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

Optimal result

Integrand size = 8, antiderivative size = 25 \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=b \operatorname {CosIntegral}(b x)-\frac {\sin (b x)}{x}-\frac {\text {Si}(b x)}{x} \]

[Out]

b*Ci(b*x)-Si(b*x)/x-sin(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 = {6638, 12, 3378, 3383} \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=b \operatorname {CosIntegral}(b x)-\frac {\text {Si}(b x)}{x}-\frac {\sin (b x)}{x} \]

[In]

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

[Out]

b*CosIntegral[b*x] - Sin[b*x]/x - SinIntegral[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 3383

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

Rule 6638

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinIntegr
al[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Sin[a + b*x]/(a + b*x)), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Si}(b x)}{x}+b \int \frac {\sin (b x)}{b x^2} \, dx \\ & = -\frac {\text {Si}(b x)}{x}+\int \frac {\sin (b x)}{x^2} \, dx \\ & = -\frac {\sin (b x)}{x}-\frac {\text {Si}(b x)}{x}+b \int \frac {\cos (b x)}{x} \, dx \\ & = b \operatorname {CosIntegral}(b x)-\frac {\sin (b x)}{x}-\frac {\text {Si}(b x)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=b \operatorname {CosIntegral}(b x)-\frac {\sin (b x)}{x}-\frac {\text {Si}(b x)}{x} \]

[In]

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

[Out]

b*CosIntegral[b*x] - Sin[b*x]/x - SinIntegral[b*x]/x

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
parts	\(-\frac {\operatorname {Si}\left (b x \right )}{x}+b \left (-\frac {\sin \left (b x \right )}{b x}+\operatorname {Ci}\left (b x \right )\right )\)	\(30\)
derivativedivides	\(b \left (-\frac {\operatorname {Si}\left (b x \right )}{b x}-\frac {\sin \left (b x \right )}{b x}+\operatorname {Ci}\left (b x \right )\right )\)	\(32\)
default	\(b \left (-\frac {\operatorname {Si}\left (b x \right )}{b x}-\frac {\sin \left (b x \right )}{b x}+\operatorname {Ci}\left (b x \right )\right )\)	\(32\)
meijerg	\(\frac {b \sqrt {\pi }\, \left (-\frac {2 b^{2} x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {3}{2}\right ], \left [2, 2, \frac {5}{2}, \frac {5}{2}\right ], -\frac {b^{2} x^{2}}{4}\right )}{9 \sqrt {\pi }}+\frac {8 \gamma -16+8 \ln \left (x \right )+8 \ln \left (b \right )}{\sqrt {\pi }}\right )}{8}\)	\(55\)

[In]

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

[Out]

-Si(b*x)/x+b*(-sin(b*x)/b/x+Ci(b*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=\frac {b x \operatorname {Ci}\left (b x\right ) - \sin \left (b x\right ) - \operatorname {Si}\left (b x\right )}{x} \]

[In]

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

[Out]

(b*x*cos_integral(b*x) - sin(b*x) - sin_integral(b*x))/x

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=- \frac {b^{3} x^{2} {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {3}{2} \\ 2, 2, \frac {5}{2}, \frac {5}{2} \end {matrix}\middle | {- \frac {b^{2} x^{2}}{4}} \right )}}{36} + \frac {b \log {\left (b^{2} x^{2} \right )}}{2} \]

[In]

integrate(Si(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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=\frac {1}{2} \, b {\left (\Gamma \left (-1, i \, b x\right ) + \Gamma \left (-1, -i \, b x\right )\right )} - \frac {\operatorname {Si}\left (b x\right )}{x} \]

[In]

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

[Out]

1/2*b*(gamma(-1, I*b*x) + gamma(-1, -I*b*x)) - sin_integral(b*x)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=\frac {b x \operatorname {Ci}\left (b x\right ) + b x \operatorname {Ci}\left (-b x\right ) - 2 \, \sin \left (b x\right )}{2 \, x} - \frac {\operatorname {Si}\left (b x\right )}{x} \]

[In]

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

[Out]

1/2*(b*x*cos_integral(b*x) + b*x*cos_integral(-b*x) - 2*sin(b*x))/x - sin_integral(b*x)/x

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {Si}(b x)}{x^2} \, dx=b\,\mathrm {cosint}\left (b\,x\right )-\frac {\mathrm {sinint}\left (b\,x\right )}{x}-\frac {\sin \left (b\,x\right )}{x} \]

[In]

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

[Out]

b*cosint(b*x) - sinint(b*x)/x - sin(b*x)/x