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

Optimal. Leaf size=218 \[ \frac {a^2 \text {Ci}(2 a+2 b x)}{2 b^3}-\frac {a^2 \log (a+b x)}{2 b^3}-\frac {\text {Ci}(2 a+2 b x)}{b^3}+\frac {a \text {Si}(2 a+2 b x)}{b^3}-\frac {2 \text {Si}(a+b x) \sin (a+b x)}{b^3}+\frac {\log (a+b x)}{b^3}-\frac {\sin ^2(a+b x)}{4 b^3}+\frac {\cos (2 a+2 b x)}{2 b^3}-\frac {a \sin (a+b x) \cos (a+b x)}{2 b^3}+\frac {2 x \text {Si}(a+b x) \cos (a+b x)}{b^2}+\frac {a x}{2 b^2}+\frac {x \sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {x^2 \text {Si}(a+b x) \sin (a+b x)}{b}-\frac {x^2}{4 b} \]

[Out]

1/2*a*x/b^2-1/4*x^2/b-Ci(2*b*x+2*a)/b^3+1/2*a^2*Ci(2*b*x+2*a)/b^3+1/2*cos(2*b*x+2*a)/b^3+ln(b*x+a)/b^3-1/2*a^2
*ln(b*x+a)/b^3+2*x*cos(b*x+a)*Si(b*x+a)/b^2+a*Si(2*b*x+2*a)/b^3-1/2*a*cos(b*x+a)*sin(b*x+a)/b^3+1/2*x*cos(b*x+
a)*sin(b*x+a)/b^2-2*Si(b*x+a)*sin(b*x+a)/b^3+x^2*Si(b*x+a)*sin(b*x+a)/b-1/4*sin(b*x+a)^2/b^3

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Rubi [A]  time = 0.66, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 14, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {6519, 6742, 2635, 8, 3310, 30, 3312, 3302, 6513, 4573, 6741, 2638, 3299, 6517} \[ \frac {a^2 \text {CosIntegral}(2 a+2 b x)}{2 b^3}-\frac {a^2 \log (a+b x)}{2 b^3}-\frac {\text {CosIntegral}(2 a+2 b x)}{b^3}+\frac {a \text {Si}(2 a+2 b x)}{b^3}-\frac {2 \text {Si}(a+b x) \sin (a+b x)}{b^3}+\frac {2 x \text {Si}(a+b x) \cos (a+b x)}{b^2}+\frac {a x}{2 b^2}+\frac {\log (a+b x)}{b^3}-\frac {\sin ^2(a+b x)}{4 b^3}+\frac {\cos (2 a+2 b x)}{2 b^3}-\frac {a \sin (a+b x) \cos (a+b x)}{2 b^3}+\frac {x \sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {x^2 \text {Si}(a+b x) \sin (a+b x)}{b}-\frac {x^2}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/(2*b^2) - x^2/(4*b) + Cos[2*a + 2*b*x]/(2*b^3) - CosIntegral[2*a + 2*b*x]/b^3 + (a^2*CosIntegral[2*a + 2
*b*x])/(2*b^3) + Log[a + b*x]/b^3 - (a^2*Log[a + b*x])/(2*b^3) - (a*Cos[a + b*x]*Sin[a + b*x])/(2*b^3) + (x*Co
s[a + b*x]*Sin[a + b*x])/(2*b^2) - Sin[a + b*x]^2/(4*b^3) + (2*x*Cos[a + b*x]*SinIntegral[a + b*x])/b^2 - (2*S
in[a + b*x]*SinIntegral[a + b*x])/b^3 + (x^2*Sin[a + b*x]*SinIntegral[a + b*x])/b + (a*SinIntegral[2*a + 2*b*x
])/b^3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3299

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

Rule 3302

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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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 4573

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

Rule 6513

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] + (Dist[d/b, Int[((e + f*x)^m*Cos[a + b*x]*Sin[c + d*x])/(c
+ d*x), x], x] + Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Cos[a + b*x]*SinIntegral[c + d*x], x], x]) /; FreeQ[{a,
b, c, d, e, f}, x] && IGtQ[m, 0]

