\(\int x \text {Si}(a+b x)^2 \, dx\) [27]

Optimal result

Integrand size = 10, antiderivative size = 154 \[ \int x \text {Si}(a+b x)^2 \, dx=\frac {\cos (2 a+2 b x)}{4 b^2}-\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}+\frac {\log (a+b x)}{2 b^2}-\frac {a \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \cos (a+b x) \text {Si}(a+b x)}{b}-\frac {\sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {a (a+b x) \text {Si}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Si}(a+b x)^2}{2 b}+\frac {a \text {Si}(2 a+2 b x)}{b^2} \] Output:

1/4*cos(2*b*x+2*a)/b^2-1/2*Ci(2*b*x+2*a)/b^2+1/2*ln(b*x+a)/b^2-a*cos(b*x+a 
)*Si(b*x+a)/b^2+x*cos(b*x+a)*Si(b*x+a)/b-sin(b*x+a)*Si(b*x+a)/b^2-1/2*a*(b 
*x+a)*Si(b*x+a)^2/b^2+1/2*x*(b*x+a)*Si(b*x+a)^2/b+a*Si(2*b*x+2*a)/b^2
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.62 \[ \int x \text {Si}(a+b x)^2 \, dx=\frac {\cos (2 (a+b x))-2 \operatorname {CosIntegral}(2 (a+b x))+2 \log (a+b x)-4 ((a-b x) \cos (a+b x)+\sin (a+b x)) \text {Si}(a+b x)-2 \left (a^2-b^2 x^2\right ) \text {Si}(a+b x)^2+4 a \text {Si}(2 (a+b x))}{4 b^2} \] Input:

Integrate[x*SinIntegral[a + b*x]^2,x]
 

Output:

(Cos[2*(a + b*x)] - 2*CosIntegral[2*(a + b*x)] + 2*Log[a + b*x] - 4*((a - 
b*x)*Cos[a + b*x] + Sin[a + b*x])*SinIntegral[a + b*x] - 2*(a^2 - b^2*x^2) 
*SinIntegral[a + b*x]^2 + 4*a*SinIntegral[2*(a + b*x)])/(4*b^2)
 

Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.600, Rules used = {7063, 7059, 7065, 4906, 27, 3042, 3780, 7067, 5084, 7071, 3042, 3793, 2009, 7292, 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.

\(\displaystyle \int x \text {Si}(a+b x)^2 \, dx\)

\(\Big \downarrow \) 7063

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

\(\Big \downarrow \) 7059

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

\(\Big \downarrow \) 7065

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 7067

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

\(\Big \downarrow \) 5084

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

\(\Big \downarrow \) 7071

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[x*SinIntegral[a + b*x]^2,x]
 

Output:

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

Defintions of rubi rules used

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 2009

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

rule 3042

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

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 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 5084

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Simp[1/2^p   Int[u*Sin[ 
2*v]^p, x], x] /; EqQ[w, v] && IntegerQ[p]
 

rule 7059

Int[SinIntegral[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(SinInte 
gral[a + b*x]^2/b), x] - Simp[2   Int[Sin[a + b*x]*SinIntegral[a + b*x], x] 
, x] /; FreeQ[{a, b}, x]
 

rule 7063

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_) + (b_.)*(x_)]^2, x_Symbol] 
:> Simp[(a + b*x)*(c + d*x)^m*(SinIntegral[a + b*x]^2/(b*(m + 1))), x] + (- 
Simp[2/(m + 1)   Int[(c + d*x)^m*Sin[a + b*x]*SinIntegral[a + b*x], x], x] 
+ Simp[(b*c - a*d)*(m/(b*(m + 1)))   Int[(c + d*x)^(m - 1)*SinIntegral[a + 
b*x]^2, x], x]) /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0]
 

rule 7065

Int[Sin[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[(-Cos[a + b*x])*(SinIntegral[c + d*x]/b), x] + Simp[d/b   Int[Cos[a + b 
*x]*(Sin[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]
 

