\(\int \frac {\text {Shi}(b x)}{x^2} \, dx\) [7]

Optimal result

Integrand size = 8, antiderivative size = 25 \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=b \text {Chi}(b x)-\frac {\sinh (b x)}{x}-\frac {\text {Shi}(b x)}{x} \]

[Out]

b*Chi(b*x)-Shi(b*x)/x-sinh(b*x)/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, 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 = {6667, 12, 3378, 3382} \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=b \text {Chi}(b x)-\frac {\text {Shi}(b x)}{x}-\frac {\sinh (b x)}{x} \]

[In]

Int[SinhIntegral[b*x]/x^2,x]

[Out]

b*CoshIntegral[b*x] - Sinh[b*x]/x - SinhIntegral[b*x]/x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3378

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m
+ 1))), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 6667

Int[((c_.) + (d_.)*(x_))^(m_.)*SinhIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinhInte
gral[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Sinh[a + b*x]/(a + b*x)), x], x] /
; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Shi}(b x)}{x}+b \int \frac {\sinh (b x)}{b x^2} \, dx \\ & = -\frac {\text {Shi}(b x)}{x}+\int \frac {\sinh (b x)}{x^2} \, dx \\ & = -\frac {\sinh (b x)}{x}-\frac {\text {Shi}(b x)}{x}+b \int \frac {\cosh (b x)}{x} \, dx \\ & = b \text {Chi}(b x)-\frac {\sinh (b x)}{x}-\frac {\text {Shi}(b x)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=b \text {Chi}(b x)-\frac {\sinh (b x)}{x}-\frac {\text {Shi}(b x)}{x} \]

[In]

Integrate[SinhIntegral[b*x]/x^2,x]

[Out]

b*CoshIntegral[b*x] - Sinh[b*x]/x - SinhIntegral[b*x]/x

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
parts	\(-\frac {\operatorname {Shi}\left (b x \right )}{x}+b \left (-\frac {\sinh \left (b x \right )}{b x}+\operatorname {Chi}\left (b x \right )\right )\)	\(30\)
derivativedivides	\(b \left (-\frac {\operatorname {Shi}\left (b x \right )}{b x}-\frac {\sinh \left (b x \right )}{b x}+\operatorname {Chi}\left (b x \right )\right )\)	\(32\)
default	\(b \left (-\frac {\operatorname {Shi}\left (b x \right )}{b x}-\frac {\sinh \left (b x \right )}{b x}+\operatorname {Chi}\left (b x \right )\right )\)	\(32\)
meijerg	\(\frac {\sqrt {\pi }\, b \left (\frac {16}{\sqrt {\pi }}-\frac {4 \,{\mathrm e}^{b x}}{\sqrt {\pi }\, b x}+\frac {4 \,{\mathrm e}^{-b x}}{\sqrt {\pi }\, b x}-\frac {4 \left (-9 b x +9\right ) \left (-\gamma -\ln \left (-b x \right )-\operatorname {Ei}_{1}\left (-b x \right )\right )}{9 \sqrt {\pi }\, b x}+\frac {4 \left (9 b x +9\right ) \left (-\gamma -\ln \left (b x \right )-\operatorname {Ei}_{1}\left (b x \right )\right )}{9 \sqrt {\pi }\, b x}+\frac {8 \gamma -16+8 \ln \left (x \right )+8 \ln \left (i b \right )}{\sqrt {\pi }}\right )}{8}\)	\(135\)

[In]

int(Shi(b*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-Shi(b*x)/x+b*(-sinh(b*x)/b/x+Chi(b*x))

Fricas [F]

\[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(Shi(b*x)/x^2,x, algorithm="fricas")

[Out]

integral(sinh_integral(b*x)/x^2, x)

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\frac {b^{3} x^{2} {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {3}{2} \\ 2, 2, \frac {5}{2}, \frac {5}{2} \end {matrix}\middle | {\frac {b^{2} x^{2}}{4}} \right )}}{36} + \frac {b \log {\left (b^{2} x^{2} \right )}}{2} \]

[In]

integrate(Shi(b*x)/x**2,x)

[Out]

b**3*x**2*hyper((1, 1, 3/2), (2, 2, 5/2, 5/2), b**2*x**2/4)/36 + b*log(b**2*x**2)/2

Maxima [F]

\[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(Shi(b*x)/x^2,x, algorithm="maxima")

[Out]

integrate(Shi(b*x)/x^2, x)

Giac [F]

\[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int { \frac {{\rm Shi}\left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(Shi(b*x)/x^2,x, algorithm="giac")

[Out]

integrate(Shi(b*x)/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {Shi}(b x)}{x^2} \, dx=\int \frac {\mathrm {sinhint}\left (b\,x\right )}{x^2} \,d x \]

[In]

int(sinhint(b*x)/x^2,x)

[Out]

int(sinhint(b*x)/x^2, x)