Integrand size = 20, antiderivative size = 20 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=-\frac {b}{40 x^4}+\frac {b \cos \left (b^2 \pi x^2\right )}{40 x^4}+\frac {1}{24} b^5 \pi ^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{15 x^3}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}-\frac {b^3 \pi \sin \left (b^2 \pi x^2\right )}{24 x^2}-\frac {1}{15} b^4 \pi ^2 \text {Int}\left (\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2},x\right ) \]
-1/40*b/x^4+1/24*b^5*Pi^2*Ci(b^2*Pi*x^2)+1/40*b*cos(b^2*Pi*x^2)/x^4-1/15*b ^2*Pi*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x^3-1/5*FresnelS(b*x)*sin(1/2*b^2* Pi*x^2)/x^5-1/24*b^3*Pi*sin(b^2*Pi*x^2)/x^2-1/15*b^4*Pi^2*Unintegrable(Fre snelS(b*x)*sin(1/2*b^2*Pi*x^2)/x^2,x)
Not integrable
Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx \]
Not integrable
Time = 1.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {7010, 3861, 3042, 3778, 25, 3042, 3778, 3042, 3783, 7018, 3860, 3042, 3778, 3042, 3783, 7012}
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.
\(\displaystyle \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 7010 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{10} b \int \frac {\cos \left (b^2 \pi x^2\right )}{x^5}dx-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3861 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \int \frac {\cos \left (b^2 \pi x^2\right )}{x^6}dx^2-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \int \frac {\sin \left (b^2 \pi x^2+\frac {\pi }{2}\right )}{x^6}dx^2-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \left (\frac {1}{2} \pi b^2 \int -\frac {\sin \left (b^2 \pi x^2\right )}{x^4}dx^2-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \int \frac {\sin \left (b^2 \pi x^2\right )}{x^4}dx^2-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \int \frac {\sin \left (b^2 \pi x^2\right )}{x^4}dx^2-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \int \frac {\cos \left (b^2 \pi x^2\right )}{x^2}dx^2-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \int \frac {\sin \left (b^2 \pi x^2+\frac {\pi }{2}\right )}{x^2}dx^2-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {1}{5} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^4}dx-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 7018 |
\(\displaystyle \frac {1}{5} \pi b^2 \left (-\frac {1}{3} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx+\frac {1}{6} b \int \frac {\sin \left (b^2 \pi x^2\right )}{x^3}dx-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3860 |
\(\displaystyle \frac {1}{5} \pi b^2 \left (-\frac {1}{3} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx+\frac {1}{12} b \int \frac {\sin \left (b^2 \pi x^2\right )}{x^4}dx^2-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \pi b^2 \left (-\frac {1}{3} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx+\frac {1}{12} b \int \frac {\sin \left (b^2 \pi x^2\right )}{x^4}dx^2-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {1}{5} \pi b^2 \left (-\frac {1}{3} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx+\frac {1}{12} b \left (\pi b^2 \int \frac {\cos \left (b^2 \pi x^2\right )}{x^2}dx^2-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \pi b^2 \left (-\frac {1}{3} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx+\frac {1}{12} b \left (\pi b^2 \int \frac {\sin \left (b^2 \pi x^2+\frac {\pi }{2}\right )}{x^2}dx^2-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {1}{5} \pi b^2 \left (-\frac {1}{3} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx+\frac {1}{12} b \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
\(\Big \downarrow \) 7012 |
\(\displaystyle \frac {1}{5} \pi b^2 \left (-\frac {1}{3} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx+\frac {1}{12} b \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {1}{20} b \left (-\frac {1}{2} \pi b^2 \left (\pi b^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x^2}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {b}{40 x^4}\) |
3.1.85.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 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[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(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[Simplify[ (m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ (m + 1)/n], 0]))
Int[FresnelS[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[x^( m + 1)*Sin[d*x^2]*(FresnelS[b*x]/(m + 1)), x] + (-Simp[d*(x^(m + 2)/(Pi*b*( m + 1)*(m + 2))), x] - Simp[2*(d/(m + 1)) Int[x^(m + 2)*Cos[d*x^2]*Fresne lS[b*x], x], x] + Simp[d/(Pi*b*(m + 1)) Int[x^(m + 1)*Cos[2*d*x^2], x], x ]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && ILtQ[m, -2]
Int[FresnelS[(a_.) + (b_.)*(x_)]^(n_.)*((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.) *(x_)^2], x_Symbol] :> Unintegrable[(e*x)^m*FresnelS[a + b*x]^n*Sin[c + d*x ^2], x] /; FreeQ[{a, b, c, d, e, m, n}, x]
Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[x^( m + 1)*Cos[d*x^2]*(FresnelS[b*x]/(m + 1)), x] + (Simp[2*(d/(m + 1)) Int[x ^(m + 2)*Sin[d*x^2]*FresnelS[b*x], x], x] - Simp[d/(Pi*b*(m + 1)) Int[x^( m + 1)*Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2/4)*b^4] && ILtQ[m, -1]
Not integrable
Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
\[\int \frac {\operatorname {FresnelS}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x^{6}}d x\]
Not integrable
Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{6}} \,d x } \]
Not integrable
Time = 5.69 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{x^{6}}\, dx \]
Not integrable
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{6}} \,d x } \]
Not integrable
Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{6}} \,d x } \]
Not integrable
Time = 4.83 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{x^6} \,d x \]