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

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

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ -\frac {a^4 \text {CosIntegral}(a+b x)}{4 b^4}+\frac {a^3 \sin (a+b x)}{4 b^4}-\frac {a^2 x \sin (a+b x)}{4 b^3}-\frac {a^2 \cos (a+b x)}{4 b^4}+\frac {a x^2 \sin (a+b x)}{4 b^2}-\frac {3 x^2 \cos (a+b x)}{4 b^2}-\frac {a \sin (a+b x)}{2 b^4}+\frac {3 x \sin (a+b x)}{2 b^3}+\frac {a x \cos (a+b x)}{2 b^3}+\frac {3 \cos (a+b x)}{2 b^4}+\frac {1}{4} x^4 \text {CosIntegral}(a+b x)-\frac {x^3 \sin (a+b x)}{4 b} \]

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 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^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*}

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Mathematica [A]  time = 0.26, size = 95, normalized size = 0.52 \[ \frac {\left (b^4 x^4-a^4\right ) \text {Ci}(a+b x)-\left (\left (a^2-2 a b x+3 b^2 x^2-6\right ) \cos (a+b x)\right )+\left (a^3-a^2 b x+a b^2 x^2-2 a-b^3 x^3+6 b x\right ) \sin (a+b x)}{4 b^4} \]

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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fricas [F]  time = 2.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \operatorname {Ci}\left (b x + a\right ), x\right ) \]

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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giac [A]  time = 0.51, size = 196, normalized size = 1.07 \[ \frac {1}{4} \, x^{4} \operatorname {Ci}\left (b x + a\right ) - \frac {a^{4} \cos \relax (a)^{2} \operatorname {Ci}\left (b x + a\right ) + a^{4} \cos \relax (a)^{2} \operatorname {Ci}\left (-b x - a\right ) + 2 \, b^{3} x^{3} \sin \left (b x + a\right ) + a^{4} \operatorname {Ci}\left (b x + a\right ) \sin \relax (a)^{2} + a^{4} \operatorname {Ci}\left (-b x - a\right ) \sin \relax (a)^{2} - 2 \, a b^{2} x^{2} \sin \left (b x + a\right ) + 6 \, b^{2} x^{2} \cos \left (b x + a\right ) + 2 \, a^{2} b x \sin \left (b x + a\right ) - 4 \, a b x \cos \left (b x + a\right ) - 2 \, a^{3} \sin \left (b x + a\right ) + 2 \, a^{2} \cos \left (b x + a\right ) - 12 \, b x \sin \left (b x + a\right ) + 4 \, a \sin \left (b x + a\right ) - 12 \, \cos \left (b x + a\right )}{8 \, b^{4}} \]

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)
+ 2*b^3*x^3*sin(b*x + a) + a^4*cos_integral(b*x + a)*sin(a)^2 + a^4*cos_integral(-b*x - a)*sin(a)^2 - 2*a*b^2*
x^2*sin(b*x + a) + 6*b^2*x^2*cos(b*x + a) + 2*a^2*b*x*sin(b*x + a) - 4*a*b*x*cos(b*x + a) - 2*a^3*sin(b*x + a)
+ 2*a^2*cos(b*x + a) - 12*b*x*sin(b*x + a) + 4*a*sin(b*x + a) - 12*cos(b*x + a))/b^4

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maple [A]  time = 0.02, size = 154, normalized size = 0.84 \[ \frac {\frac {\Ci \left (b x +a \right ) b^{4} x^{4}}{4}-\frac {\sin \left (b x +a \right ) \left (b x +a \right )^{3}}{4}-\frac {3 \left (b x +a \right )^{2} \cos \left (b x +a \right )}{4}+\frac {3 \cos \left (b x +a \right )}{2}+\frac {3 \left (b x +a \right ) \sin \left (b x +a \right )}{2}+a \left (\left (b x +a \right )^{2} \sin \left (b x +a \right )-2 \sin \left (b x +a \right )+2 \left (b x +a \right ) \cos \left (b x +a \right )\right )-\frac {3 a^{2} \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )}{2}+a^{3} \sin \left (b x +a \right )-\frac {a^{4} \Ci \left (b x +a \right )}{4}}{b^{4}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Ci}\left (b x + a\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {cosint}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosint(a + b*x),x)

[Out]

int(x^3*cosint(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {Ci}{\left (a + b x \right )}\, dx \]

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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