Integrand size = 14, antiderivative size = 79 \[ \int x^3 \sin ^3\left (a+b x^2\right ) \, dx=-\frac {x^2 \cos \left (a+b x^2\right )}{3 b}+\frac {\sin \left (a+b x^2\right )}{3 b^2}-\frac {x^2 \cos \left (a+b x^2\right ) \sin ^2\left (a+b x^2\right )}{6 b}+\frac {\sin ^3\left (a+b x^2\right )}{18 b^2} \]
-1/3*x^2*cos(b*x^2+a)/b+1/3*sin(b*x^2+a)/b^2-1/6*x^2*cos(b*x^2+a)*sin(b*x^ 2+a)^2/b+1/18*sin(b*x^2+a)^3/b^2
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int x^3 \sin ^3\left (a+b x^2\right ) \, dx=-\frac {27 b x^2 \cos \left (a+b x^2\right )-3 b x^2 \cos \left (3 \left (a+b x^2\right )\right )-27 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )}{72 b^2} \]
-1/72*(27*b*x^2*Cos[a + b*x^2] - 3*b*x^2*Cos[3*(a + b*x^2)] - 27*Sin[a + b *x^2] + Sin[3*(a + b*x^2)])/b^2
Time = 0.40 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3860, 3042, 3791, 3042, 3777, 3042, 3117}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \sin ^3\left (a+b x^2\right ) \, dx\) |
\(\Big \downarrow \) 3860 |
\(\displaystyle \frac {1}{2} \int x^2 \sin ^3\left (b x^2+a\right )dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int x^2 \sin \left (b x^2+a\right )^3dx^2\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int x^2 \sin \left (b x^2+a\right )dx^2+\frac {\sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^2 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int x^2 \sin \left (b x^2+a\right )dx^2+\frac {\sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^2 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \left (\frac {\int \cos \left (b x^2+a\right )dx^2}{b}-\frac {x^2 \cos \left (a+b x^2\right )}{b}\right )+\frac {\sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^2 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \left (\frac {\int \sin \left (b x^2+a+\frac {\pi }{2}\right )dx^2}{b}-\frac {x^2 \cos \left (a+b x^2\right )}{b}\right )+\frac {\sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^2 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{2} \left (\frac {\sin ^3\left (a+b x^2\right )}{9 b^2}+\frac {2}{3} \left (\frac {\sin \left (a+b x^2\right )}{b^2}-\frac {x^2 \cos \left (a+b x^2\right )}{b}\right )-\frac {x^2 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\) |
(-1/3*(x^2*Cos[a + b*x^2]*Sin[a + b*x^2]^2)/b + Sin[a + b*x^2]^3/(9*b^2) + (2*(-((x^2*Cos[a + b*x^2])/b) + Sin[a + b*x^2]/b^2))/3)/2
3.1.24.3.1 Defintions of rubi rules used
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
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] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ (m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ (m + 1)/n], 0]))
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {3 x^{2} \cos \left (b \,x^{2}+a \right )}{8 b}+\frac {3 \sin \left (b \,x^{2}+a \right )}{8 b^{2}}+\frac {x^{2} \cos \left (3 b \,x^{2}+3 a \right )}{24 b}-\frac {\sin \left (3 b \,x^{2}+3 a \right )}{72 b^{2}}\) | \(66\) |
risch | \(-\frac {3 x^{2} \cos \left (b \,x^{2}+a \right )}{8 b}+\frac {3 \sin \left (b \,x^{2}+a \right )}{8 b^{2}}+\frac {x^{2} \cos \left (3 b \,x^{2}+3 a \right )}{24 b}-\frac {\sin \left (3 b \,x^{2}+3 a \right )}{72 b^{2}}\) | \(66\) |
parallelrisch | \(\frac {-27 \cos \left (b \,x^{2}+a \right ) b \,x^{2}+3 x^{2} \cos \left (3 b \,x^{2}+3 a \right ) b -24 b \,x^{2}-24 i \ln \left (\tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )-i\right )+24 i \ln \left (\tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )+i\right )+27 \sin \left (b \,x^{2}+a \right )-\sin \left (3 b \,x^{2}+3 a \right )}{72 b^{2}}\) | \(103\) |
