3.431 \(\int (a+b \cos (c+d x))^3 \, dx\)

Optimal. Leaf size=76 \[ a^3 x+\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)}{d}+\frac {3 a b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} a b^2 x-\frac {b^3 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 90, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2656, 2734} \[ \frac {2 b \left (4 a^2+b^2\right ) \sin (c+d x)}{3 d}+\frac {1}{2} a x \left (2 a^2+3 b^2\right )+\frac {5 a b^2 \sin (c+d x) \cos (c+d x)}{6 d}+\frac {b \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3,x]

[Out]

(a*(2*a^2 + 3*b^2)*x)/2 + (2*b*(4*a^2 + b^2)*Sin[c + d*x])/(3*d) + (5*a*b^2*Cos[c + d*x]*Sin[c + d*x])/(6*d) +
(b*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d)

Rule 2656

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

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^3 \, dx &=\frac {b (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int (a+b \cos (c+d x)) \left (3 a^2+2 b^2+5 a b \cos (c+d x)\right ) \, dx\\ &=\frac {1}{2} a \left (2 a^2+3 b^2\right ) x+\frac {2 b \left (4 a^2+b^2\right ) \sin (c+d x)}{3 d}+\frac {5 a b^2 \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 80, normalized size = 1.05 \[ \frac {12 a^3 c+12 a^3 d x+9 b \left (4 a^2+b^2\right ) \sin (c+d x)+9 a b^2 \sin (2 (c+d x))+18 a b^2 c+18 a b^2 d x+b^3 \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3,x]

[Out]

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

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fricas [A]  time = 0.69, size = 66, normalized size = 0.87 \[ \frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} d x + {\left (2 \, b^{3} \cos \left (d x + c\right )^{2} + 9 \, a b^{2} \cos \left (d x + c\right ) + 18 \, a^{2} b + 4 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.40, size = 72, normalized size = 0.95 \[ \frac {b^{3} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {3 \, a b^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} x + \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 76, normalized size = 1.00 \[ \frac {\frac {b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 b^{2} a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \sin \left (d x +c \right )+a^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 72, normalized size = 0.95 \[ a^{3} x + \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2}}{4 \, d} - \frac {{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{3}}{3 \, d} + \frac {3 \, a^{2} b \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.60, size = 77, normalized size = 1.01 \[ a^3\,x+\frac {3\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {3\,a\,b^2\,x}{2}+\frac {3\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,a^2\,b\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^3,x)

[Out]

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

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sympy [A]  time = 0.52, size = 128, normalized size = 1.68 \[ \begin {cases} a^{3} x + \frac {3 a^{2} b \sin {\left (c + d x \right )}}{d} + \frac {3 a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*sin(c + d*x)/d + 3*a*b**2*x*sin(c + d*x)**2/2 + 3*a*b**2*x*cos(c + d*x)**2/2 + 3*
a*b**2*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*b**3*sin(c + d*x)**3/(3*d) + b**3*sin(c + d*x)*cos(c + d*x)**2/d, N
e(d, 0)), (x*(a + b*cos(c))**3, True))

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