Integrand size = 20, antiderivative size = 20 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=-\frac {b}{24 x^3}+\frac {b \cos \left (b^2 \pi x^2\right )}{24 x^3}+\frac {7 b^4 \pi ^2 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{24 \sqrt {2}}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{8 x^2}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 x^4}-\frac {7 b^3 \pi \sin \left (b^2 \pi x^2\right )}{48 x}-\frac {1}{8} b^4 \pi ^2 \text {Int}\left (\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x},x\right ) \]
-1/24*b/x^3+1/24*b*cos(b^2*Pi*x^2)/x^3-1/8*b^2*Pi*cos(1/2*b^2*Pi*x^2)*Fres nelS(b*x)/x^2-1/4*FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/x^4-7/48*b^3*Pi*sin(b^ 2*Pi*x^2)/x+7/48*b^4*Pi^2*FresnelC(b*x*2^(1/2))*2^(1/2)-1/8*b^4*Pi^2*Unint egrable(FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/x,x)
Not integrable
Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx \]
Not integrable
Time = 0.80 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {7010, 3869, 3868, 3833, 7018, 3868, 3833, 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.
\(\displaystyle \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 7010 |
\(\displaystyle \frac {1}{4} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^3}dx-\frac {1}{8} b \int \frac {\cos \left (b^2 \pi x^2\right )}{x^4}dx-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
\(\Big \downarrow \) 3869 |
\(\displaystyle \frac {1}{4} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^3}dx-\frac {1}{8} b \left (-\frac {2}{3} \pi b^2 \int \frac {\sin \left (b^2 \pi x^2\right )}{x^2}dx-\frac {\cos \left (\pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
\(\Big \downarrow \) 3868 |
\(\displaystyle \frac {1}{4} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^3}dx-\frac {1}{8} b \left (-\frac {2}{3} \pi b^2 \left (2 \pi b^2 \int \cos \left (b^2 \pi x^2\right )dx-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle \frac {1}{4} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{x^3}dx-\frac {1}{8} b \left (-\frac {2}{3} \pi b^2 \left (\sqrt {2} \pi b \operatorname {FresnelC}\left (\sqrt {2} b x\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
\(\Big \downarrow \) 7018 |
\(\displaystyle \frac {1}{4} \pi b^2 \left (-\frac {1}{2} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x}dx+\frac {1}{4} b \int \frac {\sin \left (b^2 \pi x^2\right )}{x^2}dx-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^2}\right )-\frac {1}{8} b \left (-\frac {2}{3} \pi b^2 \left (\sqrt {2} \pi b \operatorname {FresnelC}\left (\sqrt {2} b x\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
\(\Big \downarrow \) 3868 |
\(\displaystyle \frac {1}{4} \pi b^2 \left (-\frac {1}{2} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x}dx+\frac {1}{4} b \left (2 \pi b^2 \int \cos \left (b^2 \pi x^2\right )dx-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^2}\right )-\frac {1}{8} b \left (-\frac {2}{3} \pi b^2 \left (\sqrt {2} \pi b \operatorname {FresnelC}\left (\sqrt {2} b x\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle \frac {1}{4} \pi b^2 \left (-\frac {1}{2} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x}dx+\frac {1}{4} b \left (\sqrt {2} \pi b \operatorname {FresnelC}\left (\sqrt {2} b x\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^2}\right )-\frac {1}{8} b \left (-\frac {2}{3} \pi b^2 \left (\sqrt {2} \pi b \operatorname {FresnelC}\left (\sqrt {2} b x\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
\(\Big \downarrow \) 7012 |
\(\displaystyle \frac {1}{4} \pi b^2 \left (-\frac {1}{2} \pi b^2 \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x}dx+\frac {1}{4} b \left (\sqrt {2} \pi b \operatorname {FresnelC}\left (\sqrt {2} b x\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\operatorname {FresnelS}(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^2}\right )-\frac {1}{8} b \left (-\frac {2}{3} \pi b^2 \left (\sqrt {2} \pi b \operatorname {FresnelC}\left (\sqrt {2} b x\right )-\frac {\sin \left (\pi b^2 x^2\right )}{x}\right )-\frac {\cos \left (\pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac {b}{24 x^3}\) |
3.1.84.3.1 Defintions of rubi rules used
Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(e*x) ^(m + 1)*(Sin[c + d*x^n]/(e*(m + 1))), x] - Simp[d*(n/(e^n*(m + 1))) Int[ (e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & & LtQ[m, -1]
Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*x) ^(m + 1)*(Cos[c + d*x^n]/(e*(m + 1))), x] + Simp[d*(n/(e^n*(m + 1))) Int[ (e*x)^(m + n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & & LtQ[m, -1]
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 = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
\[\int \frac {\operatorname {FresnelS}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x^{5}}d x\]
Not integrable
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}} \,d x } \]
Not integrable
Time = 3.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{x^{5}}\, dx \]
Not integrable
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}} \,d x } \]
Not integrable
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}} \,d x } \]
Not integrable
Time = 4.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{x^5} \,d x \]