norman | \(\frac {\frac {x^{2} \left (\tan ^{4}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{b}+\frac {2 \tan \left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )}{3 b^{2}}+\frac {16 \left (\tan ^{3}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{9 b^{2}}+\frac {2 \left (\tan ^{5}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{3 b^{2}}-\frac {x^{2}}{3 b}-\frac {x^{2} \left (\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{b}+\frac {x^{2} \left (\tan ^{6}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )}{3 b}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {b \,x^{2}}{2}\right )\right )^{3}}\) | \(142\) |
-3/8*x^2*cos(b*x^2+a)/b+3/8*sin(b*x^2+a)/b^2+1/24/b*x^2*cos(3*b*x^2+3*a)-1 /72/b^2*sin(3*b*x^2+3*a)
Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int x^3 \sin ^3\left (a+b x^2\right ) \, dx=\frac {3 \, b x^{2} \cos \left (b x^{2} + a\right )^{3} - 9 \, b x^{2} \cos \left (b x^{2} + a\right ) - {\left (\cos \left (b x^{2} + a\right )^{2} - 7\right )} \sin \left (b x^{2} + a\right )}{18 \, b^{2}} \]
1/18*(3*b*x^2*cos(b*x^2 + a)^3 - 9*b*x^2*cos(b*x^2 + a) - (cos(b*x^2 + a)^ 2 - 7)*sin(b*x^2 + a))/b^2
Time = 0.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int x^3 \sin ^3\left (a+b x^2\right ) \, dx=\begin {cases} - \frac {x^{2} \sin ^{2}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{2 b} - \frac {x^{2} \cos ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {7 \sin ^{3}{\left (a + b x^{2} \right )}}{18 b^{2}} + \frac {\sin {\left (a + b x^{2} \right )} \cos ^{2}{\left (a + b x^{2} \right )}}{3 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sin ^{3}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \]
Piecewise((-x**2*sin(a + b*x**2)**2*cos(a + b*x**2)/(2*b) - x**2*cos(a + b *x**2)**3/(3*b) + 7*sin(a + b*x**2)**3/(18*b**2) + sin(a + b*x**2)*cos(a + b*x**2)**2/(3*b**2), Ne(b, 0)), (x**4*sin(a)**3/4, True))
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.76 \[ \int x^3 \sin ^3\left (a+b x^2\right ) \, dx=\frac {3 \, b x^{2} \cos \left (3 \, b x^{2} + 3 \, a\right ) - 27 \, b x^{2} \cos \left (b x^{2} + a\right ) - \sin \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, \sin \left (b x^{2} + a\right )}{72 \, b^{2}} \]
1/72*(3*b*x^2*cos(3*b*x^2 + 3*a) - 27*b*x^2*cos(b*x^2 + a) - sin(3*b*x^2 + 3*a) + 27*sin(b*x^2 + a))/b^2
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19 \[ \int x^3 \sin ^3\left (a+b x^2\right ) \, dx=-\frac {{\left (\cos \left (b x^{2} + a\right )^{3} - 3 \, \cos \left (b x^{2} + a\right )\right )} a}{6 \, b^{2}} + \frac {3 \, {\left (b x^{2} + a\right )} \cos \left (3 \, b x^{2} + 3 \, a\right ) - 27 \, {\left (b x^{2} + a\right )} \cos \left (b x^{2} + a\right ) - \sin \left (3 \, b x^{2} + 3 \, a\right ) + 27 \, \sin \left (b x^{2} + a\right )}{72 \, b^{2}} \]
-1/6*(cos(b*x^2 + a)^3 - 3*cos(b*x^2 + a))*a/b^2 + 1/72*(3*(b*x^2 + a)*cos (3*b*x^2 + 3*a) - 27*(b*x^2 + a)*cos(b*x^2 + a) - sin(3*b*x^2 + 3*a) + 27* sin(b*x^2 + a))/b^2
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int x^3 \sin ^3\left (a+b x^2\right ) \, dx=\frac {\frac {7\,\sin \left (b\,x^2+a\right )}{18}-\frac {{\cos \left (b\,x^2+a\right )}^2\,\sin \left (b\,x^2+a\right )}{18}+b\,\left (\frac {x^2\,{\cos \left (b\,x^2+a\right )}^3}{6}-\frac {x^2\,\cos \left (b\,x^2+a\right )}{2}\right )}{b^2} \]