Rule 6517

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

Rule 6519

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {align*} \int x^2 \cos (a+b x) \text {Si}(a+b x) \, dx &=\frac {x^2 \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {2 \int x \sin (a+b x) \text {Si}(a+b x) \, dx}{b}-\int \frac {x^2 \sin ^2(a+b x)}{a+b x} \, dx\\ &=\frac {2 x \cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x^2 \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {2 \int \cos (a+b x) \text {Si}(a+b x) \, dx}{b^2}-\frac {2 \int \frac {x \cos (a+b x) \sin (a+b x)}{a+b x} \, dx}{b}-\int \left (-\frac {a \sin ^2(a+b x)}{b^2}+\frac {x \sin ^2(a+b x)}{b}+\frac {a^2 \sin ^2(a+b x)}{b^2 (a+b x)}\right ) \, dx\\ &=\frac {2 x \cos (a+b x) \text {Si}(a+b x)}{b^2}-\frac {2 \sin (a+b x) \text {Si}(a+b x)}{b^3}+\frac {x^2 \sin (a+b x) \text {Si}(a+b x)}{b}+\frac {2 \int \frac {\sin ^2(a+b x)}{a+b x} \, dx}{b^2}+\frac {a \int \sin ^2(a+b x) \, dx}{b^2}-\frac {a^2 \int \frac {\sin ^2(a+b x)}{a+b x} \, dx}{b^2}-\frac {\int x \sin ^2(a+b x) \, dx}{b}-\frac {\int \frac {x \sin (2 (a+b x))}{a+b x} \, dx}{b}\\ &=-\frac {a \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {x \cos (a+b x) \sin (a+b x)}{2 b^2}-\frac {\sin ^2(a+b x)}{4 b^3}+\frac {2 x \cos (a+b x) \text {Si}(a+b x)}{b^2}-\frac {2 \sin (a+b x) \text {Si}(a+b x)}{b^3}+\frac {x^2 \sin (a+b x) \text {Si}(a+b x)}{b}+\frac {2 \int \left (\frac {1}{2 (a+b x)}-\frac {\cos (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b^2}+\frac {a \int 1 \, dx}{2 b^2}-\frac {a^2 \int \left (\frac {1}{2 (a+b x)}-\frac {\cos (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b^2}-\frac {\int x \, dx}{2 b}-\frac {\int \frac {x \sin (2 a+2 b x)}{a+b x} \, dx}{b}\\ &=\frac {a x}{2 b^2}-\frac {x^2}{4 b}+\frac {\log (a+b x)}{b^3}-\frac {a^2 \log (a+b x)}{2 b^3}-\frac {a \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {x \cos (a+b x) \sin (a+b x)}{2 b^2}-\frac {\sin ^2(a+b x)}{4 b^3}+\frac {2 x \cos (a+b x) \text {Si}(a+b x)}{b^2}-\frac {2 \sin (a+b x) \text {Si}(a+b x)}{b^3}+\frac {x^2 \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\int \frac {\cos (2 a+2 b x)}{a+b x} \, dx}{b^2}+\frac {a^2 \int \frac {\cos (2 a+2 b x)}{a+b x} \, dx}{2 b^2}-\frac {\int \left (\frac {\sin (2 a+2 b x)}{b}+\frac {a \sin (2 a+2 b x)}{b (-a-b x)}\right ) \, dx}{b}\\ &=\frac {a x}{2 b^2}-\frac {x^2}{4 b}-\frac {\text {Ci}(2 a+2 b x)}{b^3}+\frac {a^2 \text {Ci}(2 a+2 b x)}{2 b^3}+\frac {\log (a+b x)}{b^3}-\frac {a^2 \log (a+b x)}{2 b^3}-\frac {a \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {x \cos (a+b x) \sin (a+b x)}{2 b^2}-\frac {\sin ^2(a+b x)}{4 b^3}+\frac {2 x \cos (a+b x) \text {Si}(a+b x)}{b^2}-\frac {2 \sin (a+b x) \text {Si}(a+b x)}{b^3}+\frac {x^2 \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\int \sin (2 a+2 b x) \, dx}{b^2}-\frac {a \int \frac {\sin (2 a+2 b x)}{-a-b x} \, dx}{b^2}\\ &=\frac {a x}{2 b^2}-\frac {x^2}{4 b}+\frac {\cos (2 a+2 b x)}{2 b^3}-\frac {\text {Ci}(2 a+2 b x)}{b^3}+\frac {a^2 \text {Ci}(2 a+2 b x)}{2 b^3}+\frac {\log (a+b x)}{b^3}-\frac {a^2 \log (a+b x)}{2 b^3}-\frac {a \cos (a+b x) \sin (a+b x)}{2 b^3}+\frac {x \cos (a+b x) \sin (a+b x)}{2 b^2}-\frac {\sin ^2(a+b x)}{4 b^3}+\frac {2 x \cos (a+b x) \text {Si}(a+b x)}{b^2}-\frac {2 \sin (a+b x) \text {Si}(a+b x)}{b^3}+\frac {x^2 \sin (a+b x) \text {Si}(a+b x)}{b}+\frac {a \text {Si}(2 a+2 b x)}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 134, normalized size = 0.61 \[ \frac {4 \left (a^2-2\right ) \text {Ci}(2 (a+b x))-4 a^2 \log (a+b x)+8 \text {Si}(a+b x) \left (\left (b^2 x^2-2\right ) \sin (a+b x)+2 b x \cos (a+b x)\right )+8 a \text {Si}(2 (a+b x))+4 a b x+8 \log (a+b x)-2 a \sin (2 (a+b x))+2 b x \sin (2 (a+b x))+5 \cos (2 (a+b x))-2 b^2 x^2}{8 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*b*x - 2*b^2*x^2 + 5*Cos[2*(a + b*x)] + 4*(-2 + a^2)*CosIntegral[2*(a + b*x)] + 8*Log[a + b*x] - 4*a^2*Log
[a + b*x] - 2*a*Sin[2*(a + b*x)] + 2*b*x*Sin[2*(a + b*x)] + 8*(2*b*x*Cos[a + b*x] + (-2 + b^2*x^2)*Sin[a + b*x
])*SinIntegral[a + b*x] + 8*a*SinIntegral[2*(a + b*x)])/(8*b^3)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*cos(b*x + a)*sin_integral(b*x + a), x)

