3.2.18 \(\int \frac {\operatorname {FresnelC}(b x)}{x} \, dx\) [118]

3.2.18.1 Optimal result

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)
 

3.2.18.2 Mathematica [F]

\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int \frac {\operatorname {FresnelC}(b x)}{x} \, dx \]

input

Integrate[FresnelC[b*x]/x,x]
 

output

Integrate[FresnelC[b*x]/x, x]
 

3.2.18.3 Rubi [A] (verified)

Time = 0.31 (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
 

3.2.18.3.1 Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 6912

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]
 

rule 6914

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]
 

rule 6979

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]
 

3.2.18.4 Maple [A] (verified)

Time = 0.46 (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)
 

3.2.18.5 Fricas [F]

\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x} \,d x } \]

input

integrate(fresnel_cos(b*x)/x,x, algorithm="fricas")
 

output

integral(fresnel_cos(b*x)/x, x)
 

3.2.18.6 Sympy [A] (verification not implemented)

Time = 0.34 (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)
 

3.2.18.7 Maxima [F]

\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x} \,d x } \]

input

integrate(fresnel_cos(b*x)/x,x, algorithm="maxima")
 

output

integrate(fresnel_cos(b*x)/x, x)
 

3.2.18.8 Giac [F]

\[ \int \frac {\operatorname {FresnelC}(b x)}{x} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x} \,d x } \]

input

integrate(fresnel_cos(b*x)/x,x, algorithm="giac")
 

output

integrate(fresnel_cos(b*x)/x, x)
 

3.2.18.9 Mupad [F(-1)]

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)