3.11 \(\int x^2 \text {Si}(b x)^2 \, dx\)

Optimal. Leaf size=112 \[ \frac {2 \text {Si}(2 b x)}{3 b^3}-\frac {4 \text {Si}(b x) \cos (b x)}{3 b^3}-\frac {5 \sin (b x) \cos (b x)}{6 b^3}-\frac {4 x \text {Si}(b x) \sin (b x)}{3 b^2}+\frac {5 x}{6 b^2}-\frac {x \sin ^2(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {2 x^2 \text {Si}(b x) \cos (b x)}{3 b} \]

[Out]

5/6*x/b^2-4/3*cos(b*x)*Si(b*x)/b^3+2/3*x^2*cos(b*x)*Si(b*x)/b+1/3*x^3*Si(b*x)^2+2/3*Si(2*b*x)/b^3-5/6*cos(b*x)
*sin(b*x)/b^3-4/3*x*Si(b*x)*sin(b*x)/b^2-1/3*x*sin(b*x)^2/b^2

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Rubi [A]  time = 0.13, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6507, 6513, 12, 3443, 2635, 8, 6519, 6511, 4406, 3299} \[ \frac {2 \text {Si}(2 b x)}{3 b^3}-\frac {4 x \text {Si}(b x) \sin (b x)}{3 b^2}-\frac {4 \text {Si}(b x) \cos (b x)}{3 b^3}+\frac {5 x}{6 b^2}-\frac {x \sin ^2(b x)}{3 b^2}-\frac {5 \sin (b x) \cos (b x)}{6 b^3}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {2 x^2 \text {Si}(b x) \cos (b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(6*b^2) - (5*Cos[b*x]*Sin[b*x])/(6*b^3) - (x*Sin[b*x]^2)/(3*b^2) - (4*Cos[b*x]*SinIntegral[b*x])/(3*b^3)
+ (2*x^2*Cos[b*x]*SinIntegral[b*x])/(3*b) - (4*x*Sin[b*x]*SinIntegral[b*x])/(3*b^2) + (x^3*SinIntegral[b*x]^2
)/3 + (2*SinIntegral[2*b*x])/(3*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 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 3443

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m - n
+ 1)*Sin[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sin[a + b*x^n]^
(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -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 6507

Int[(x_)^(m_.)*SinIntegral[(b_.)*(x_)]^2, x_Symbol] :> Simp[(x^(m + 1)*SinIntegral[b*x]^2)/(m + 1), x] - Dist[
2/(m + 1), Int[x^m*Sin[b*x]*SinIntegral[b*x], x], x] /; FreeQ[b, x] && IGtQ[m, 0]

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]

Rubi steps

\begin {align*} \int x^2 \text {Si}(b x)^2 \, dx &=\frac {1}{3} x^3 \text {Si}(b x)^2-\frac {2}{3} \int x^2 \sin (b x) \text {Si}(b x) \, dx\\ &=\frac {2 x^2 \cos (b x) \text {Si}(b x)}{3 b}+\frac {1}{3} x^3 \text {Si}(b x)^2-\frac {2}{3} \int \frac {x \cos (b x) \sin (b x)}{b} \, dx-\frac {4 \int x \cos (b x) \text {Si}(b x) \, dx}{3 b}\\ &=\frac {2 x^2 \cos (b x) \text {Si}(b x)}{3 b}-\frac {4 x \sin (b x) \text {Si}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {4 \int \sin (b x) \text {Si}(b x) \, dx}{3 b^2}-\frac {2 \int x \cos (b x) \sin (b x) \, dx}{3 b}+\frac {4 \int \frac {\sin ^2(b x)}{b} \, dx}{3 b}\\ &=-\frac {x \sin ^2(b x)}{3 b^2}-\frac {4 \cos (b x) \text {Si}(b x)}{3 b^3}+\frac {2 x^2 \cos (b x) \text {Si}(b x)}{3 b}-\frac {4 x \sin (b x) \text {Si}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {\int \sin ^2(b x) \, dx}{3 b^2}+\frac {4 \int \frac {\cos (b x) \sin (b x)}{b x} \, dx}{3 b^2}+\frac {4 \int \sin ^2(b x) \, dx}{3 b^2}\\ &=-\frac {5 \cos (b x) \sin (b x)}{6 b^3}-\frac {x \sin ^2(b x)}{3 b^2}-\frac {4 \cos (b x) \text {Si}(b x)}{3 b^3}+\frac {2 x^2 \cos (b x) \text {Si}(b x)}{3 b}-\frac {4 x \sin (b x) \text {Si}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {4 \int \frac {\cos (b x) \sin (b x)}{x} \, dx}{3 b^3}+\frac {\int 1 \, dx}{6 b^2}+\frac {2 \int 1 \, dx}{3 b^2}\\ &=\frac {5 x}{6 b^2}-\frac {5 \cos (b x) \sin (b x)}{6 b^3}-\frac {x \sin ^2(b x)}{3 b^2}-\frac {4 \cos (b x) \text {Si}(b x)}{3 b^3}+\frac {2 x^2 \cos (b x) \text {Si}(b x)}{3 b}-\frac {4 x \sin (b x) \text {Si}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {4 \int \frac {\sin (2 b x)}{2 x} \, dx}{3 b^3}\\ &=\frac {5 x}{6 b^2}-\frac {5 \cos (b x) \sin (b x)}{6 b^3}-\frac {x \sin ^2(b x)}{3 b^2}-\frac {4 \cos (b x) \text {Si}(b x)}{3 b^3}+\frac {2 x^2 \cos (b x) \text {Si}(b x)}{3 b}-\frac {4 x \sin (b x) \text {Si}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {2 \int \frac {\sin (2 b x)}{x} \, dx}{3 b^3}\\ &=\frac {5 x}{6 b^2}-\frac {5 \cos (b x) \sin (b x)}{6 b^3}-\frac {x \sin ^2(b x)}{3 b^2}-\frac {4 \cos (b x) \text {Si}(b x)}{3 b^3}+\frac {2 x^2 \cos (b x) \text {Si}(b x)}{3 b}-\frac {4 x \sin (b x) \text {Si}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)^2+\frac {2 \text {Si}(2 b x)}{3 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 78, normalized size = 0.70 \[ \frac {4 b^3 x^3 \text {Si}(b x)^2+8 \text {Si}(b x) \left (\left (b^2 x^2-2\right ) \cos (b x)-2 b x \sin (b x)\right )+8 \text {Si}(2 b x)+8 b x-5 \sin (2 b x)+2 b x \cos (2 b x)}{12 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*x + 2*b*x*Cos[2*b*x] - 5*Sin[2*b*x] + 8*((-2 + b^2*x^2)*Cos[b*x] - 2*b*x*Sin[b*x])*SinIntegral[b*x] + 4*b
^3*x^3*SinIntegral[b*x]^2 + 8*SinIntegral[2*b*x])/(12*b^3)

