\(\int \frac {\text {Si}(a+b x)}{x^3} \, dx\) [24]

Optimal result

Integrand size = 10, antiderivative size = 111 \[ \int \frac {\text {Si}(a+b x)}{x^3} \, dx=\frac {b^2 \cos (a) \operatorname {CosIntegral}(b x)}{2 a}-\frac {b^2 \operatorname {CosIntegral}(b x) \sin (a)}{2 a^2}-\frac {b \sin (a+b x)}{2 a x}-\frac {b^2 \cos (a) \text {Si}(b x)}{2 a^2}-\frac {b^2 \sin (a) \text {Si}(b x)}{2 a}+\frac {b^2 \text {Si}(a+b x)}{2 a^2}-\frac {\text {Si}(a+b x)}{2 x^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \frac {\text {Si}(a+b x)}{x^3} \, dx=-\frac {-b^2 x^2 \operatorname {CosIntegral}(b x) (a \cos (a)-\sin (a))+a b x \sin (a+b x)+b^2 x^2 (\cos (a)+a \sin (a)) \text {Si}(b x)+a^2 \text {Si}(a+b x)-b^2 x^2 \text {Si}(a+b x)}{2 a^2 x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {7057, 7293, 2009}

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.

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

\(\Big \downarrow \) 7057

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75

method	result	size
parts	\(-\frac {\operatorname {Si}\left (b x +a \right )}{2 x^{2}}+\frac {b^{2} \left (-\frac {\operatorname {Si}\left (b x \right ) \cos \left (a \right )+\operatorname {Ci}\left (b x \right ) \sin \left (a \right )}{a^{2}}+\frac {\operatorname {Si}\left (b x +a \right )}{a^{2}}+\frac {-\frac {\sin \left (b x +a \right )}{b x}-\operatorname {Si}\left (b x \right ) \sin \left (a \right )+\operatorname {Ci}\left (b x \right ) \cos \left (a \right )}{a}\right )}{2}\)	\(83\)
derivativedivides	\(b^{2} \left (-\frac {\operatorname {Si}\left (b x +a \right )}{2 b^{2} x^{2}}-\frac {\operatorname {Si}\left (b x \right ) \cos \left (a \right )+\operatorname {Ci}\left (b x \right ) \sin \left (a \right )}{2 a^{2}}+\frac {\operatorname {Si}\left (b x +a \right )}{2 a^{2}}+\frac {-\frac {\sin \left (b x +a \right )}{b x}-\operatorname {Si}\left (b x \right ) \sin \left (a \right )+\operatorname {Ci}\left (b x \right ) \cos \left (a \right )}{2 a}\right )\)	\(86\)
default	\(b^{2} \left (-\frac {\operatorname {Si}\left (b x +a \right )}{2 b^{2} x^{2}}-\frac {\operatorname {Si}\left (b x \right ) \cos \left (a \right )+\operatorname {Ci}\left (b x \right ) \sin \left (a \right )}{2 a^{2}}+\frac {\operatorname {Si}\left (b x +a \right )}{2 a^{2}}+\frac {-\frac {\sin \left (b x +a \right )}{b x}-\operatorname {Si}\left (b x \right ) \sin \left (a \right )+\operatorname {Ci}\left (b x \right ) \cos \left (a \right )}{2 a}\right )\)	\(86\)

Input:

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

Output:

-1/2*Si(b*x+a)/x^2+1/2*b^2*(-1/a^2*(Si(b*x)*cos(a)+Ci(b*x)*sin(a))+1/a^2*S 
i(b*x+a)+1/a*(-sin(b*x+a)/b/x-Si(b*x)*sin(a)+Ci(b*x)*cos(a)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86 \[ \int \frac {\text {Si}(a+b x)}{x^3} \, dx=-\frac {a b x \sin \left (b x + a\right ) - {\left (a b^{2} x^{2} \operatorname {Ci}\left (b x\right ) - b^{2} x^{2} \operatorname {Si}\left (b x\right )\right )} \cos \left (a\right ) + {\left (a b^{2} x^{2} \operatorname {Si}\left (b x\right ) + b^{2} x^{2} \operatorname {Ci}\left (b x\right )\right )} \sin \left (a\right ) - {\left (b^{2} x^{2} - a^{2}\right )} \operatorname {Si}\left (b x + a\right )}{2 \, a^{2} x^{2}} \] Input:

