\(\int x^3 \cos ^3(a+b x^2) \, dx\) [15]

Optimal result

Integrand size = 14, antiderivative size = 79 \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\frac {\cos \left (a+b x^2\right )}{3 b^2}+\frac {\cos ^3\left (a+b x^2\right )}{18 b^2}+\frac {x^2 \sin \left (a+b x^2\right )}{3 b}+\frac {x^2 \cos ^2\left (a+b x^2\right ) \sin \left (a+b x^2\right )}{6 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3461, 3391, 3377, 2718} \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\frac {\cos ^3\left (a+b x^2\right )}{18 b^2}+\frac {\cos \left (a+b x^2\right )}{3 b^2}+\frac {x^2 \sin \left (a+b x^2\right )}{3 b}+\frac {x^2 \sin \left (a+b x^2\right ) \cos ^2\left (a+b x^2\right )}{6 b} \]

[In]

Int[x^3*Cos[a + b*x^2]^3,x]

[Out]

Cos[a + b*x^2]/(3*b^2) + Cos[a + b*x^2]^3/(18*b^2) + (x^2*Sin[a + b*x^2])/(3*b) + (x^2*Cos[a + b*x^2]^2*Sin[a
+ b*x^2])/(6*b)

Rule 2718

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

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3391

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

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]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x \cos ^3(a+b x) \, dx,x,x^2\right ) \\ & = \frac {\cos ^3\left (a+b x^2\right )}{18 b^2}+\frac {x^2 \cos ^2\left (a+b x^2\right ) \sin \left (a+b x^2\right )}{6 b}+\frac {1}{3} \text {Subst}\left (\int x \cos (a+b x) \, dx,x,x^2\right ) \\ & = \frac {\cos ^3\left (a+b x^2\right )}{18 b^2}+\frac {x^2 \sin \left (a+b x^2\right )}{3 b}+\frac {x^2 \cos ^2\left (a+b x^2\right ) \sin \left (a+b x^2\right )}{6 b}-\frac {\text {Subst}\left (\int \sin (a+b x) \, dx,x,x^2\right )}{3 b} \\ & = \frac {\cos \left (a+b x^2\right )}{3 b^2}+\frac {\cos ^3\left (a+b x^2\right )}{18 b^2}+\frac {x^2 \sin \left (a+b x^2\right )}{3 b}+\frac {x^2 \cos ^2\left (a+b x^2\right ) \sin \left (a+b x^2\right )}{6 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\frac {27 \cos \left (a+b x^2\right )+\cos \left (3 \left (a+b x^2\right )\right )+3 b x^2 \left (9 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )\right )}{72 b^2} \]

[In]

Integrate[x^3*Cos[a + b*x^2]^3,x]

[Out]

(27*Cos[a + b*x^2] + Cos[3*(a + b*x^2)] + 3*b*x^2*(9*Sin[a + b*x^2] + Sin[3*(a + b*x^2)]))/(72*b^2)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {3 x^{2} \sin \left (b \,x^{2}+a \right )}{8 b}+\frac {3 \cos \left (b \,x^{2}+a \right )}{8 b^{2}}+\frac {x^{2} \sin \left (3 b \,x^{2}+3 a \right )}{24 b}+\frac {\cos \left (3 b \,x^{2}+3 a \right )}{72 b^{2}}\)	\(66\)
risch	\(\frac {3 x^{2} \sin \left (b \,x^{2}+a \right )}{8 b}+\frac {3 \cos \left (b \,x^{2}+a \right )}{8 b^{2}}+\frac {x^{2} \sin \left (3 b \,x^{2}+3 a \right )}{24 b}+\frac {\cos \left (3 b \,x^{2}+3 a \right )}{72 b^{2}}\)	\(66\)
parallelrisch	\(\frac {7+9 \left (\tan ^{5}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right ) x^{2} b +6 \left (\tan ^{3}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right ) x^{2} b +9 \tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right ) x^{2} b +9 \left (\tan ^{4}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )+12 \left (\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{9 b^{2} \left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )^{3}}\)	\(110\)
norman	\(\frac {\frac {x^{2} \tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )}{b}+\frac {x^{2} \left (\tan ^{5}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{b}+\frac {\tan ^{4}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )}{b^{2}}+\frac {7}{9 b^{2}}+\frac {2 x^{2} \left (\tan ^{3}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{3 b}+\frac {4 \left (\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{3 b^{2}}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )^{3}}\)	\(119\)

