\(\int \cos (\frac {1}{2} (1+3 x)) \sin ^3(\frac {1}{2} (1+3 x)) \, dx\) [823]

Optimal result

Integrand size = 23, antiderivative size = 16 \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {1}{6} \sin ^4\left (\frac {1}{2}+\frac {3 x}{2}\right ) \]

[Out]

1/6*sin(1/2+3/2*x)^4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2644, 30} \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {1}{6} \sin ^4\left (\frac {3 x}{2}+\frac {1}{2}\right ) \]

[In]

Int[Cos[(1 + 3*x)/2]*Sin[(1 + 3*x)/2]^3,x]

[Out]

Sin[1/2 + (3*x)/2]^4/6

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} \text {Subst}\left (\int x^3 \, dx,x,\sin \left (\frac {1}{2}+\frac {3 x}{2}\right )\right ) \\ & = \frac {1}{6} \sin ^4\left (\frac {1}{2}+\frac {3 x}{2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {1}{2} \left (-\frac {1}{6} \cos (1+3 x)+\frac {1}{24} \cos (2+6 x)\right ) \]

[In]

Integrate[Cos[(1 + 3*x)/2]*Sin[(1 + 3*x)/2]^3,x]

[Out]

(-1/6*Cos[1 + 3*x] + Cos[2 + 6*x]/24)/2

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {\sin \left (\frac {1}{2}+\frac {3 x}{2}\right )^{4}}{6}\)	\(11\)
default	\(\frac {\sin \left (\frac {1}{2}+\frac {3 x}{2}\right )^{4}}{6}\)	\(11\)
risch	\(-\frac {\cos \left (1+3 x \right )}{12}+\frac {\cos \left (2+6 x \right )}{48}\)	\(18\)
parallelrisch	\(\frac {\cos \left (2+6 x \right )}{48}+\frac {1}{16}-\frac {\cos \left (1+3 x \right )}{12}\)	\(19\)
norman	\(\frac {8 \tan \left (\frac {1}{4}+\frac {3 x}{4}\right )^{4}}{3 \left (1+\tan \left (\frac {1}{4}+\frac {3 x}{4}\right )^{2}\right )^{4}}\)	\(23\)

[In]

int(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/6*sin(1/2+3/2*x)^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 21 vs. \(2 (10) = 20\).

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {1}{6} \, \cos \left (\frac {3}{2} \, x + \frac {1}{2}\right )^{4} - \frac {1}{3} \, \cos \left (\frac {3}{2} \, x + \frac {1}{2}\right )^{2} \]

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x, algorithm="fricas")

[Out]

1/6*cos(3/2*x + 1/2)^4 - 1/3*cos(3/2*x + 1/2)^2

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {\sin ^{4}{\left (\frac {3 x}{2} + \frac {1}{2} \right )}}{6} \]

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)**3,x)

[Out]

sin(3*x/2 + 1/2)**4/6

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {1}{6} \, \sin \left (\frac {3}{2} \, x + \frac {1}{2}\right )^{4} \]

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x, algorithm="maxima")

[Out]

1/6*sin(3/2*x + 1/2)^4

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {1}{6} \, \sin \left (\frac {3}{2} \, x + \frac {1}{2}\right )^{4} \]

[In]

integrate(cos(1/2+3/2*x)*sin(1/2+3/2*x)^3,x, algorithm="giac")

[Out]

1/6*sin(3/2*x + 1/2)^4

Mupad [B] (verification not implemented)

Time = 26.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \cos \left (\frac {1}{2} (1+3 x)\right ) \sin ^3\left (\frac {1}{2} (1+3 x)\right ) \, dx=\frac {{\left (\frac {\cos \left (3\,x+1\right )}{2}-\frac {1}{2}\right )}^2}{6} \]

[In]

int(cos((3*x)/2 + 1/2)*sin((3*x)/2 + 1/2)^3,x)

[Out]

(cos(3*x + 1)/2 - 1/2)^2/6