3.1.51 \(\int x^2 \cos (b x) \text {Si}(b x) \, dx\) [51]

Optimal. Leaf size=98 \[ -\frac {x^2}{4 b}-\frac {\text {CosIntegral}(2 b x)}{b^3}+\frac {\log (x)}{b^3}+\frac {x \cos (b x) \sin (b x)}{2 b^2}-\frac {5 \sin ^2(b x)}{4 b^3}+\frac {2 x \cos (b x) \text {Si}(b x)}{b^2}-\frac {2 \sin (b x) \text {Si}(b x)}{b^3}+\frac {x^2 \sin (b x) \text {Si}(b x)}{b} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6654, 12, 3391, 30, 6648, 2644, 6652, 3393, 3383} \begin {gather*} -\frac {\text {CosIntegral}(2 b x)}{b^3}-\frac {2 \text {Si}(b x) \sin (b x)}{b^3}+\frac {\log (x)}{b^3}-\frac {5 \sin ^2(b x)}{4 b^3}+\frac {2 x \text {Si}(b x) \cos (b x)}{b^2}+\frac {x \sin (b x) \cos (b x)}{2 b^2}+\frac {x^2 \text {Si}(b x) \sin (b x)}{b}-\frac {x^2}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*x^2/b - CosIntegral[2*b*x]/b^3 + Log[x]/b^3 + (x*Cos[b*x]*Sin[b*x])/(2*b^2) - (5*Sin[b*x]^2)/(4*b^3) + (2
*x*Cos[b*x]*SinIntegral[b*x])/b^2 - (2*Sin[b*x]*SinIntegral[b*x])/b^3 + (x^2*Sin[b*x]*SinIntegral[b*x])/b

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

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

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 3383

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 3391

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 3393

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 6648

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 6652

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 6654

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

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Mathematica [A]
time = 0.07, size = 72, normalized size = 0.73 \begin {gather*} \frac {-2 b^2 x^2+5 \cos (2 b x)-8 \text {CosIntegral}(2 b x)+8 \log (x)+2 b x \sin (2 b x)+8 \left (2 b x \cos (b x)+\left (-2+b^2 x^2\right ) \sin (b x)\right ) \text {Si}(b x)}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.37, size = 89, normalized size = 0.91

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x \right ) \left (b^{2} x^{2} \sin \left (b x \right )-2 \sin \left (b x \right )+2 b x \cos \left (b x \right )\right )-b x \left (-\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )+\frac {b^{2} x^{2}}{4}-\frac {\left (\sin ^{2}\left (b x \right )\right )}{4}+\ln \left (b x \right )-\cosineIntegral \left (2 b x \right )+\cos ^{2}\left (b x \right )}{b^{3}}\)	\(89\)
default	\(\frac {\sinIntegral \left (b x \right ) \left (b^{2} x^{2} \sin \left (b x \right )-2 \sin \left (b x \right )+2 b x \cos \left (b x \right )\right )-b x \left (-\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )+\frac {b^{2} x^{2}}{4}-\frac {\left (\sin ^{2}\left (b x \right )\right )}{4}+\ln \left (b x \right )-\cosineIntegral \left (2 b x \right )+\cos ^{2}\left (b x \right )}{b^{3}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*cos(b*x)*sin_integral(b*x), x)

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Fricas [A]
time = 0.35, size = 80, normalized size = 0.82 \begin {gather*} -\frac {b^{2} x^{2} - 8 \, b x \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) - 5 \, \cos \left (b x\right )^{2} - 2 \, {\left (b x \cos \left (b x\right ) + 2 \, {\left (b^{2} x^{2} - 2\right )} \operatorname {Si}\left (b x\right )\right )} \sin \left (b x\right ) + 2 \, \operatorname {Ci}\left (2 \, b x\right ) + 2 \, \operatorname {Ci}\left (-2 \, b x\right ) - 4 \, \log \left (x\right )}{4 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*x^2 - 8*b*x*cos(b*x)*sin_integral(b*x) - 5*cos(b*x)^2 - 2*(b*x*cos(b*x) + 2*(b^2*x^2 - 2)*sin_integr
al(b*x))*sin(b*x) + 2*cos_integral(2*b*x) + 2*cos_integral(-2*b*x) - 4*log(x))/b^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \cos {\left (b x \right )} \operatorname {Si}{\left (b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*cos(b*x)*Si(b*x), x)

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Giac [A]
time = 0.42, size = 82, normalized size = 0.84 \begin {gather*} {\left (\frac {2 \, x \cos \left (b x\right )}{b^{2}} + \frac {{\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{b^{3}}\right )} \operatorname {Si}\left (b x\right ) - \frac {2 \, b^{2} x^{2} - 2 \, b x \sin \left (2 \, b x\right ) - 5 \, \cos \left (2 \, b x\right ) + 4 \, \operatorname {Ci}\left (2 \, b x\right ) + 4 \, \operatorname {Ci}\left (-2 \, b x\right ) - 8 \, \log \left (x\right )}{8 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*cos(b*x)/b^2 + (b^2*x^2 - 2)*sin(b*x)/b^3)*sin_integral(b*x) - 1/8*(2*b^2*x^2 - 2*b*x*sin(2*b*x) - 5*cos(
2*b*x) + 4*cos_integral(2*b*x) + 4*cos_integral(-2*b*x) - 8*log(x))/b^3

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\mathrm {sinint}\left (b\,x\right )\,\cos \left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*sinint(b*x)*cos(b*x), x)

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