\(\int x^2 \cos (4 x^3) \cos (5 x^3) \, dx\) [775]

Optimal result

Integrand size = 16, antiderivative size = 19 \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=\frac {\sin \left (x^3\right )}{6}+\frac {1}{54} \sin \left (9 x^3\right ) \]

[Out]

1/6*sin(x^3)+1/54*sin(9*x^3)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {4668, 3461, 2717} \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=\frac {\sin \left (x^3\right )}{6}+\frac {1}{54} \sin \left (9 x^3\right ) \]

[In]

Int[x^2*Cos[4*x^3]*Cos[5*x^3],x]

[Out]

Sin[x^3]/6 + Sin[9*x^3]/54

Rule 2717

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

Rule 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 4668

Int[Cos[v_]^(p_.)*Cos[w_]^(q_.)*(x_)^(m_.), x_Symbol] :> Int[ExpandTrigReduce[x^m, Cos[v]^p*Cos[w]^q, x], x] /
; IGtQ[m, 0] && IGtQ[p, 0] && IGtQ[q, 0] && ((PolynomialQ[v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x]
&& IndependentQ[Cancel[v/w], x]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} x^2 \cos \left (x^3\right )+\frac {1}{2} x^2 \cos \left (9 x^3\right )\right ) \, dx \\ & = \frac {1}{2} \int x^2 \cos \left (x^3\right ) \, dx+\frac {1}{2} \int x^2 \cos \left (9 x^3\right ) \, dx \\ & = \frac {1}{6} \text {Subst}\left (\int \cos (x) \, dx,x,x^3\right )+\frac {1}{6} \text {Subst}\left (\int \cos (9 x) \, dx,x,x^3\right ) \\ & = \frac {\sin \left (x^3\right )}{6}+\frac {1}{54} \sin \left (9 x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=\frac {\sin \left (x^3\right )}{6}+\frac {1}{54} \sin \left (9 x^3\right ) \]

[In]

Integrate[x^2*Cos[4*x^3]*Cos[5*x^3],x]

[Out]

Sin[x^3]/6 + Sin[9*x^3]/54

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {\sin \left (x^{3}\right )}{6}+\frac {\sin \left (9 x^{3}\right )}{54}\)	\(16\)
risch	\(\frac {\sin \left (x^{3}\right )}{6}+\frac {\sin \left (9 x^{3}\right )}{54}\)	\(16\)
parallelrisch	\(\frac {\sin \left (x^{3}\right )}{6}+\frac {\sin \left (9 x^{3}\right )}{54}\)	\(16\)
norman	\(\frac {\frac {8 \tan \left (2 x^{3}\right ) \tan \left (\frac {5 x^{3}}{2}\right )^{2}}{27}-\frac {10 \tan \left (2 x^{3}\right )^{2} \tan \left (\frac {5 x^{3}}{2}\right )}{27}-\frac {8 \tan \left (2 x^{3}\right )}{27}+\frac {10 \tan \left (\frac {5 x^{3}}{2}\right )}{27}}{\left (1+\tan \left (2 x^{3}\right )^{2}\right ) \left (1+\tan \left (\frac {5 x^{3}}{2}\right )^{2}\right )}\)	\(75\)

[In]

int(x^2*cos(4*x^3)*cos(5*x^3),x,method=_RETURNVERBOSE)

[Out]

1/6*sin(x^3)+1/54*sin(9*x^3)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (15) = 30\).

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=\frac {1}{27} \, {\left (128 \, \cos \left (x^{3}\right )^{8} - 224 \, \cos \left (x^{3}\right )^{6} + 120 \, \cos \left (x^{3}\right )^{4} - 20 \, \cos \left (x^{3}\right )^{2} + 5\right )} \sin \left (x^{3}\right ) \]

[In]

integrate(x^2*cos(4*x^3)*cos(5*x^3),x, algorithm="fricas")

[Out]

1/27*(128*cos(x^3)^8 - 224*cos(x^3)^6 + 120*cos(x^3)^4 - 20*cos(x^3)^2 + 5)*sin(x^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (14) = 28\).

Time = 0.60 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=- \frac {4 \sin {\left (4 x^{3} \right )} \cos {\left (5 x^{3} \right )}}{27} + \frac {5 \sin {\left (5 x^{3} \right )} \cos {\left (4 x^{3} \right )}}{27} \]

[In]

integrate(x**2*cos(4*x**3)*cos(5*x**3),x)

[Out]

-4*sin(4*x**3)*cos(5*x**3)/27 + 5*sin(5*x**3)*cos(4*x**3)/27

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=\frac {1}{54} \, \sin \left (9 \, x^{3}\right ) + \frac {1}{6} \, \sin \left (x^{3}\right ) \]

[In]

integrate(x^2*cos(4*x^3)*cos(5*x^3),x, algorithm="maxima")

[Out]

1/54*sin(9*x^3) + 1/6*sin(x^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (15) = 30\).

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05 \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=\frac {128}{27} \, \sin \left (x^{3}\right )^{9} - \frac {32}{3} \, \sin \left (x^{3}\right )^{7} + 8 \, \sin \left (x^{3}\right )^{5} - \frac {20}{9} \, \sin \left (x^{3}\right )^{3} + \frac {1}{3} \, \sin \left (x^{3}\right ) \]

[In]

integrate(x^2*cos(4*x^3)*cos(5*x^3),x, algorithm="giac")

[Out]

128/27*sin(x^3)^9 - 32/3*sin(x^3)^7 + 8*sin(x^3)^5 - 20/9*sin(x^3)^3 + 1/3*sin(x^3)

Mupad [B] (verification not implemented)

Time = 26.65 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^2 \cos \left (4 x^3\right ) \cos \left (5 x^3\right ) \, dx=\frac {\sin \left (x^3\right )}{6}+\frac {\sin \left (9\,x^3\right )}{54} \]

[In]

int(x^2*cos(4*x^3)*cos(5*x^3),x)

[Out]

sin(x^3)/6 + sin(9*x^3)/54