Integrand size = 8, antiderivative size = 69 \[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\frac {1}{2} b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )+\frac {1}{2} b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {1}{2} i b^2 \pi x^2\right ) \] Output:
1/2*b*x*hypergeom([1/2, 1/2],[3/2, 3/2],-1/2*I*b^2*Pi*x^2)+1/2*b*x*hyperge om([1/2, 1/2],[3/2, 3/2],1/2*I*b^2*Pi*x^2)
\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int \frac {\operatorname {FresnelC}(b x)}{x} \, dx \] Input:
Integrate[FresnelC[b*x]/x,x]
Output:
Integrate[FresnelC[b*x]/x, x]
Time = 0.32 (sec) , antiderivative size = 69, 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 = {6979, 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.
\(\displaystyle \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx\) |
\(\Big \downarrow \) 6979 |
\(\displaystyle \left (\frac {1}{4}-\frac {i}{4}\right ) \int \frac {\text {erf}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx+\left (\frac {1}{4}+\frac {i}{4}\right ) \int -\frac {i \text {erfi}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \left (\frac {1}{4}-\frac {i}{4}\right ) \int \frac {\text {erfi}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x}dx+\frac {1}{2} b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )\) |
\(\Big \downarrow \) 6914 |
\(\displaystyle \frac {1}{2} b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )+\frac {1}{2} b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {1}{2} i b^2 \pi x^2\right )\) |
Input:
Int[FresnelC[b*x]/x,x]
Output:
(b*x*HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2}, (-1/2*I)*b^2*Pi*x^2])/2 + ( b*x*HypergeometricPFQ[{1/2, 1/2}, {3/2, 3/2}, (I/2)*b^2*Pi*x^2])/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[FresnelC[(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 = 0.51 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.33
method | result | size |
meijerg | \(b x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {1}{2}, \frac {5}{4}, \frac {5}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )\) | \(23\) |
Input:
int(FresnelC(b*x)/x,x,method=_RETURNVERBOSE)
Output:
b*x*hypergeom([1/4,1/4],[1/2,5/4,5/4],-1/16*x^4*Pi^2*b^4)
\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x} \,d x } \] Input:
integrate(fresnel_cos(b*x)/x,x, algorithm="fricas")
Output:
integral(fresnel_cos(b*x)/x, x)
Time = 0.35 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.59 \[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\frac {b x \Gamma ^{2}\left (\frac {1}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {1}{2}, \frac {5}{4}, \frac {5}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 \Gamma ^{2}\left (\frac {5}{4}\right )} \] Input:
integrate(fresnelc(b*x)/x,x)
Output:
b*x*gamma(1/4)**2*hyper((1/4, 1/4), (1/2, 5/4, 5/4), -pi**2*b**4*x**4/16)/ (16*gamma(5/4)**2)
\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x} \,d x } \] Input:
integrate(fresnel_cos(b*x)/x,x, algorithm="maxima")
Output:
integrate(fresnel_cos(b*x)/x, x)
\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x} \,d x } \] Input:
integrate(fresnel_cos(b*x)/x,x, algorithm="giac")
Output:
integrate(fresnel_cos(b*x)/x, x)
Timed out. \[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int \frac {\mathrm {FresnelC}\left (b\,x\right )}{x} \,d x \] Input:
int(FresnelC(b*x)/x,x)
Output:
int(FresnelC(b*x)/x, x)
\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int \frac {\mathrm {FresnelC}\left (b x \right )}{x}d x \] Input:
int(FresnelC(b*x)/x,x)
Output:
int(FresnelC(b*x)/x,x)