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

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

[Out]

a^3*x + (3*a*b^2*x)/2 + (b*(3*a^2 + b^2)*Sin[c + d*x])/d + (3*a*b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (b^3*Si
n[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0691565, 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*}

Mathematica [A]  time = 0.12404, size = 80, normalized size = 1.05 \[ \frac{9 b \left (4 a^2+b^2\right ) \sin (c+d x)+12 a^3 c+12 a^3 d x+9 a b^2 \sin (2 (c+d x))+18 a b^2 c+18 a b^2 d x+b^3 \sin (3 (c+d x))}{12 d} \]

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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Maple [A]  time = 0.032, size = 76, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+3\,a{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{2}b\sin \left ( dx+c \right ) +{a}^{3} \left ( dx+c \right ) \right ) } \end{align*}

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*a*b^2*(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 = 0.944881, size = 97, normalized size = 1.28 \begin{align*} 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} \end{align*}

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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Fricas [A]  time = 1.95134, size = 153, normalized size = 2.01 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 0.666712, size = 128, normalized size = 1.68 \begin{align*} \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{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.34079, size = 97, normalized size = 1.28 \begin{align*} \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} \end{align*}

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