Integrand size = 20, antiderivative size = 136 \[ \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\frac {\cos \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac {\operatorname {FresnelC}(b x) \operatorname {FresnelS}(b x)}{2 b^3 \pi }-\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {i x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )}{8 b \pi }+\frac {x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{b^2 \pi } \]
1/4*cos(b^2*Pi*x^2)/b^3/Pi^2-1/2*FresnelC(b*x)*FresnelS(b*x)/b^3/Pi-1/8*I* x^2*hypergeom([1, 1],[3/2, 2],-1/2*I*b^2*Pi*x^2)/b/Pi+1/8*I*x^2*hypergeom( [1, 1],[3/2, 2],1/2*I*b^2*Pi*x^2)/b/Pi+x*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2) /b^2/Pi
\[ \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx \]
Time = 0.43 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {7009, 3860, 3042, 3118, 7001}
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.
| \(\displaystyle \int x^2 \operatorname {FresnelC}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right ) \, dx\) |
| \(\Big \downarrow \) 7009 |
| \(\displaystyle -\frac {\int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\int x \sin \left (b^2 \pi x^2\right )dx}{2 \pi b}+\frac {x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\) |
| \(\Big \downarrow \) 3860 |
| \(\displaystyle -\frac {\int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\int \sin \left (b^2 \pi x^2\right )dx^2}{4 \pi b}+\frac {x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {\int \sin \left (b^2 \pi x^2\right )dx^2}{4 \pi b}+\frac {x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\) |
| \(\Big \downarrow \) 3118 |
| \(\displaystyle -\frac {\int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}+\frac {x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}\) |
| \(\Big \downarrow \) 7001 |
| \(\displaystyle -\frac {\frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )+\frac {\operatorname {FresnelC}(b x) \operatorname {FresnelS}(b x)}{2 b}}{\pi b^2}+\frac {x \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac {\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}\) |
Cos[b^2*Pi*x^2]/(4*b^3*Pi^2) - ((FresnelC[b*x]*FresnelS[b*x])/(2*b) + (I/8 )*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-1/2*I)*b^2*Pi*x^2] - (I/8)*b *x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2])/(b^2*Pi) + (x* FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)
3.2.86.3.1 Defintions of rubi rules used
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[FresnelC[(b_.)*(x_)]*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[b*Pi*FresnelC [b*x]*(FresnelS[b*x]/(4*d)), x] + (Simp[(1/8)*I*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I)*d*x^2], x] - Simp[(1/8)*I*b*x^2*HypergeometricPFQ[{1, 1 }, {3/2, 2}, I*d*x^2], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4]
Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[x^( m - 1)*Sin[d*x^2]*(FresnelC[b*x]/(2*d)), x] + (-Simp[(m - 1)/(2*d) Int[x^ (m - 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Simp[b/(4*d) Int[x^(m - 1)*Sin [2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && IGtQ[m, 1]
\[\int x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) \operatorname {FresnelC}\left (b x \right )d x\]
\[ \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int { x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) \,d x } \]
\[ \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )\, dx \]
\[ \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int { x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) \,d x } \]
\[ \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int { x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) \,d x } \]
Timed out. \[ \int x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelC}(b x) \, dx=\int x^2\,\mathrm {FresnelC}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \]