\(\int \operatorname {FresnelC}(b x) \sin (\frac {1}{2} b^2 \pi x^2) \, dx\) [208]

Optimal result

Integrand size = 17, antiderivative size = 80 \[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\frac {\operatorname {FresnelC}(b x) \operatorname {FresnelS}(b x)}{2 b}+\frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right ) \] Output:

1/2*FresnelC(b*x)*FresnelS(b*x)/b+1/8*I*b*x^2*hypergeom([1, 1],[3/2, 2],-1 
/2*I*b^2*Pi*x^2)-1/8*I*b*x^2*hypergeom([1, 1],[3/2, 2],1/2*I*b^2*Pi*x^2)
 

Mathematica [F]

\[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \] Input:

Integrate[FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]
 

Output:

Integrate[FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2], x]
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {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.

\(\displaystyle \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right ) \, dx\)

\(\Big \downarrow \) 7001

\(\displaystyle \frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{8} i b x^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i b^2 \pi x^2\right )+\frac {\operatorname {FresnelC}(b x) \operatorname {FresnelS}(b x)}{2 b}\)

Input:

Int[FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]
 

Output:

(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]
 

Defintions of rubi rules used

rule 7001

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]
 

Maple [F]

\[\int \operatorname {FresnelC}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )d x\]

Input:

int(FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)
 

Output:

int(FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)
 

Fricas [F]

\[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \] Input:

integrate(fresnel_cos(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="fricas")
 

Output:

integral(fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2), x)
 

Sympy [F]

\[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )\, dx \] Input:

integrate(fresnelc(b*x)*sin(1/2*b**2*pi*x**2),x)
 

Output:

Integral(sin(pi*b**2*x**2/2)*fresnelc(b*x), x)
 

Maxima [F]

\[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \] Input:

integrate(fresnel_cos(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="maxima")
 

Output:

integrate(fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2), x)
 

Giac [F]

\[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int { \operatorname {C}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \,d x } \] Input:

integrate(fresnel_cos(b*x)*sin(1/2*b^2*pi*x^2),x, algorithm="giac")
 

Output:

integrate(fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int \mathrm {FresnelC}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right ) \,d x \] Input:

int(FresnelC(b*x)*sin((Pi*b^2*x^2)/2),x)
 

Output:

int(FresnelC(b*x)*sin((Pi*b^2*x^2)/2), x)
 

Reduce [F]

\[ \int \operatorname {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx=\int \mathrm {FresnelC}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )d x \] Input:

int(FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)
 

Output:

int(FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)