3.1.10 \(\int x^3 \text {Si}(b x)^2 \, dx\) [10]

Optimal. Leaf size=149 \[ \frac {x^2}{2 b^2}+\frac {3 \text {CosIntegral}(2 b x)}{2 b^4}-\frac {3 \log (x)}{2 b^4}-\frac {x \cos (b x) \sin (b x)}{b^3}+\frac {2 \sin ^2(b x)}{b^4}-\frac {x^2 \sin ^2(b x)}{4 b^2}-\frac {3 x \cos (b x) \text {Si}(b x)}{b^3}+\frac {x^3 \cos (b x) \text {Si}(b x)}{2 b}+\frac {3 \sin (b x) \text {Si}(b x)}{b^4}-\frac {3 x^2 \sin (b x) \text {Si}(b x)}{2 b^2}+\frac {1}{4} x^4 \text {Si}(b x)^2 \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 11, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, Rules used = {6642, 6648, 12, 3524, 3391, 30, 6654, 2644, 6652, 3393, 3383} \begin {gather*} \frac {3 \text {CosIntegral}(2 b x)}{2 b^4}+\frac {3 \text {Si}(b x) \sin (b x)}{b^4}-\frac {3 \log (x)}{2 b^4}+\frac {2 \sin ^2(b x)}{b^4}-\frac {3 x \text {Si}(b x) \cos (b x)}{b^3}-\frac {x \sin (b x) \cos (b x)}{b^3}-\frac {3 x^2 \text {Si}(b x) \sin (b x)}{2 b^2}+\frac {x^2}{2 b^2}-\frac {x^2 \sin ^2(b x)}{4 b^2}+\frac {1}{4} x^4 \text {Si}(b x)^2+\frac {x^3 \text {Si}(b x) \cos (b x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*SinIntegral[b*x]^2,x]

[Out]

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

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 3524

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 6642

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

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Mathematica [A]
time = 0.09, size = 107, normalized size = 0.72 \begin {gather*} \frac {3 b^2 x^2-8 \cos (2 b x)+b^2 x^2 \cos (2 b x)+12 \text {CosIntegral}(2 b x)-12 \log (x)-4 b x \sin (2 b x)+4 \left (b x \left (-6+b^2 x^2\right ) \cos (b x)-3 \left (-2+b^2 x^2\right ) \sin (b x)\right ) \text {Si}(b x)+2 b^4 x^4 \text {Si}(b x)^2}{8 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*SinIntegral[b*x]^2,x]

[Out]

(3*b^2*x^2 - 8*Cos[2*b*x] + b^2*x^2*Cos[2*b*x] + 12*CosIntegral[2*b*x] - 12*Log[x] - 4*b*x*Sin[2*b*x] + 4*(b*x
*(-6 + b^2*x^2)*Cos[b*x] - 3*(-2 + b^2*x^2)*Sin[b*x])*SinIntegral[b*x] + 2*b^4*x^4*SinIntegral[b*x]^2)/(8*b^4)

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Maple [A]
time = 0.45, size = 154, normalized size = 1.03

