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

Optimal. Leaf size=187 \[ \frac {a^2 \text {Si}(2 a+2 b x)}{2 b^3}-\frac {a \text {Ci}(2 a+2 b x)}{b^3}-\frac {\text {Si}(2 a+2 b x)}{b^3}+\frac {2 \text {Si}(a+b x) \cos (a+b x)}{b^3}+\frac {a \log (a+b x)}{b^3}+\frac {\sin (2 a+2 b x)}{8 b^3}+\frac {a \cos (2 a+2 b x)}{4 b^3}+\frac {\sin (a+b x) \cos (a+b x)}{b^3}+\frac {2 x \text {Si}(a+b x) \sin (a+b x)}{b^2}-\frac {x \cos (2 a+2 b x)}{4 b^2}-\frac {x^2 \text {Si}(a+b x) \cos (a+b x)}{b}-\frac {x}{b^2} \]

[Out]

-x/b^2-a*Ci(2*b*x+2*a)/b^3+1/4*a*cos(2*b*x+2*a)/b^3-1/4*x*cos(2*b*x+2*a)/b^2+a*ln(b*x+a)/b^3+2*cos(b*x+a)*Si(b
*x+a)/b^3-x^2*cos(b*x+a)*Si(b*x+a)/b-Si(2*b*x+2*a)/b^3+1/2*a^2*Si(2*b*x+2*a)/b^3+cos(b*x+a)*sin(b*x+a)/b^3+2*x
*Si(b*x+a)*sin(b*x+a)/b^2+1/8*sin(2*b*x+2*a)/b^3

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Rubi [A]  time = 0.71, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6513, 4573, 6741, 6742, 2638, 3296, 2637, 3299, 6519, 2635, 8, 3312, 3302, 6511, 4406, 12} \[ \frac {a^2 \text {Si}(2 a+2 b x)}{2 b^3}-\frac {a \text {CosIntegral}(2 a+2 b x)}{b^3}-\frac {\text {Si}(2 a+2 b x)}{b^3}+\frac {2 x \text {Si}(a+b x) \sin (a+b x)}{b^2}+\frac {2 \text {Si}(a+b x) \cos (a+b x)}{b^3}+\frac {a \log (a+b x)}{b^3}+\frac {\sin (2 a+2 b x)}{8 b^3}+\frac {a \cos (2 a+2 b x)}{4 b^3}-\frac {x \cos (2 a+2 b x)}{4 b^2}+\frac {\sin (a+b x) \cos (a+b x)}{b^3}-\frac {x^2 \text {Si}(a+b x) \cos (a+b x)}{b}-\frac {x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

Rule 3296

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

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 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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(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]
&& IGtQ[p, 0]