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fricas [F]  time = 1.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {Si}\left (b x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*sin_integral(b*x)^2, x)

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giac [C]  time = 0.60, size = 150, normalized size = 1.34 \[ \frac {1}{3} \, x^{3} \operatorname {Si}\left (b x\right )^{2} - \frac {2}{3} \, {\left (\frac {2 \, x \sin \left (b x\right )}{b^{2}} - \frac {{\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{b^{3}}\right )} \operatorname {Si}\left (b x\right ) + \frac {3 \, b x \tan \left (b x\right )^{2} + 2 \, \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) \tan \left (b x\right )^{2} - 2 \, \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) \tan \left (b x\right )^{2} + 4 \, \operatorname {Si}\left (2 \, b x\right ) \tan \left (b x\right )^{2} + 5 \, b x + 2 \, \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) - 2 \, \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) + 4 \, \operatorname {Si}\left (2 \, b x\right ) - 5 \, \tan \left (b x\right )}{6 \, {\left (b^{3} \tan \left (b x\right )^{2} + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*sin_integral(b*x)^2 - 2/3*(2*x*sin(b*x)/b^2 - (b^2*x^2 - 2)*cos(b*x)/b^3)*sin_integral(b*x) + 1/6*(3*b
*x*tan(b*x)^2 + 2*imag_part(cos_integral(2*b*x))*tan(b*x)^2 - 2*imag_part(cos_integral(-2*b*x))*tan(b*x)^2 + 4
*sin_integral(2*b*x)*tan(b*x)^2 + 5*b*x + 2*imag_part(cos_integral(2*b*x)) - 2*imag_part(cos_integral(-2*b*x))
+ 4*sin_integral(2*b*x) - 5*tan(b*x))/(b^3*tan(b*x)^2 + b^3)

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maple [A]  time = 0.05, size = 84, normalized size = 0.75 \[ \frac {\frac {b^{3} x^{3} \Si \left (b x \right )^{2}}{3}-2 \Si \left (b x \right ) \left (-\frac {b^{2} x^{2} \cos \left (b x \right )}{3}+\frac {2 \cos \left (b x \right )}{3}+\frac {2 b x \sin \left (b x \right )}{3}\right )+\frac {b x \left (\cos ^{2}\left (b x \right )\right )}{3}-\frac {5 \sin \left (b x \right ) \cos \left (b x \right )}{6}+\frac {b x}{2}+\frac {2 \Si \left (2 b x \right )}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/3*b^3*x^3*Si(b*x)^2-2*Si(b*x)*(-1/3*b^2*x^2*cos(b*x)+2/3*cos(b*x)+2/3*b*x*sin(b*x))+1/3*b*x*cos(b*x)^
2-5/6*sin(b*x)*cos(b*x)+1/2*b*x+2/3*Si(2*b*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*Si(b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*sinint(b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*Si(b*x)**2, x)

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