\(\int \frac {\text {Si}(b x)}{x^3} \, dx\) [8]

Optimal result

Integrand size = 8, antiderivative size = 46 \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=-\frac {b \cos (b x)}{4 x}-\frac {\sin (b x)}{4 x^2}-\frac {1}{4} b^2 \text {Si}(b x)-\frac {\text {Si}(b x)}{2 x^2} \] Output:

-1/4*b*cos(b*x)/x-1/4*sin(b*x)/x^2-1/4*b^2*Si(b*x)-1/2*Si(b*x)/x^2
 

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=-\frac {b \cos (b x)}{4 x}-\frac {\sin (b x)}{4 x^2}-\frac {1}{4} b^2 \text {Si}(b x)-\frac {\text {Si}(b x)}{2 x^2} \] Input:

Integrate[SinIntegral[b*x]/x^3,x]
 

Output:

-1/4*(b*Cos[b*x])/x - Sin[b*x]/(4*x^2) - (b^2*SinIntegral[b*x])/4 - SinInt 
egral[b*x]/(2*x^2)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {7057, 27, 3042, 3778, 3042, 3778, 25, 3042, 3780}

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.

\(\displaystyle \int \frac {\text {Si}(b x)}{x^3} \, dx\)

\(\Big \downarrow \) 7057

\(\displaystyle \frac {1}{2} b \int \frac {\sin (b x)}{b x^3}dx-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {\sin (b x)}{x^3}dx-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {\sin (b x)}{x^3}dx-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 3778

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \int \frac {\cos (b x)}{x^2}dx-\frac {\sin (b x)}{2 x^2}\right )-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \int \frac {\sin \left (b x+\frac {\pi }{2}\right )}{x^2}dx-\frac {\sin (b x)}{2 x^2}\right )-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 3778

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (b \int -\frac {\sin (b x)}{x}dx-\frac {\cos (b x)}{x}\right )-\frac {\sin (b x)}{2 x^2}\right )-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (-b \int \frac {\sin (b x)}{x}dx-\frac {\cos (b x)}{x}\right )-\frac {\sin (b x)}{2 x^2}\right )-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (-b \int \frac {\sin (b x)}{x}dx-\frac {\cos (b x)}{x}\right )-\frac {\sin (b x)}{2 x^2}\right )-\frac {\text {Si}(b x)}{2 x^2}\)

\(\Big \downarrow \) 3780

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} b \left (-b \text {Si}(b x)-\frac {\cos (b x)}{x}\right )-\frac {\sin (b x)}{2 x^2}\right )-\frac {\text {Si}(b x)}{2 x^2}\)

Input:

Int[SinIntegral[b*x]/x^3,x]
 

Output:

-1/2*SinIntegral[b*x]/x^2 + (-1/2*Sin[b*x]/x^2 + (b*(-(Cos[b*x]/x) - b*Sin 
Integral[b*x]))/2)/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778

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] - Simp[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 3780

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

rule 7057

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

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02

method	result	size
parts	\(-\frac {\operatorname {Si}\left (b x \right )}{2 x^{2}}+\frac {b^{2} \left (-\frac {\sin \left (b x \right )}{2 b^{2} x^{2}}-\frac {\cos \left (b x \right )}{2 b x}-\frac {\operatorname {Si}\left (b x \right )}{2}\right )}{2}\)	\(47\)
derivativedivides	\(b^{2} \left (-\frac {\operatorname {Si}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\sin \left (b x \right )}{4 b^{2} x^{2}}-\frac {\cos \left (b x \right )}{4 b x}-\frac {\operatorname {Si}\left (b x \right )}{4}\right )\)	\(48\)
default	\(b^{2} \left (-\frac {\operatorname {Si}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\sin \left (b x \right )}{4 b^{2} x^{2}}-\frac {\cos \left (b x \right )}{4 b x}-\frac {\operatorname {Si}\left (b x \right )}{4}\right )\)	\(48\)
meijerg	\(\frac {\sqrt {\pi }\, b^{2} \left (-\frac {4 \cos \left (b x \right )}{x b \sqrt {\pi }}-\frac {4 \sin \left (b x \right )}{b^{2} x^{2} \sqrt {\pi }}-\frac {4 \left (b^{2} x^{2}+2\right ) \operatorname {Si}\left (b x \right )}{b^{2} x^{2} \sqrt {\pi }}\right )}{16}\)	\(64\)
orering	\(\frac {\left (-\frac {1}{4} b^{2} x^{3}-\frac {7}{2} x \right ) \operatorname {Si}\left (b x \right )}{x^{3}}-2 x^{2} \left (\frac {\sin \left (b x \right )}{x^{4}}-\frac {3 \,\operatorname {Si}\left (b x \right )}{x^{4}}\right )-\frac {x^{3} \left (\frac {b \cos \left (b x \right )}{x^{4}}-\frac {7 \sin \left (b x \right )}{x^{5}}+\frac {12 \,\operatorname {Si}\left (b x \right )}{x^{5}}\right )}{4}\)	\(78\)