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 6511

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

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 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 \sin (a+b x) \text {Si}(a+b x) \, dx &=-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 \int x \cos (a+b x) \text {Si}(a+b x) \, dx}{b}+\int \frac {x^2 \cos (a+b x) \sin (a+b x)}{a+b x} \, dx\\ &=-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 x \sin (a+b x) \text {Si}(a+b x)}{b^2}+\frac {1}{2} \int \frac {x^2 \sin (2 (a+b x))}{a+b x} \, dx-\frac {2 \int \sin (a+b x) \text {Si}(a+b x) \, dx}{b^2}-\frac {2 \int \frac {x \sin ^2(a+b x)}{a+b x} \, dx}{b}\\ &=\frac {2 \cos (a+b x) \text {Si}(a+b x)}{b^3}-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 x \sin (a+b x) \text {Si}(a+b x)}{b^2}+\frac {1}{2} \int \frac {x^2 \sin (2 a+2 b x)}{a+b x} \, dx-\frac {2 \int \frac {\cos (a+b x) \sin (a+b x)}{a+b x} \, dx}{b^2}-\frac {2 \int \left (\frac {\sin ^2(a+b x)}{b}-\frac {a \sin ^2(a+b x)}{b (a+b x)}\right ) \, dx}{b}\\ &=\frac {2 \cos (a+b x) \text {Si}(a+b x)}{b^3}-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 x \sin (a+b x) \text {Si}(a+b x)}{b^2}+\frac {1}{2} \int \left (-\frac {a \sin (2 a+2 b x)}{b^2}+\frac {x \sin (2 a+2 b x)}{b}+\frac {a^2 \sin (2 a+2 b x)}{b^2 (a+b x)}\right ) \, dx-\frac {2 \int \sin ^2(a+b x) \, dx}{b^2}-\frac {2 \int \frac {\sin (2 a+2 b x)}{2 (a+b x)} \, dx}{b^2}+\frac {(2 a) \int \frac {\sin ^2(a+b x)}{a+b x} \, dx}{b^2}\\ &=\frac {\cos (a+b x) \sin (a+b x)}{b^3}+\frac {2 \cos (a+b x) \text {Si}(a+b x)}{b^3}-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 x \sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {\int 1 \, dx}{b^2}-\frac {\int \frac {\sin (2 a+2 b x)}{a+b x} \, dx}{b^2}-\frac {a \int \sin (2 a+2 b x) \, dx}{2 b^2}+\frac {(2 a) \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^2 \int \frac {\sin (2 a+2 b x)}{a+b x} \, dx}{2 b^2}+\frac {\int x \sin (2 a+2 b x) \, dx}{2 b}\\ &=-\frac {x}{b^2}+\frac {a \cos (2 a+2 b x)}{4 b^3}-\frac {x \cos (2 a+2 b x)}{4 b^2}+\frac {a \log (a+b x)}{b^3}+\frac {\cos (a+b x) \sin (a+b x)}{b^3}+\frac {2 \cos (a+b x) \text {Si}(a+b x)}{b^3}-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 x \sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{b^3}+\frac {a^2 \text {Si}(2 a+2 b x)}{2 b^3}+\frac {\int \cos (2 a+2 b x) \, dx}{4 b^2}-\frac {a \int \frac {\cos (2 a+2 b x)}{a+b x} \, dx}{b^2}\\ &=-\frac {x}{b^2}+\frac {a \cos (2 a+2 b x)}{4 b^3}-\frac {x \cos (2 a+2 b x)}{4 b^2}-\frac {a \text {Ci}(2 a+2 b x)}{b^3}+\frac {a \log (a+b x)}{b^3}+\frac {\cos (a+b x) \sin (a+b x)}{b^3}+\frac {\sin (2 a+2 b x)}{8 b^3}+\frac {2 \cos (a+b x) \text {Si}(a+b x)}{b^3}-\frac {x^2 \cos (a+b x) \text {Si}(a+b x)}{b}+\frac {2 x \sin (a+b x) \text {Si}(a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{b^3}+\frac {a^2 \text {Si}(2 a+2 b x)}{2 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 123, normalized size = 0.66 \[ \frac {4 a^2 \text {Si}(2 (a+b x))-8 \text {Si}(a+b x) \left (\left (b^2 x^2-2\right ) \cos (a+b x)-2 b x \sin (a+b x)\right )-8 a \text {Ci}(2 (a+b x))-8 \text {Si}(2 (a+b x))+8 a \log (a+b x)+5 \sin (2 (a+b x))+2 a \cos (2 (a+b x))-2 b x \cos (2 (a+b x))-8 b x}{8 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \sin \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.44, size = 398, normalized size = 2.13 \[ {\left (\frac {2 \, x \sin \left (b x + a\right )}{b^{2}} - \frac {{\left (b^{2} x^{2} - 2\right )} \cos \left (b x + a\right )}{b^{3}}\right )} \operatorname {Si}\left (b x + a\right ) + \frac {a^{2} \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - a^{2} \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + 2 \, a^{2} \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )^{2} - 3 \, b x \tan \left (b x + a\right )^{2} + 4 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x + a\right )^{2} - 2 \, a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - 2 \, a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + a^{2} \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - a^{2} \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, a^{2} \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - a \tan \left (b x + a\right )^{2} - 2 \, \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} + 2 \, \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x + a\right )^{2} - 4 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )^{2} - 5 \, b x + 4 \, a \log \left ({\left | b x + a \right |}\right ) - 2 \, a \Re \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - 2 \, a \Re \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + a - 2 \, \Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) + 2 \, \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) - 4 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) + 5 \, \tan \left (b x + a\right )}{4 \, {\left (b^{3} \tan \left (b x + a\right )^{2} + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*sin(b*x + a)/b^2 - (b^2*x^2 - 2)*cos(b*x + a)/b^3)*sin_integral(b*x + a) + 1/4*(a^2*imag_part(cos_integra
l(2*b*x + 2*a))*tan(b*x + a)^2 - a^2*imag_part(cos_integral(-2*b*x - 2*a))*tan(b*x + a)^2 + 2*a^2*sin_integral
(2*b*x + 2*a)*tan(b*x + a)^2 - 3*b*x*tan(b*x + a)^2 + 4*a*log(abs(b*x + a))*tan(b*x + a)^2 - 2*a*real_part(cos
_integral(2*b*x + 2*a))*tan(b*x + a)^2 - 2*a*real_part(cos_integral(-2*b*x - 2*a))*tan(b*x + a)^2 + a^2*imag_p
art(cos_integral(2*b*x + 2*a)) - a^2*imag_part(cos_integral(-2*b*x - 2*a)) + 2*a^2*sin_integral(2*b*x + 2*a) -
a*tan(b*x + a)^2 - 2*imag_part(cos_integral(2*b*x + 2*a))*tan(b*x + a)^2 + 2*imag_part(cos_integral(-2*b*x -
2*a))*tan(b*x + a)^2 - 4*sin_integral(2*b*x + 2*a)*tan(b*x + a)^2 - 5*b*x + 4*a*log(abs(b*x + a)) - 2*a*real_p
art(cos_integral(2*b*x + 2*a)) - 2*a*real_part(cos_integral(-2*b*x - 2*a)) + a - 2*imag_part(cos_integral(2*b*
x + 2*a)) + 2*imag_part(cos_integral(-2*b*x - 2*a)) - 4*sin_integral(2*b*x + 2*a) + 5*tan(b*x + a))/(b^3*tan(b
*x + a)^2 + b^3)