[In]

int(x^3*cos(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

3/8*x^2*sin(b*x^2+a)/b+3/8*cos(b*x^2+a)/b^2+1/24/b*x^2*sin(3*b*x^2+3*a)+1/72/b^2*cos(3*b*x^2+3*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\frac {\cos \left (b x^{2} + a\right )^{3} + 3 \, {\left (b x^{2} \cos \left (b x^{2} + a\right )^{2} + 2 \, b x^{2}\right )} \sin \left (b x^{2} + a\right ) + 6 \, \cos \left (b x^{2} + a\right )}{18 \, b^{2}} \]

[In]

integrate(x^3*cos(b*x^2+a)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\begin {cases} \frac {x^{2} \sin ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {x^{2} \sin {\left (a + b x^{2} \right )} \cos ^{2}{\left (a + b x^{2} \right )}}{2 b} + \frac {\sin ^{2}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{3 b^{2}} + \frac {7 \cos ^{3}{\left (a + b x^{2} \right )}}{18 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \cos ^{3}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*cos(b*x**2+a)**3,x)

[Out]

Piecewise((x**2*sin(a + b*x**2)**3/(3*b) + x**2*sin(a + b*x**2)*cos(a + b*x**2)**2/(2*b) + sin(a + b*x**2)**2*
cos(a + b*x**2)/(3*b**2) + 7*cos(a + b*x**2)**3/(18*b**2), Ne(b, 0)), (x**4*cos(a)**3/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\frac {3 \, b x^{2} \sin \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, b x^{2} \sin \left (b x^{2} + a\right ) + \cos \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, \cos \left (b x^{2} + a\right )}{72 \, b^{2}} \]

[In]

integrate(x^3*cos(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/72*(3*b*x^2*sin(3*b*x^2 + 3*a) + 27*b*x^2*sin(b*x^2 + a) + cos(3*b*x^2 + 3*a) + 27*cos(b*x^2 + a))/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\frac {{\left (\sin \left (b x^{2} + a\right )^{3} - 3 \, \sin \left (b x^{2} + a\right )\right )} a}{6 \, b^{2}} + \frac {3 \, {\left (b x^{2} + a\right )} \sin \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, {\left (b x^{2} + a\right )} \sin \left (b x^{2} + a\right ) + \cos \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, \cos \left (b x^{2} + a\right )}{72 \, b^{2}} \]

[In]

integrate(x^3*cos(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/6*(sin(b*x^2 + a)^3 - 3*sin(b*x^2 + a))*a/b^2 + 1/72*(3*(b*x^2 + a)*sin(3*b*x^2 + 3*a) + 27*(b*x^2 + a)*sin(
b*x^2 + a) + cos(3*b*x^2 + 3*a) + 27*cos(b*x^2 + a))/b^2

Mupad [B] (verification not implemented)

Time = 13.93 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int x^3 \cos ^3\left (a+b x^2\right ) \, dx=\frac {\frac {\cos \left (b\,x^2+a\right )}{3}+\frac {{\cos \left (b\,x^2+a\right )}^3}{18}+b\,\left (\frac {x^2\,\sin \left (b\,x^2+a\right )}{3}+\frac {x^2\,{\cos \left (b\,x^2+a\right )}^2\,\sin \left (b\,x^2+a\right )}{6}\right )}{b^2} \]

[In]

int(x^3*cos(a + b*x^2)^3,x)

[Out]

(cos(a + b*x^2)/3 + cos(a + b*x^2)^3/18 + b*((x^2*sin(a + b*x^2))/3 + (x^2*cos(a + b*x^2)^2*sin(a + b*x^2))/6)
)/b^2