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giac [C]  time = 0.39, size = 431, normalized size = 1.98 \[ {\left (\frac {2 \, x \cos \left (b x + a\right )}{b^{2}} + \frac {{\left (b^{2} x^{2} - 2\right )} \sin \left (b x + a\right )}{b^{3}}\right )} \operatorname {Si}\left (b x + a\right ) - \frac {2 \, b^{2} x^{2} \tan \left (b x + a\right )^{2} - 4 \, a b x \tan \left (b x + a\right )^{2} + 4 \, a^{2} \log \left ({\left | b x + a \right |}\right ) \tan \left (b x + a\right )^{2} - 2 \, a^{2} \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - 2 \, a^{2} \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} - 4 \, a \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + 4 \, a \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - 8 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )^{2} - 4 \, a b x + 4 \, a^{2} \log \left ({\left | b x + a \right |}\right ) - 2 \, a^{2} \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - 2 \, a^{2} \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) - 4 \, b x \tan \left (b x + a\right ) - 8 \, \log \left ({\left | b x + a \right |}\right ) \tan \left (b x + a\right )^{2} + 4 \, \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + 4 \, \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - 4 \, a \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) + 4 \, a \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) - 8 \, a \operatorname {Si}\left (2 \, b x + 2 \, a\right ) + 4 \, a \tan \left (b x + a\right ) + 5 \, \tan \left (b x + a\right )^{2} - 8 \, \log \left ({\left | b x + a \right |}\right ) + 4 \, \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) + 4 \, \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) - 5}{8 \, {\left (b^{3} \tan \left (b x + a\right )^{2} + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*cos(b*x + a)/b^2 + (b^2*x^2 - 2)*sin(b*x + a)/b^3)*sin_integral(b*x + a) - 1/8*(2*b^2*x^2*tan(b*x + a)^2
- 4*a*b*x*tan(b*x + a)^2 + 4*a^2*log(abs(b*x + a))*tan(b*x + a)^2 - 2*a^2*real_part(cos_integral(2*b*x + 2*a))
*tan(b*x + a)^2 - 2*a^2*real_part(cos_integral(-2*b*x - 2*a))*tan(b*x + a)^2 + 2*b^2*x^2 - 4*a*imag_part(cos_i
ntegral(2*b*x + 2*a))*tan(b*x + a)^2 + 4*a*imag_part(cos_integral(-2*b*x - 2*a))*tan(b*x + a)^2 - 8*a*sin_inte
gral(2*b*x + 2*a)*tan(b*x + a)^2 - 4*a*b*x + 4*a^2*log(abs(b*x + a)) - 2*a^2*real_part(cos_integral(2*b*x + 2*
a)) - 2*a^2*real_part(cos_integral(-2*b*x - 2*a)) - 4*b*x*tan(b*x + a) - 8*log(abs(b*x + a))*tan(b*x + a)^2 +
4*real_part(cos_integral(2*b*x + 2*a))*tan(b*x + a)^2 + 4*real_part(cos_integral(-2*b*x - 2*a))*tan(b*x + a)^2
- 4*a*imag_part(cos_integral(2*b*x + 2*a)) + 4*a*imag_part(cos_integral(-2*b*x - 2*a)) - 8*a*sin_integral(2*b
*x + 2*a) + 4*a*tan(b*x + a) + 5*tan(b*x + a)^2 - 8*log(abs(b*x + a)) + 4*real_part(cos_integral(2*b*x + 2*a))
+ 4*real_part(cos_integral(-2*b*x - 2*a)) - 5)/(b^3*tan(b*x + a)^2 + b^3)

