3.86 \(\int x^3 \text{CosIntegral}(a+b x) \, dx\)

Optimal. Leaf size=184 \[ -\frac{a^4 \text{CosIntegral}(a+b x)}{4 b^4}+\frac{a^3 \sin (a+b x)}{4 b^4}-\frac{a^2 x \sin (a+b x)}{4 b^3}-\frac{a^2 \cos (a+b x)}{4 b^4}+\frac{a x^2 \sin (a+b x)}{4 b^2}-\frac{3 x^2 \cos (a+b x)}{4 b^2}-\frac{a \sin (a+b x)}{2 b^4}+\frac{3 x \sin (a+b x)}{2 b^3}+\frac{a x \cos (a+b x)}{2 b^3}+\frac{3 \cos (a+b x)}{2 b^4}+\frac{1}{4} x^4 \text{CosIntegral}(a+b x)-\frac{x^3 \sin (a+b x)}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.374411, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6504, 6742, 2637, 3296, 2638, 3302} \[ -\frac{a^4 \text{CosIntegral}(a+b x)}{4 b^4}+\frac{a^3 \sin (a+b x)}{4 b^4}-\frac{a^2 x \sin (a+b x)}{4 b^3}-\frac{a^2 \cos (a+b x)}{4 b^4}+\frac{a x^2 \sin (a+b x)}{4 b^2}-\frac{3 x^2 \cos (a+b x)}{4 b^2}-\frac{a \sin (a+b x)}{2 b^4}+\frac{3 x \sin (a+b x)}{2 b^3}+\frac{a x \cos (a+b x)}{2 b^3}+\frac{3 \cos (a+b x)}{2 b^4}+\frac{1}{4} x^4 \text{CosIntegral}(a+b x)-\frac{x^3 \sin (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*CosIntegral[a + b*x],x]

[Out]

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

Rule 6504

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2637

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

Rule 3296

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 2638

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

Rule 3302

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]

Rubi steps

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

Mathematica [A]  time = 0.217777, size = 95, normalized size = 0.52 \[ \frac{\left (b^4 x^4-a^4\right ) \text{CosIntegral}(a+b x)+\left (-a^2 b x+a^3+a b^2 x^2-2 a-b^3 x^3+6 b x\right ) \sin (a+b x)-\left (a^2-2 a b x+3 b^2 x^2-6\right ) \cos (a+b x)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*CosIntegral[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.052, size = 154, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{\it Ci} \left ( bx+a \right ){b}^{4}{x}^{4}}{4}}-{\frac{\sin \left ( bx+a \right ) \left ( bx+a \right ) ^{3}}{4}}-{\frac{3\, \left ( bx+a \right ) ^{2}\cos \left ( bx+a \right ) }{4}}+{\frac{3\,\cos \left ( bx+a \right ) }{2}}+{\frac{3\,\sin \left ( bx+a \right ) \left ( bx+a \right ) }{2}}+a \left ( \left ( bx+a \right ) ^{2}\sin \left ( bx+a \right ) -2\,\sin \left ( bx+a \right ) +2\, \left ( bx+a \right ) \cos \left ( bx+a \right ) \right ) -{\frac{3\,{a}^{2} \left ( \sin \left ( bx+a \right ) \left ( bx+a \right ) +\cos \left ( bx+a \right ) \right ) }{2}}+{a}^{3}\sin \left ( bx+a \right ) -{\frac{{a}^{4}{\it Ci} \left ( bx+a \right ) }{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*Ci(b*x+a),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm Ci}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*Ci(b*x+a),x, algorithm="maxima")

[Out]

integrate(x^3*Ci(b*x + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} \operatorname{Ci}\left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*Ci(b*x+a),x, algorithm="fricas")

[Out]

integral(x^3*cos_integral(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*Ci(b*x+a),x)

[Out]

Integral(x**3*Ci(a + b*x), x)

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Giac [A]  time = 1.20939, size = 108, normalized size = 0.59 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{Ci}\left (b x + a\right ) - \frac{a^{4} \cos \left (a\right )^{2} \operatorname{Ci}\left (b x + a\right ) + a^{4} \cos \left (a\right )^{2} \operatorname{Ci}\left (-b x - a\right ) + a^{4} \operatorname{Ci}\left (b x + a\right ) \sin \left (a\right )^{2} + a^{4} \operatorname{Ci}\left (-b x - a\right ) \sin \left (a\right )^{2}}{8 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*Ci(b*x+a),x, algorithm="giac")

[Out]

1/4*x^4*cos_integral(b*x + a) - 1/8*(a^4*cos(a)^2*cos_integral(b*x + a) + a^4*cos(a)^2*cos_integral(-b*x - a)
+ a^4*cos_integral(b*x + a)*sin(a)^2 + a^4*cos_integral(-b*x - a)*sin(a)^2)/b^4