Integrand size = 8, antiderivative size = 59 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2} \] Output:
1/3*x^2*cos(1/2*b^2*Pi*x^2)/b/Pi+1/3*x^3*FresnelS(b*x)-2/3*sin(1/2*b^2*Pi* x^2)/b^3/Pi^2
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b \pi }+\frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{3 b^3 \pi ^2} \] Input:
Integrate[x^2*FresnelS[b*x],x]
Output:
(x^2*Cos[(b^2*Pi*x^2)/2])/(3*b*Pi) + (x^3*FresnelS[b*x])/3 - (2*Sin[(b^2*P i*x^2)/2])/(3*b^3*Pi^2)
Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6980, 3860, 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.
\(\displaystyle \int x^2 \operatorname {FresnelS}(b x) \, dx\) |
\(\Big \downarrow \) 6980 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{3} b \int x^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx\) |
\(\Big \downarrow \) 3860 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{6} b \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{6} b \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx^2\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{6} b \left (\frac {2 \int \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx^2}{\pi b^2}-\frac {2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{6} b \left (\frac {2 \int \sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )dx^2}{\pi b^2}-\frac {2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{3} x^3 \operatorname {FresnelS}(b x)-\frac {1}{6} b \left (\frac {4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac {2 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
Input:
Int[x^2*FresnelS[b*x],x]
Output:
(x^3*FresnelS[b*x])/3 - (b*((-2*x^2*Cos[(b^2*Pi*x^2)/2])/(b^2*Pi) + (4*Sin [(b^2*Pi*x^2)/2])/(b^4*Pi^2)))/6
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[(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]))
Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1 )*(FresnelS[b*x]/(d*(m + 1))), x] - Simp[b/(d*(m + 1)) Int[(d*x)^(m + 1)* Sin[(Pi/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]
Time = 0.53 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{3} x^{3}}{3}+\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\) | \(54\) |
default | \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{3} x^{3}}{3}+\frac {b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\) | \(54\) |
parts | \(\frac {x^{3} \operatorname {FresnelS}\left (b x \right )}{3}-\frac {b \left (-\frac {x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}\right )}{3}\) | \(54\) |
meijerg | \(\frac {\frac {\sqrt {\pi }\, x^{2} b^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3}-\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \sqrt {\pi }}+\frac {\pi ^{\frac {3}{2}} x^{3} b^{3} \operatorname {FresnelS}\left (b x \right )}{3}}{b^{3} \pi ^{\frac {3}{2}}}\) | \(60\) |
Input:
int(x^2*FresnelS(b*x),x,method=_RETURNVERBOSE)
Output:
1/b^3*(1/3*FresnelS(b*x)*b^3*x^3+1/3/Pi*b^2*x^2*cos(1/2*b^2*Pi*x^2)-2/3/Pi ^2*sin(1/2*b^2*Pi*x^2))
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {\pi ^{2} b^{3} x^{3} \operatorname {S}\left (b x\right ) + \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \] Input:
integrate(x^2*fresnel_sin(b*x),x, algorithm="fricas")
Output:
1/3*(pi^2*b^3*x^3*fresnel_sin(b*x) + pi*b^2*x^2*cos(1/2*pi*b^2*x^2) - 2*si n(1/2*pi*b^2*x^2))/(pi^2*b^3)
Time = 0.59 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {x^{3} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{4 \pi b \Gamma \left (\frac {7}{4}\right )} - \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{2 \pi ^{2} b^{3} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate(x**2*fresnels(b*x),x)
Output:
x**3*fresnels(b*x)*gamma(3/4)/(4*gamma(7/4)) + x**2*cos(pi*b**2*x**2/2)*ga mma(3/4)/(4*pi*b*gamma(7/4)) - sin(pi*b**2*x**2/2)*gamma(3/4)/(2*pi**2*b** 3*gamma(7/4))
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {S}\left (b x\right ) + \frac {\pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{3 \, \pi ^{2} b^{3}} \] Input:
integrate(x^2*fresnel_sin(b*x),x, algorithm="maxima")
Output:
1/3*x^3*fresnel_sin(b*x) + 1/3*(pi*b^2*x^2*cos(1/2*pi*b^2*x^2) - 2*sin(1/2 *pi*b^2*x^2))/(pi^2*b^3)
\[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\int { x^{2} \operatorname {S}\left (b x\right ) \,d x } \] Input:
integrate(x^2*fresnel_sin(b*x),x, algorithm="giac")
Output:
integrate(x^2*fresnel_sin(b*x), x)
Timed out. \[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\int x^2\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \] Input:
int(x^2*FresnelS(b*x),x)
Output:
int(x^2*FresnelS(b*x), x)
\[ \int x^2 \operatorname {FresnelS}(b x) \, dx=\int x^{2} \mathrm {FresnelS}\left (b x \right )d x \] Input:
int(x^2*FresnelS(b*x),x)
Output:
int(x^2*FresnelS(b*x),x)