3.18 \(\int x^3 \text {Si}(a+b x) \, dx\)

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

[Out]

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

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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 = {6503, 6742, 2638, 3296, 2637, 3299} \[ -\frac {a^4 \text {Si}(a+b x)}{4 b^4}-\frac {a^2 \sin (a+b x)}{4 b^4}-\frac {a^3 \cos (a+b x)}{4 b^4}+\frac {a^2 x \cos (a+b x)}{4 b^3}-\frac {3 x^2 \sin (a+b x)}{4 b^2}-\frac {a x^2 \cos (a+b x)}{4 b^2}+\frac {a x \sin (a+b x)}{2 b^3}+\frac {3 \sin (a+b x)}{2 b^4}+\frac {a \cos (a+b x)}{2 b^4}-\frac {3 x \cos (a+b x)}{2 b^3}+\frac {1}{4} x^4 \text {Si}(a+b x)+\frac {x^3 \cos (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*SinIntegral[a + b*x],x]

[Out]

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

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 3299

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 6503

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]

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 96, normalized size = 0.52 \[ \frac {\left (b^4 x^4-a^4\right ) \text {Si}(a+b x)-\left (a^2-2 a b x+3 b^2 x^2-6\right ) \sin (a+b x)+\left (-a^3+a^2 b x-a b^2 x^2+2 a+b^3 x^3-6 b x\right ) \cos (a+b x)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*SinIntegral[a + b*x],x]

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \operatorname {Si}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.75, size = 338, normalized size = 1.84 \[ \frac {1}{4} \, x^{4} \operatorname {Si}\left (b x + a\right ) - \frac {{\left (2 \, b^{3} x^{3} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, a b^{2} x^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + a^{4} \Im \left (\operatorname {Ci}\left (b x + a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - a^{4} \Im \left (\operatorname {Ci}\left (-b x - a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a^{4} \operatorname {Si}\left (b x + a\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, b^{3} x^{3} + 2 \, a^{2} b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a b^{2} x^{2} + a^{4} \Im \left (\operatorname {Ci}\left (b x + a\right ) \right ) - a^{4} \Im \left (\operatorname {Ci}\left (-b x - a\right ) \right ) + 2 \, a^{4} \operatorname {Si}\left (b x + a\right ) + 12 \, b^{2} x^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 2 \, a^{3} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, a^{2} b x - 8 \, a b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 12 \, b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a^{3} + 4 \, a^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) + 4 \, a \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 12 \, b x - 4 \, a - 24 \, \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )} b}{8 \, {\left (b^{5} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*sin_integral(b*x + a) - 1/8*(2*b^3*x^3*tan(1/2*b*x + 1/2*a)^2 - 2*a*b^2*x^2*tan(1/2*b*x + 1/2*a)^2 + a
^4*imag_part(cos_integral(b*x + a))*tan(1/2*b*x + 1/2*a)^2 - a^4*imag_part(cos_integral(-b*x - a))*tan(1/2*b*x
+ 1/2*a)^2 + 2*a^4*sin_integral(b*x + a)*tan(1/2*b*x + 1/2*a)^2 - 2*b^3*x^3 + 2*a^2*b*x*tan(1/2*b*x + 1/2*a)^
2 + 2*a*b^2*x^2 + a^4*imag_part(cos_integral(b*x + a)) - a^4*imag_part(cos_integral(-b*x - a)) + 2*a^4*sin_int
egral(b*x + a) + 12*b^2*x^2*tan(1/2*b*x + 1/2*a) - 2*a^3*tan(1/2*b*x + 1/2*a)^2 - 2*a^2*b*x - 8*a*b*x*tan(1/2*
b*x + 1/2*a) - 12*b*x*tan(1/2*b*x + 1/2*a)^2 + 2*a^3 + 4*a^2*tan(1/2*b*x + 1/2*a) + 4*a*tan(1/2*b*x + 1/2*a)^2
+ 12*b*x - 4*a - 24*tan(1/2*b*x + 1/2*a))*b/(b^5*tan(1/2*b*x + 1/2*a)^2 + b^5)

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maple [A]  time = 0.02, size = 157, normalized size = 0.85 \[ \frac {\frac {\Si \left (b x +a \right ) b^{4} x^{4}}{4}+\frac {\left (b x +a \right )^{3} \cos \left (b x +a \right )}{4}-\frac {3 \left (b x +a \right )^{2} \sin \left (b x +a \right )}{4}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \left (b x +a \right ) \cos \left (b x +a \right )}{2}+a \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )-\frac {3 a^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{2}-a^{3} \cos \left (b x +a \right )-\frac {a^{4} \Si \left (b x +a \right )}{4}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*Si(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinint(a + b*x),x)

[Out]

int(x^3*sinint(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*Si(a + b*x), x)

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