method	result	size
derivativedivides	\(\frac {\frac {b^{4} x^{4} \sinIntegral \left (b x \right )^{2}}{4}-2 \sinIntegral \left (b x \right ) \left (-\frac {b^{3} x^{3} \cos \left (b x \right )}{4}+\frac {3 b^{2} x^{2} \sin \left (b x \right )}{4}-\frac {3 \sin \left (b x \right )}{2}+\frac {3 b x \cos \left (b x \right )}{2}\right )+\frac {b^{2} x^{2} \left (\cos ^{2}\left (b x \right )\right )}{4}-\frac {b x \left (\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )}{2}-\frac {b^{2} x^{2}}{4}+\frac {\left (\sin ^{2}\left (b x \right )\right )}{2}+\frac {3 b x \left (-\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )}{2}-\frac {3 \left (\cos ^{2}\left (b x \right )\right )}{2}-\frac {3 \ln \left (b x \right )}{2}+\frac {3 \cosineIntegral \left (2 b x \right )}{2}}{b^{4}}\)	\(154\)
default	\(\frac {\frac {b^{4} x^{4} \sinIntegral \left (b x \right )^{2}}{4}-2 \sinIntegral \left (b x \right ) \left (-\frac {b^{3} x^{3} \cos \left (b x \right )}{4}+\frac {3 b^{2} x^{2} \sin \left (b x \right )}{4}-\frac {3 \sin \left (b x \right )}{2}+\frac {3 b x \cos \left (b x \right )}{2}\right )+\frac {b^{2} x^{2} \left (\cos ^{2}\left (b x \right )\right )}{4}-\frac {b x \left (\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )}{2}-\frac {b^{2} x^{2}}{4}+\frac {\left (\sin ^{2}\left (b x \right )\right )}{2}+\frac {3 b x \left (-\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )}{2}-\frac {3 \left (\cos ^{2}\left (b x \right )\right )}{2}-\frac {3 \ln \left (b x \right )}{2}+\frac {3 \cosineIntegral \left (2 b x \right )}{2}}{b^{4}}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*Si(b*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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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^3*sin_integral(b*x)^2,x, algorithm="maxima")

[Out]

integrate(x^3*sin_integral(b*x)^2, x)

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Fricas [A]
time = 0.39, size = 112, normalized size = 0.75 \begin {gather*} \frac {b^{4} x^{4} \operatorname {Si}\left (b x\right )^{2} + b^{2} x^{2} + {\left (b^{2} x^{2} - 8\right )} \cos \left (b x\right )^{2} + 2 \, {\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) - 2 \, {\left (2 \, b x \cos \left (b x\right ) + 3 \, {\left (b^{2} x^{2} - 2\right )} \operatorname {Si}\left (b x\right )\right )} \sin \left (b x\right ) + 3 \, \operatorname {Ci}\left (2 \, b x\right ) + 3 \, \operatorname {Ci}\left (-2 \, b x\right ) - 6 \, \log \left (x\right )}{4 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin_integral(b*x)^2,x, algorithm="fricas")

[Out]

1/4*(b^4*x^4*sin_integral(b*x)^2 + b^2*x^2 + (b^2*x^2 - 8)*cos(b*x)^2 + 2*(b^3*x^3 - 6*b*x)*cos(b*x)*sin_integ
ral(b*x) - 2*(2*b*x*cos(b*x) + 3*(b^2*x^2 - 2)*sin_integral(b*x))*sin(b*x) + 3*cos_integral(2*b*x) + 3*cos_int
egral(-2*b*x) - 6*log(x))/b^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \operatorname {Si}^{2}{\left (b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*Si(b*x)**2,x)

[Out]

Integral(x**3*Si(b*x)**2, x)

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Giac [A]
time = 0.41, size = 117, normalized size = 0.79 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {Si}\left (b x\right )^{2} + \frac {1}{2} \, {\left (\frac {{\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right )}{b^{4}} - \frac {3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{b^{4}}\right )} \operatorname {Si}\left (b x\right ) + \frac {b^{2} x^{2} \cos \left (2 \, b x\right ) + 3 \, b^{2} x^{2} - 4 \, b x \sin \left (2 \, b x\right ) - 8 \, \cos \left (2 \, b x\right ) + 6 \, \operatorname {Ci}\left (2 \, b x\right ) + 6 \, \operatorname {Ci}\left (-2 \, b x\right ) - 12 \, \log \left (x\right )}{8 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sin_integral(b*x)^2,x, algorithm="giac")

[Out]

1/4*x^4*sin_integral(b*x)^2 + 1/2*((b^3*x^3 - 6*b*x)*cos(b*x)/b^4 - 3*(b^2*x^2 - 2)*sin(b*x)/b^4)*sin_integral
(b*x) + 1/8*(b^2*x^2*cos(2*b*x) + 3*b^2*x^2 - 4*b*x*sin(2*b*x) - 8*cos(2*b*x) + 6*cos_integral(2*b*x) + 6*cos_
integral(-2*b*x) - 12*log(x))/b^4

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\mathrm {sinint}\left (b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinint(b*x)^2,x)

[Out]

int(x^3*sinint(b*x)^2, x)

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