rule 7067

Int[((e_.) + (f_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + 
(d_.)*(x_)], x_Symbol] :> Simp[(-(e + f*x)^m)*Cos[a + b*x]*(SinIntegral[c + 
 d*x]/b), x] + (Simp[d/b   Int[(e + f*x)^m*Cos[a + b*x]*(Sin[c + d*x]/(c + 
d*x)), x], x] + Simp[f*(m/b)   Int[(e + f*x)^(m - 1)*Cos[a + b*x]*SinIntegr 
al[c + d*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0]
 

rule 7071

Int[Cos[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[Sin[a + b*x]*(SinIntegral[c + d*x]/b), x] - Simp[d/b   Int[Sin[a + b*x] 
*(Sin[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]
 

rule 7292

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

rule 7293

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

Maple [A] (verified)

Time = 16.53 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {\operatorname {Si}\left (b x +a \right )^{2} \left (\frac {\left (b x +a \right )^{2}}{2}-\left (b x +a \right ) a \right )-2 \,\operatorname {Si}\left (b x +a \right ) \left (\frac {\sin \left (b x +a \right )}{2}-\frac {\left (b x +a \right ) \cos \left (b x +a \right )}{2}+a \cos \left (b x +a \right )\right )+\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}+\frac {\cos \left (b x +a \right )^{2}}{2}+a \,\operatorname {Si}\left (2 b x +2 a \right )}{b^{2}}\)	\(111\)
default	\(\frac {\operatorname {Si}\left (b x +a \right )^{2} \left (\frac {\left (b x +a \right )^{2}}{2}-\left (b x +a \right ) a \right )-2 \,\operatorname {Si}\left (b x +a \right ) \left (\frac {\sin \left (b x +a \right )}{2}-\frac {\left (b x +a \right ) \cos \left (b x +a \right )}{2}+a \cos \left (b x +a \right )\right )+\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}+\frac {\cos \left (b x +a \right )^{2}}{2}+a \,\operatorname {Si}\left (2 b x +2 a \right )}{b^{2}}\)	\(111\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.65 \[ \int x \text {Si}(a+b x)^2 \, dx=\frac {2 \, {\left (b x - a\right )} \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \operatorname {Si}\left (b x + a\right )^{2} + \cos \left (b x + a\right )^{2} + 2 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - 2 \, \sin \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) - \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) + \log \left (b x + a\right )}{2 \, b^{2}} \] Input:

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

Output:

1/2*(2*(b*x - a)*cos(b*x + a)*sin_integral(b*x + a) + (b^2*x^2 - a^2)*sin_ 
integral(b*x + a)^2 + cos(b*x + a)^2 + 2*a*sin_integral(2*b*x + 2*a) - 2*s 
in(b*x + a)*sin_integral(b*x + a) - cos_integral(2*b*x + 2*a) + log(b*x + 
a))/b^2
 

Sympy [F]

\[ \int x \text {Si}(a+b x)^2 \, dx=\int x \operatorname {Si}^{2}{\left (a + b x \right )}\, dx \] Input:

integrate(x*Si(b*x+a)**2,x)
 

Output:

Integral(x*Si(a + b*x)**2, x)
 

Maxima [F]

\[ \int x \text {Si}(a+b x)^2 \, dx=\int { x \operatorname {Si}\left (b x + a\right )^{2} \,d x } \] Input:

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

Output:

integrate(x*sin_integral(b*x + a)^2, x)
 

Giac [F]

\[ \int x \text {Si}(a+b x)^2 \, dx=\int { x \operatorname {Si}\left (b x + a\right )^{2} \,d x } \] Input:

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

Output:

integrate(x*sin_integral(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int x \text {Si}(a+b x)^2 \, dx=\int x\,{\mathrm {sinint}\left (a+b\,x\right )}^2 \,d x \] Input:

int(x*sinint(a + b*x)^2,x)
 

Output:

int(x*sinint(a + b*x)^2, x)
 

Reduce [F]

\[ \int x \text {Si}(a+b x)^2 \, dx=\int \mathit {si} \left (b x +a \right )^{2} x d x \] Input:

int(x*Si(b*x+a)^2,x)
 

Output:

int(si(a + b*x)**2*x,x)