3.87 \(\int x^2 \text {Ci}(a+b x) \, dx\)

Optimal. Leaf size=118 \[ \frac {a^3 \text {Ci}(a+b x)}{3 b^3}-\frac {a^2 \sin (a+b x)}{3 b^3}+\frac {2 \sin (a+b x)}{3 b^3}+\frac {a \cos (a+b x)}{3 b^3}+\frac {a x \sin (a+b x)}{3 b^2}-\frac {2 x \cos (a+b x)}{3 b^2}+\frac {1}{3} x^3 \text {Ci}(a+b x)-\frac {x^2 \sin (a+b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6504, 6742, 2637, 3296, 2638, 3302} \[ \frac {a^3 \text {CosIntegral}(a+b x)}{3 b^3}-\frac {a^2 \sin (a+b x)}{3 b^3}+\frac {a x \sin (a+b x)}{3 b^2}+\frac {2 \sin (a+b x)}{3 b^3}+\frac {a \cos (a+b x)}{3 b^3}-\frac {2 x \cos (a+b x)}{3 b^2}+\frac {1}{3} x^3 \text {CosIntegral}(a+b x)-\frac {x^2 \sin (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*CosIntegral[a + b*x],x]

[Out]

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

Rule 2637

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 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]

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]]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 64, normalized size = 0.54 \[ \frac {\left (a^3+b^3 x^3\right ) \text {Ci}(a+b x)-\left (a^2-a b x+b^2 x^2-2\right ) \sin (a+b x)+(a-2 b x) \cos (a+b x)}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*CosIntegral[a + b*x],x]

[Out]

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

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {Ci}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Ci(b*x+a),x, algorithm="fricas")

[Out]

integral(x^2*cos_integral(b*x + a), x)

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giac [A]  time = 0.22, size = 143, normalized size = 1.21 \[ \frac {1}{3} \, x^{3} \operatorname {Ci}\left (b x + a\right ) + \frac {a^{3} \cos \relax (a)^{2} \operatorname {Ci}\left (b x + a\right ) + a^{3} \cos \relax (a)^{2} \operatorname {Ci}\left (-b x - a\right ) + a^{3} \operatorname {Ci}\left (b x + a\right ) \sin \relax (a)^{2} + a^{3} \operatorname {Ci}\left (-b x - a\right ) \sin \relax (a)^{2} - 2 \, b^{2} x^{2} \sin \left (b x + a\right ) + 2 \, a b x \sin \left (b x + a\right ) - 4 \, b x \cos \left (b x + a\right ) - 2 \, a^{2} \sin \left (b x + a\right ) + 2 \, a \cos \left (b x + a\right ) + 4 \, \sin \left (b x + a\right )}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Ci(b*x+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 99, normalized size = 0.84 \[ \frac {\frac {b^{3} x^{3} \Ci \left (b x +a \right )}{3}-\frac {\left (b x +a \right )^{2} \sin \left (b x +a \right )}{3}+\frac {2 \sin \left (b x +a \right )}{3}-\frac {2 \left (b x +a \right ) \cos \left (b x +a \right )}{3}+a \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )-a^{2} \sin \left (b x +a \right )+\frac {a^{3} \Ci \left (b x +a \right )}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Ci(b*x+a),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Ci}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Ci(b*x+a),x, algorithm="maxima")

[Out]

integrate(x^2*Ci(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {cosint}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosint(a + b*x),x)

[Out]

int(x^2*cosint(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {Ci}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*Ci(b*x+a),x)

[Out]

Integral(x**2*Ci(a + b*x), x)

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