3.2.13 \(\int x^3 \text {CosIntegral}(b x) \sin (b x) \, dx\) [113]

Optimal. Leaf size=147 \[ -\frac {5 x}{2 b^3}+\frac {x^3}{6 b}+\frac {x \cos ^2(b x)}{2 b^3}+\frac {6 x \cos (b x) \text {CosIntegral}(b x)}{b^3}-\frac {x^3 \cos (b x) \text {CosIntegral}(b x)}{b}-\frac {4 \cos (b x) \sin (b x)}{b^4}+\frac {x^2 \cos (b x) \sin (b x)}{2 b^2}-\frac {6 \text {CosIntegral}(b x) \sin (b x)}{b^4}+\frac {3 x^2 \text {CosIntegral}(b x) \sin (b x)}{b^2}-\frac {3 x \sin ^2(b x)}{2 b^3}+\frac {3 \text {Si}(2 b x)}{b^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 11, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {6655, 12, 3392, 30, 2715, 8, 6649, 3524, 6647, 4491, 3380} \begin {gather*} -\frac {6 \text {CosIntegral}(b x) \sin (b x)}{b^4}+\frac {3 \text {Si}(2 b x)}{b^4}-\frac {4 \sin (b x) \cos (b x)}{b^4}+\frac {6 x \text {CosIntegral}(b x) \cos (b x)}{b^3}-\frac {5 x}{2 b^3}-\frac {3 x \sin ^2(b x)}{2 b^3}+\frac {x \cos ^2(b x)}{2 b^3}+\frac {3 x^2 \text {CosIntegral}(b x) \sin (b x)}{b^2}+\frac {x^2 \sin (b x) \cos (b x)}{2 b^2}-\frac {x^3 \text {CosIntegral}(b x) \cos (b x)}{b}+\frac {x^3}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*CosIntegral[b*x]*Sin[b*x],x]

[Out]

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

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 30

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 3380

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 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[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 4491

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 6647

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

Rule 6649

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

Rule 6655

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 94, normalized size = 0.64 \begin {gather*} \frac {-36 b x+2 b^3 x^3+12 b x \cos (2 b x)-12 \text {CosIntegral}(b x) \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 )-24 \sin (2 b x)+3 b^2 x^2 \sin (2 b x)+36 \text {Si}(2 b x)}{12 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*CosIntegral[b*x]*Sin[b*x],x]

[Out]

(-36*b*x + 2*b^3*x^3 + 12*b*x*Cos[2*b*x] - 12*CosIntegral[b*x]*(b*x*(-6 + b^2*x^2)*Cos[b*x] - 3*(-2 + b^2*x^2)
*Sin[b*x]) - 24*Sin[2*b*x] + 3*b^2*x^2*Sin[2*b*x] + 36*SinIntegral[2*b*x])/(12*b^4)

________________________________________________________________________________________

Maple [A]
time = 0.40, size = 111, normalized size = 0.76

method	result	size
derivativedivides	\(\frac {\cosineIntegral \left (b x \right ) \left (-b^{3} x^{3} \cos \left (b x \right )+3 b^{2} x^{2} \sin \left (b x \right )-6 \sin \left (b x \right )+6 b x \cos \left (b x \right )\right )+b^{2} x^{2} \left (\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )+2 b x \left (\cos ^{2}\left (b x \right )\right )-4 \sin \left (b x \right ) \cos \left (b x \right )-4 b x -\frac {b^{3} x^{3}}{3}+3 \sinIntegral \left (2 b x \right )}{b^{4}}\)	\(111\)
default	\(\frac {\cosineIntegral \left (b x \right ) \left (-b^{3} x^{3} \cos \left (b x \right )+3 b^{2} x^{2} \sin \left (b x \right )-6 \sin \left (b x \right )+6 b x \cos \left (b x \right )\right )+b^{2} x^{2} \left (\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )+2 b x \left (\cos ^{2}\left (b x \right )\right )-4 \sin \left (b x \right ) \cos \left (b x \right )-4 b x -\frac {b^{3} x^{3}}{3}+3 \sinIntegral \left (2 b x \right )}{b^{4}}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*fresnel_cos(b*x)*sin(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*fresnel_cos(b*x)*sin(b*x), x)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (137) = 274\).
time = 0.44, size = 361, normalized size = 2.46 \begin {gather*} -\frac {2 \, \pi b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \cos \left (b x\right ) + 2 \, {\left (\pi ^{3} b^{4} x^{3} - 6 \, \pi ^{3} b^{2} x\right )} \cos \left (b x\right ) \operatorname {C}\left (b x\right ) + {\left (6 \, \pi ^{3} \sin \left (\frac {1}{2 \, \pi }\right ) - {\left (3 \, \pi ^{2} - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right )\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - {\left (6 \, \pi ^{3} \sin \left (\frac {1}{2 \, \pi }\right ) - {\left (3 \, \pi ^{2} - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right )\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - {\left (6 \, \pi ^{3} \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (3 \, \pi ^{2} - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + {\left (6 \, \pi ^{3} \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (3 \, \pi ^{2} - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + 2 \, {\left (3 \, \pi ^{2} b^{2} x \sin \left (b x\right ) - {\left (\pi ^{2} b^{3} x^{2} - 6 \, \pi ^{2} b + b\right )} \cos \left (b x\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, {\left (\pi b^{2} x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 3 \, {\left (\pi ^{3} b^{3} x^{2} - 2 \, \pi ^{3} b\right )} \operatorname {C}\left (b x\right )\right )} \sin \left (b x\right )}{2 \, \pi ^{3} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnel_cos(b*x)*sin(b*x),x, algorithm="fricas")

[Out]

-1/2*(2*pi*b*cos(1/2*pi*b^2*x^2)*cos(b*x) + 2*(pi^3*b^4*x^3 - 6*pi^3*b^2*x)*cos(b*x)*fresnel_cos(b*x) + (6*pi^
3*sin(1/2/pi) - (3*pi^2 - 1)*cos(1/2/pi))*sqrt(b^2)*fresnel_cos((pi*b*x + 1)*sqrt(b^2)/(pi*b)) - (6*pi^3*sin(1
/2/pi) - (3*pi^2 - 1)*cos(1/2/pi))*sqrt(b^2)*fresnel_cos((pi*b*x - 1)*sqrt(b^2)/(pi*b)) - (6*pi^3*cos(1/2/pi)
+ (3*pi^2 - 1)*sin(1/2/pi))*sqrt(b^2)*fresnel_sin((pi*b*x + 1)*sqrt(b^2)/(pi*b)) + (6*pi^3*cos(1/2/pi) + (3*pi
^2 - 1)*sin(1/2/pi))*sqrt(b^2)*fresnel_sin((pi*b*x - 1)*sqrt(b^2)/(pi*b)) + 2*(3*pi^2*b^2*x*sin(b*x) - (pi^2*b
^3*x^2 - 6*pi^2*b + b)*cos(b*x))*sin(1/2*pi*b^2*x^2) + 2*(pi*b^2*x*cos(1/2*pi*b^2*x^2) - 3*(pi^3*b^3*x^2 - 2*p
i^3*b)*fresnel_cos(b*x))*sin(b*x))/(pi^3*b^5)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*Ci(b*x)*sin(b*x),x)

[Out]

Integral(x**3*sin(b*x)*Ci(b*x), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*fresnel_cos(b*x)*sin(b*x),x, algorithm="giac")

[Out]

integrate(x^3*fresnel_cos(b*x)*sin(b*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosint(b*x)*sin(b*x),x)

[Out]

int(x^3*cosint(b*x)*sin(b*x), x)

________________________________________________________________________________________