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

Optimal. Leaf size=63 \[ -\frac {3 x \cos (b x)}{2 b^3}+\frac {x^3 \cos (b x)}{4 b}+\frac {3 \sin (b x)}{2 b^4}-\frac {3 x^2 \sin (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Si}(b x) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6638, 12, 3377, 2717} \begin {gather*} \frac {3 \sin (b x)}{2 b^4}-\frac {3 x \cos (b x)}{2 b^3}-\frac {3 x^2 \sin (b x)}{4 b^2}+\frac {1}{4} x^4 \text {Si}(b x)+\frac {x^3 \cos (b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*Cos[b*x])/(2*b^3) + (x^3*Cos[b*x])/(4*b) + (3*Sin[b*x])/(2*b^4) - (3*x^2*Sin[b*x])/(4*b^2) + (x^4*SinInt
egral[b*x])/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 2717

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

Rule 3377

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 6638

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinIntegr
al[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Sin[a + b*x]/(a + b*x)), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 50, normalized size = 0.79 \begin {gather*} \frac {b x \left (-6+b^2 x^2\right ) \cos (b x)-3 \left (-2+b^2 x^2\right ) \sin (b x)+b^4 x^4 \text {Si}(b x)}{4 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.27, size = 56, normalized size = 0.89

method	result	size
meijerg	\(\frac {b \,x^{5} \hypergeom \left (\left [\frac {1}{2}, \frac {5}{2}\right ], \left [\frac {3}{2}, \frac {3}{2}, \frac {7}{2}\right ], -\frac {b^{2} x^{2}}{4}\right )}{5}\)	\(23\)
derivativedivides	\(\frac {\frac {b^{4} x^{4} \sinIntegral \left (b x \right )}{4}+\frac {b^{3} x^{3} \cos \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \sin \left (b x \right )}{4}+\frac {3 \sin \left (b x \right )}{2}-\frac {3 b x \cos \left (b x \right )}{2}}{b^{4}}\)	\(56\)
default	\(\frac {\frac {b^{4} x^{4} \sinIntegral \left (b x \right )}{4}+\frac {b^{3} x^{3} \cos \left (b x \right )}{4}-\frac {3 b^{2} x^{2} \sin \left (b x \right )}{4}+\frac {3 \sin \left (b x \right )}{2}-\frac {3 b x \cos \left (b x \right )}{2}}{b^{4}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^4*(1/4*b^4*x^4*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))

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Maxima [A]
time = 0.26, size = 48, normalized size = 0.76 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {Si}\left (b x\right ) + \frac {{\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right ) - 3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 49, normalized size = 0.78 \begin {gather*} \frac {b^{4} x^{4} \operatorname {Si}\left (b x\right ) + {\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right ) - 3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b 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),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.68, size = 61, normalized size = 0.97 \begin {gather*} \frac {x^{4} \operatorname {Si}{\left (b x \right )}}{4} + \frac {x^{3} \cos {\left (b x \right )}}{4 b} - \frac {3 x^{2} \sin {\left (b x \right )}}{4 b^{2}} - \frac {3 x \cos {\left (b x \right )}}{2 b^{3}} + \frac {3 \sin {\left (b x \right )}}{2 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*Si(b*x)/4 + x**3*cos(b*x)/(4*b) - 3*x**2*sin(b*x)/(4*b**2) - 3*x*cos(b*x)/(2*b**3) + 3*sin(b*x)/(2*b**4)

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Giac [A]
time = 0.41, size = 49, normalized size = 0.78 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {Si}\left (b x\right ) + \frac {{\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right )}{4 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b 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),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \frac {\sin \left (b\,x\right )\,\left (\frac {6}{b^4}-\frac {3\,x^2}{b^2}\right )}{4}+\frac {x^4\,\mathrm {sinint}\left (b\,x\right )}{4}-\frac {\cos \left (b\,x\right )\,\left (\frac {6\,x}{b^3}-\frac {x^3}{b}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(b*x)*(6/b^4 - (3*x^2)/b^2))/4 + (x^4*sinint(b*x))/4 - (cos(b*x)*((6*x)/b^3 - x^3/b))/4

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