Integrand size = 8, antiderivative size = 73 \[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {1}{2} i b^2 \pi x^2\right ) \] Output:
1/2*I*b*x*hypergeom([1/2, 1/2],[3/2, 3/2],-1/2*I*b^2*Pi*x^2)-1/2*I*b*x*hyp ergeom([1/2, 1/2],[3/2, 3/2],1/2*I*b^2*Pi*x^2)
\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int \frac {\operatorname {FresnelS}(b x)}{x} \, dx \] Input:
Integrate[FresnelS[b*x]/x,x]
Output:
Integrate[FresnelS[b*x]/x, x]
Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6978, 26, 6912, 6914}
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.
\(\displaystyle \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx\) |
\(\Big \downarrow \) 6978 |
\(\displaystyle \left (\frac {1}{4}+\frac {i}{4}\right ) \int \frac {\text {erf}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx+\left (\frac {1}{4}-\frac {i}{4}\right ) \int -\frac {i \text {erfi}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \left (\frac {1}{4}+\frac {i}{4}\right ) \int \frac {\text {erf}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx-\left (\frac {1}{4}+\frac {i}{4}\right ) \int \frac {\text {erfi}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx\) |
\(\Big \downarrow \) 6912 |
\(\displaystyle \frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )-\left (\frac {1}{4}+\frac {i}{4}\right ) \int \frac {\text {erfi}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx\) |
\(\Big \downarrow \) 6914 |
\(\displaystyle \frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {1}{2} i b^2 \pi x^2\right )\) |
Input:
Int[FresnelS[b*x]/x,x]
Output:
(I/2)*b*x*HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2}, (-1/2*I)*b^2*Pi*x^2] - (I/2)*b*x*HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2}, (I/2)*b^2*Pi*x^2]
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[Erf[(b_.)*(x_)]/(x_), x_Symbol] :> Simp[2*b*(x/Sqrt[Pi])*Hypergeometric PFQ[{1/2, 1/2}, {3/2, 3/2}, (-b^2)*x^2], x] /; FreeQ[b, x]
Int[Erfi[(b_.)*(x_)]/(x_), x_Symbol] :> Simp[2*b*(x/Sqrt[Pi])*Hypergeometri cPFQ[{1/2, 1/2}, {3/2, 3/2}, b^2*x^2], x] /; FreeQ[b, x]
Int[FresnelS[(b_.)*(x_)]/(x_), x_Symbol] :> Simp[(1 + I)/4 Int[Erf[(Sqrt[ Pi]/2)*(1 + I)*b*x]/x, x], x] + Simp[(1 - I)/4 Int[Erf[(Sqrt[Pi]/2)*(1 - I)*b*x]/x, x], x] /; FreeQ[b, x]
Time = 0.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.40
method | result | size |
meijerg | \(\frac {\pi \,x^{3} b^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {3}{2}, \frac {7}{4}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{18}\) | \(29\) |
Input:
int(FresnelS(b*x)/x,x,method=_RETURNVERBOSE)
Output:
1/18*Pi*x^3*b^3*hypergeom([3/4,3/4],[3/2,7/4,7/4],-1/16*x^4*Pi^2*b^4)
\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x} \,d x } \] Input:
integrate(fresnel_sin(b*x)/x,x, algorithm="fricas")
Output:
integral(fresnel_sin(b*x)/x, x)
Time = 0.37 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63 \[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\frac {\pi b^{3} x^{3} \Gamma ^{2}\left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {3}{2}, \frac {7}{4}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma ^{2}\left (\frac {7}{4}\right )} \] Input:
integrate(fresnels(b*x)/x,x)
Output:
pi*b**3*x**3*gamma(3/4)**2*hyper((3/4, 3/4), (3/2, 7/4, 7/4), -pi**2*b**4* x**4/16)/(32*gamma(7/4)**2)
\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x} \,d x } \] Input:
integrate(fresnel_sin(b*x)/x,x, algorithm="maxima")
Output:
integrate(fresnel_sin(b*x)/x, x)
\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x} \,d x } \] Input:
integrate(fresnel_sin(b*x)/x,x, algorithm="giac")
Output:
integrate(fresnel_sin(b*x)/x, x)
Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x} \,d x \] Input:
int(FresnelS(b*x)/x,x)
Output:
int(FresnelS(b*x)/x, x)
\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int \frac {\mathrm {FresnelS}\left (b x \right )}{x}d x \] Input:
int(FresnelS(b*x)/x,x)
Output:
int(FresnelS(b*x)/x,x)