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

Optimal result

Integrand size = 8, antiderivative size = 38 \[ \int \frac {\text {Shi}(b x)}{x} \, dx=\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;-b x)+\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;b x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, 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 = {6665} \[ \int \frac {\text {Shi}(b x)}{x} \, dx=\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;-b x)+\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;b x) \]

[In]

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

[Out]

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

Rule 6665

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;-b x)+\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Shi}(b x)}{x} \, dx=\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;-b x)+\frac {1}{2} b x \, _3F_3(1,1,1;2,2,2;b x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.53

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

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\text {Shi}(b x)}{x} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x} \,d x } \]

[In]

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

[Out]

integral(sinh_integral(b*x)/x, x)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.53 \[ \int \frac {\text {Shi}(b x)}{x} \, dx=b x {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{2}, \frac {1}{2} \\ \frac {3}{2}, \frac {3}{2}, \frac {3}{2} \end {matrix}\middle | {\frac {b^{2} x^{2}}{4}} \right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\text {Shi}(b x)}{x} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x} \,d x } \]

[In]

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

[Out]

integrate(Shi(b*x)/x, x)

Giac [F]

\[ \int \frac {\text {Shi}(b x)}{x} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x} \,d x } \]

[In]

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

[Out]

integrate(Shi(b*x)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {Shi}(b x)}{x} \, dx=\int \frac {\mathrm {sinhint}\left (b\,x\right )}{x} \,d x \]

[In]

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

[Out]

int(sinhint(b*x)/x, x)