\(\int (a+a \cos (c+d x))^3 \, dx\) [26]

Optimal result

Integrand size = 12, antiderivative size = 63 \[ \int (a+a \cos (c+d x))^3 \, dx=\frac {5 a^3 x}{2}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2724, 2717, 2715, 8, 2713} \[ \int (a+a \cos (c+d x))^3 \, dx=-\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {5 a^3 x}{2} \]

[In]

Int[(a + a*Cos[c + d*x])^3,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

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

Rule 2717

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

Rule 2724

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx \\ & = a^3 x+a^3 \int \cos ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos (c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \, dx \\ & = a^3 x+\frac {3 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{2} \left (3 a^3\right ) \int 1 \, dx-\frac {a^3 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d} \\ & = \frac {5 a^3 x}{2}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int (a+a \cos (c+d x))^3 \, dx=\frac {a^3 (30 c+30 d x+45 \sin (c+d x)+9 \sin (2 (c+d x))+\sin (3 (c+d x)))}{12 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^3,x]

[Out]

(a^3*(30*c + 30*d*x + 45*Sin[c + d*x] + 9*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(12*d)

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {a^{3} \left (30 d x +\sin \left (3 d x +3 c \right )+9 \sin \left (2 d x +2 c \right )+45 \sin \left (d x +c \right )\right )}{12 d}\)	\(42\)
risch	\(\frac {5 a^{3} x}{2}+\frac {15 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 a^{3} \sin \left (2 d x +2 c \right )}{4 d}\)	\(56\)
derivativedivides	\(\frac {\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} \sin \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\)	\(74\)
default	\(\frac {\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} \sin \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\)	\(74\)
parts	\(a^{3} x +\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {3 a^{3} \sin \left (d x +c \right )}{d}+\frac {3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(75\)
norman	\(\frac {\frac {5 a^{3} x}{2}+\frac {11 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {40 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {5 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {15 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(130\)

[In]

int((a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/12*a^3*(30*d*x+sin(3*d*x+3*c)+9*sin(2*d*x+2*c)+45*sin(d*x+c))/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.79 \[ \int (a+a \cos (c+d x))^3 \, dx=\frac {15 \, a^{3} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 22 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(15*a^3*d*x + (2*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c) + 22*a^3)*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (58) = 116\).

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.92 \[ \int (a+a \cos (c+d x))^3 \, dx=\begin {cases} \frac {3 a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + a^{3} x + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 a^{3} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11 \[ \int (a+a \cos (c+d x))^3 \, dx=a^{3} x - \frac {{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3}}{3 \, d} + \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{4 \, d} + \frac {3 \, a^{3} \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int (a+a \cos (c+d x))^3 \, dx=\frac {5}{2} \, a^{3} x + \frac {a^{3} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {3 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {15 \, a^{3} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

5/2*a^3*x + 1/12*a^3*sin(3*d*x + 3*c)/d + 3/4*a^3*sin(2*d*x + 2*c)/d + 15/4*a^3*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 14.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^3 \, dx=\frac {5\,a^3\,x}{2}+\frac {11\,a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {3\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

[In]

int((a + a*cos(c + d*x))^3,x)

[Out]

(5*a^3*x)/2 + (11*a^3*sin(c + d*x))/(3*d) + (a^3*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (3*a^3*cos(c + d*x)*sin(
c + d*x))/(2*d)