Input:

int(Si(b*x)/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*Si(b*x)/x^2+1/2*b^2*(-1/2*sin(b*x)/b^2/x^2-1/2*cos(b*x)/b/x-1/2*Si(b* 
x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67 \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=-\frac {b x \cos \left (b x\right ) + {\left (b^{2} x^{2} + 2\right )} \operatorname {Si}\left (b x\right ) + \sin \left (b x\right )}{4 \, x^{2}} \] Input:

integrate(sin_integral(b*x)/x^3,x, algorithm="fricas")
 

Output:

-1/4*(b*x*cos(b*x) + (b^2*x^2 + 2)*sin_integral(b*x) + sin(b*x))/x^2
 

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=- \frac {b^{2} \operatorname {Si}{\left (b x \right )}}{4} - \frac {b \cos {\left (b x \right )}}{4 x} - \frac {\sin {\left (b x \right )}}{4 x^{2}} - \frac {\operatorname {Si}{\left (b x \right )}}{2 x^{2}} \] Input:

integrate(Si(b*x)/x**3,x)
 

Output:

-b**2*Si(b*x)/4 - b*cos(b*x)/(4*x) - sin(b*x)/(4*x**2) - Si(b*x)/(2*x**2)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.70 \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=-\frac {1}{4} \, b^{2} {\left (-i \, \Gamma \left (-2, i \, b x\right ) + i \, \Gamma \left (-2, -i \, b x\right )\right )} - \frac {\operatorname {Si}\left (b x\right )}{2 \, x^{2}} \] Input:

integrate(sin_integral(b*x)/x^3,x, algorithm="maxima")
 

Output:

-1/4*b^2*(-I*gamma(-2, I*b*x) + I*gamma(-2, -I*b*x)) - 1/2*sin_integral(b* 
x)/x^2
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 3.24 \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=-\frac {b^{2} x^{2} \Im \left ( \operatorname {Ci}\left (b x\right ) \right ) \tan \left (\frac {1}{2} \, b x\right )^{2} - b^{2} x^{2} \Im \left ( \operatorname {Ci}\left (-b x\right ) \right ) \tan \left (\frac {1}{2} \, b x\right )^{2} + 2 \, b^{2} x^{2} \operatorname {Si}\left (b x\right ) \tan \left (\frac {1}{2} \, b x\right )^{2} + b^{2} x^{2} \Im \left ( \operatorname {Ci}\left (b x\right ) \right ) - b^{2} x^{2} \Im \left ( \operatorname {Ci}\left (-b x\right ) \right ) + 2 \, b^{2} x^{2} \operatorname {Si}\left (b x\right ) - 2 \, b x \tan \left (\frac {1}{2} \, b x\right )^{2} + 2 \, b x + 4 \, \tan \left (\frac {1}{2} \, b x\right )}{8 \, {\left (x^{2} \tan \left (\frac {1}{2} \, b x\right )^{2} + x^{2}\right )}} - \frac {\operatorname {Si}\left (b x\right )}{2 \, x^{2}} \] Input:

integrate(sin_integral(b*x)/x^3,x, algorithm="giac")
 

Output:

-1/8*(b^2*x^2*imag_part(cos_integral(b*x))*tan(1/2*b*x)^2 - b^2*x^2*imag_p 
art(cos_integral(-b*x))*tan(1/2*b*x)^2 + 2*b^2*x^2*sin_integral(b*x)*tan(1 
/2*b*x)^2 + b^2*x^2*imag_part(cos_integral(b*x)) - b^2*x^2*imag_part(cos_i 
ntegral(-b*x)) + 2*b^2*x^2*sin_integral(b*x) - 2*b*x*tan(1/2*b*x)^2 + 2*b* 
x + 4*tan(1/2*b*x))/(x^2*tan(1/2*b*x)^2 + x^2) - 1/2*sin_integral(b*x)/x^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=-\frac {\frac {\sin \left (b\,x\right )}{2}+\frac {b\,x\,\cos \left (b\,x\right )}{2}}{2\,x^2}-\frac {b^2\,\mathrm {sinint}\left (b\,x\right )}{4}-\frac {\mathrm {sinint}\left (b\,x\right )}{2\,x^2} \] Input:

int(sinint(b*x)/x^3,x)
 

Output:

- (sin(b*x)/2 + (b*x*cos(b*x))/2)/(2*x^2) - (b^2*sinint(b*x))/4 - sinint(b 
*x)/(2*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {\text {Si}(b x)}{x^3} \, dx=\frac {-\cos \left (b x \right ) b x -\mathit {si} \left (b x \right ) b^{2} x^{2}-2 \mathit {si} \left (b x \right )-\sin \left (b x \right )}{4 x^{2}} \] Input:

int(Si(b*x)/x^3,x)
 

Output:

( - cos(b*x)*b*x - si(b*x)*b**2*x**2 - 2*si(b*x) - sin(b*x))/(4*x**2)