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maple [A]  time = 0.05, size = 209, normalized size = 0.96 \[ \frac {x^{2} \Si \left (b x +a \right ) \sin \left (b x +a \right )}{b}+\frac {2 x \cos \left (b x +a \right ) \Si \left (b x +a \right )}{b^{2}}-\frac {2 \Si \left (b x +a \right ) \sin \left (b x +a \right )}{b^{3}}+\frac {x \cos \left (b x +a \right ) \sin \left (b x +a \right )}{2 b^{2}}-\frac {a \cos \left (b x +a \right ) \sin \left (b x +a \right )}{2 b^{3}}-\frac {x^{2}}{4 b}+\frac {a x}{2 b^{2}}+\frac {3 a^{2}}{4 b^{3}}-\frac {\sin ^{2}\left (b x +a \right )}{4 b^{3}}-\frac {a^{2} \ln \left (b x +a \right )}{2 b^{3}}+\frac {a^{2} \Ci \left (2 b x +2 a \right )}{2 b^{3}}+\frac {\cos ^{2}\left (b x +a \right )}{b^{3}}+\frac {a \Si \left (2 b x +2 a \right )}{b^{3}}+\frac {\ln \left (b x +a \right )}{b^{3}}-\frac {\Ci \left (2 b x +2 a \right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*Si(b*x+a)*sin(b*x+a)/b+2*x*cos(b*x+a)*Si(b*x+a)/b^2-2*Si(b*x+a)*sin(b*x+a)/b^3+1/2*x*cos(b*x+a)*sin(b*x+a)
/b^2-1/2*a*cos(b*x+a)*sin(b*x+a)/b^3-1/4*x^2/b+1/2*a*x/b^2+3/4/b^3*a^2-1/4*sin(b*x+a)^2/b^3-1/2*a^2*ln(b*x+a)/
b^3+1/2*a^2*Ci(2*b*x+2*a)/b^3+cos(b*x+a)^2/b^3+a*Si(2*b*x+2*a)/b^3+ln(b*x+a)/b^3-Ci(2*b*x+2*a)/b^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Si}\left (b x + a\right ) \cos \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\mathrm {sinint}\left (a+b\,x\right )\,\cos \left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cos {\left (a + b x \right )} \operatorname {Si}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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