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

Output:

-1/2*(a*b*x*sin(b*x + a) - (a*b^2*x^2*cos_integral(b*x) - b^2*x^2*sin_inte 
gral(b*x))*cos(a) + (a*b^2*x^2*sin_integral(b*x) + b^2*x^2*cos_integral(b* 
x))*sin(a) - (b^2*x^2 - a^2)*sin_integral(b*x + a))/(a^2*x^2)
 

Sympy [F]

\[ \int \frac {\text {Si}(a+b x)}{x^3} \, dx=\int \frac {\operatorname {Si}{\left (a + b x \right )}}{x^{3}}\, dx \] Input:

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

Output:

Integral(Si(a + b*x)/x**3, x)
 

Maxima [F]

\[ \int \frac {\text {Si}(a+b x)}{x^3} \, dx=\int { \frac {\operatorname {Si}\left (b x + a\right )}{x^{3}} \,d x } \] Input:

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

Output:

integrate(sin_integral(b*x + a)/x^3, x)
 

Giac [C] (verification not implemented)

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

Time = 0.14 (sec) , antiderivative size = 809, normalized size of antiderivative = 7.29 \[ \int \frac {\text {Si}(a+b x)}{x^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(a*b*x*real_part(cos_integral(b*x))*tan(1/2*b*x)^2*tan(1/2*a)^2 + a*b 
*x*real_part(cos_integral(-b*x))*tan(1/2*b*x)^2*tan(1/2*a)^2 + 2*a*b*x*ima 
g_part(cos_integral(b*x))*tan(1/2*b*x)^2*tan(1/2*a) - 2*a*b*x*imag_part(co 
s_integral(-b*x))*tan(1/2*b*x)^2*tan(1/2*a) + 4*a*b*x*sin_integral(b*x)*ta 
n(1/2*b*x)^2*tan(1/2*a) - b*x*imag_part(cos_integral(b*x + a))*tan(1/2*b*x 
)^2*tan(1/2*a)^2 - b*x*imag_part(cos_integral(b*x))*tan(1/2*b*x)^2*tan(1/2 
*a)^2 + b*x*imag_part(cos_integral(-b*x - a))*tan(1/2*b*x)^2*tan(1/2*a)^2 
+ b*x*imag_part(cos_integral(-b*x))*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*b*x*si 
n_integral(b*x + a)*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*b*x*sin_integral(b*x)* 
tan(1/2*b*x)^2*tan(1/2*a)^2 - a*b*x*real_part(cos_integral(b*x))*tan(1/2*b 
*x)^2 - a*b*x*real_part(cos_integral(-b*x))*tan(1/2*b*x)^2 + 2*b*x*real_pa 
rt(cos_integral(b*x))*tan(1/2*b*x)^2*tan(1/2*a) + 2*b*x*real_part(cos_inte 
gral(-b*x))*tan(1/2*b*x)^2*tan(1/2*a) + a*b*x*real_part(cos_integral(b*x)) 
*tan(1/2*a)^2 + a*b*x*real_part(cos_integral(-b*x))*tan(1/2*a)^2 - b*x*ima 
g_part(cos_integral(b*x + a))*tan(1/2*b*x)^2 + b*x*imag_part(cos_integral( 
b*x))*tan(1/2*b*x)^2 + b*x*imag_part(cos_integral(-b*x - a))*tan(1/2*b*x)^ 
2 - b*x*imag_part(cos_integral(-b*x))*tan(1/2*b*x)^2 - 2*b*x*sin_integral( 
b*x + a)*tan(1/2*b*x)^2 + 2*b*x*sin_integral(b*x)*tan(1/2*b*x)^2 + 2*a*b*x 
*imag_part(cos_integral(b*x))*tan(1/2*a) - 2*a*b*x*imag_part(cos_integral( 
-b*x))*tan(1/2*a) + 4*a*b*x*sin_integral(b*x)*tan(1/2*a) - b*x*imag_par...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\text {Si}(a+b x)}{x^3} \, dx=\int \frac {\mathit {si} \left (b x +a \right )}{x^{3}}d x \] Input:

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

Output:

int(si(a + b*x)/x**3,x)