\(\int \frac {\text {Si}(b x)}{x} \, dx\) [6]

Optimal result

Integrand size = 8, antiderivative size = 43 \[ \int \frac {\text {Si}(b x)}{x} \, dx=\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;-i b x)+\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;i b x) \] Output:

1/2*b*x*hypergeom([1, 1, 1],[2, 2, 2],-I*b*x)+1/2*b*x*hypergeom([1, 1, 1], 
[2, 2, 2],I*b*x)
 

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Si}(b x)}{x} \, dx=\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;-i b x)+\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;i b x) \] Input:

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

Output:

(b*x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, (-I)*b*x])/2 + (b*x*Hypergeom 
etricPFQ[{1, 1, 1}, {2, 2, 2}, I*b*x])/2
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {7055}

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.

\(\displaystyle \int \frac {\text {Si}(b x)}{x} \, dx\)

\(\Big \downarrow \) 7055

\(\displaystyle \frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;-i b x)+\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;i b x)\)

Input:

Int[SinIntegral[b*x]/x,x]
 

Output:

(b*x*HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, (-I)*b*x])/2 + (b*x*Hypergeom 
etricPFQ[{1, 1, 1}, {2, 2, 2}, I*b*x])/2
 

Defintions of rubi rules used

rule 7055

Int[SinIntegral[(b_.)*(x_)]/(x_), x_Symbol] :> Simp[(1/2)*b*x*Hypergeometri 
cPFQ[{1, 1, 1}, {2, 2, 2}, (-I)*b*x], x] + Simp[(1/2)*b*x*HypergeometricPFQ 
[{1, 1, 1}, {2, 2, 2}, I*b*x], x] /; FreeQ[b, x]
 

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.47

method	result	size
meijerg	\(b x \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {1}{2}\right ], \left [\frac {3}{2}, \frac {3}{2}, \frac {3}{2}\right ], -\frac {b^{2} x^{2}}{4}\right )\)	\(20\)

Input:

int(Si(b*x)/x,x,method=_RETURNVERBOSE)
 

Output:

b*x*hypergeom([1/2,1/2],[3/2,3/2,3/2],-1/4*b^2*x^2)
 

Fricas [F]

\[ \int \frac {\text {Si}(b x)}{x} \, dx=\int { \frac {\operatorname {Si}\left (b x\right )}{x} \,d x } \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.51 \[ \int \frac {\text {Si}(b x)}{x} \, dx=b x {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{2}, \frac {1}{2} \\ \frac {3}{2}, \frac {3}{2}, \frac {3}{2} \end {matrix}\middle | {- \frac {b^{2} x^{2}}{4}} \right )} \] Input:

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

Output:

b*x*hyper((1/2, 1/2), (3/2, 3/2, 3/2), -b**2*x**2/4)
 

Maxima [F]

\[ \int \frac {\text {Si}(b x)}{x} \, dx=\int { \frac {\operatorname {Si}\left (b x\right )}{x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\text {Si}(b x)}{x} \, dx=\int { \frac {\operatorname {Si}\left (b x\right )}{x} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {Si}(b x)}{x} \, dx=\int \frac {\mathrm {sinint}\left (b\,x\right )}{x} \,d x \] Input:

int(sinint(b*x)/x,x)
 

Output:

int(sinint(b*x)/x, x)
 

Reduce [F]

\[ \int \frac {\text {Si}(b x)}{x} \, dx=\int \frac {\mathit {si} \left (b x \right )}{x}d x \] Input:

int(Si(b*x)/x,x)
 

Output:

int(si(b*x)/x,x)