c^{3} \sin ^{2}{\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {2 c^{3} \cos ^{3}{\left (d + e x \right )}}{3 e} & \text {for}\: e \neq 0 \\x \left (a + b \cos {\relax (d )} + c \sin {\relax (d )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*sin(d + e*x)/e - 3*a**2*c*cos(d + e*x)/e + 3*a*b**2*x*sin(d + e*x)**2/2 + 3*a*b**
2*x*cos(d + e*x)**2/2 + 3*a*b**2*sin(d + e*x)*cos(d + e*x)/(2*e) - 3*a*b*c*cos(d + e*x)**2/e + 3*a*c**2*x*sin(
d + e*x)**2/2 + 3*a*c**2*x*cos(d + e*x)**2/2 - 3*a*c**2*sin(d + e*x)*cos(d + e*x)/(2*e) + 2*b**3*sin(d + e*x)*
*3/(3*e) + b**3*sin(d + e*x)*cos(d + e*x)**2/e - b**2*c*cos(d + e*x)**3/e + b*c**2*sin(d + e*x)**3/e - c**3*si
n(d + e*x)**2*cos(d + e*x)/e - 2*c**3*cos(d + e*x)**3/(3*e), Ne(e, 0)), (x*(a + b*cos(d) + c*sin(d))**3, True)
)

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