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maple [A]  time = 0.04, size = 171, normalized size = 0.91 \[ -\frac {x^{2} \cos \left (b x +a \right ) \Si \left (b x +a \right )}{b}+\frac {2 x \Si \left (b x +a \right ) \sin \left (b x +a \right )}{b^{2}}+\frac {2 \cos \left (b x +a \right ) \Si \left (b x +a \right )}{b^{3}}-\frac {\left (\cos ^{2}\left (b x +a \right )\right ) x}{2 b^{2}}+\frac {a \left (\cos ^{2}\left (b x +a \right )\right )}{2 b^{3}}+\frac {5 \cos \left (b x +a \right ) \sin \left (b x +a \right )}{4 b^{3}}-\frac {3 x}{4 b^{2}}-\frac {3 a}{4 b^{3}}+\frac {a^{2} \Si \left (2 b x +2 a \right )}{2 b^{3}}+\frac {a \ln \left (b x +a \right )}{b^{3}}-\frac {a \Ci \left (2 b x +2 a \right )}{b^{3}}-\frac {\Si \left (2 b x +2 a \right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*cos(b*x+a)*Si(b*x+a)/b+2*x*Si(b*x+a)*sin(b*x+a)/b^2+2*cos(b*x+a)*Si(b*x+a)/b^3-1/2/b^2*cos(b*x+a)^2*x+1/2
/b^3*a*cos(b*x+a)^2+5/4*cos(b*x+a)*sin(b*x+a)/b^3-3/4*x/b^2-3/4/b^3*a+1/2*a^2*Si(2*b*x+2*a)/b^3+a*ln(b*x+a)/b^
3-a*Ci(2*b*x+2*a)/b^3-Si(2*b*x+2*a)/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {sinint}\left (a+b\,x\right )\,\sin \left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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