3.1.19 \(\int x^2 \text {Si}(a+b x) \, dx\) [19]

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

[Out]

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

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Rubi [A]
time = 0.19, 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 = {6638, 6874, 2718, 3377, 2717, 3380} \begin {gather*} \frac {a^3 \text {Si}(a+b x)}{3 b^3}+\frac {a^2 \cos (a+b x)}{3 b^3}+\frac {a \sin (a+b x)}{3 b^3}-\frac {2 \cos (a+b x)}{3 b^3}-\frac {2 x \sin (a+b x)}{3 b^2}-\frac {a x \cos (a+b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(a+b x)+\frac {x^2 \cos (a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*SinIntegral[a + b*x],x]

[Out]

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

Rule 2717

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

Rule 2718

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

Rule 3377

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 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 6638

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 6874

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

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 63, normalized size = 0.53 \begin {gather*} \frac {\left (-2+a^2-a b x+b^2 x^2\right ) \cos (a+b x)+(a-2 b x) \sin (a+b x)+\left (a^3+b^3 x^3\right ) \text {Si}(a+b x)}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*SinIntegral[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.30, size = 99, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {\frac {\sinIntegral \left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \sinIntegral \left (b x +a \right )}{3}+a^{2} \cos \left (b x +a \right )+a \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )+\frac {\left (b x +a \right )^{2} \cos \left (b x +a \right )}{3}-\frac {2 \cos \left (b x +a \right )}{3}-\frac {2 \left (b x +a \right ) \sin \left (b x +a \right )}{3}}{b^{3}}\)	\(99\)
default	\(\frac {\frac {\sinIntegral \left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \sinIntegral \left (b x +a \right )}{3}+a^{2} \cos \left (b x +a \right )+a \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )+\frac {\left (b x +a \right )^{2} \cos \left (b x +a \right )}{3}-\frac {2 \cos \left (b x +a \right )}{3}-\frac {2 \left (b x +a \right ) \sin \left (b x +a \right )}{3}}{b^{3}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Si(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.31, size = 91, normalized size = 0.77 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {Si}\left (b x + a\right ) - \frac {a^{3} {\left (i \, {\rm Ei}\left (i \, b x + i \, a\right ) - i \, {\rm Ei}\left (-i \, b x - i \, a\right )\right )} - 2 \, {\left ({\left (b x + a\right )}^{2} - 3 \, {\left (b x + a\right )} a + 3 \, a^{2} - 2\right )} \cos \left (b x + a\right ) + 2 \, {\left (2 \, b x - a\right )} \sin \left (b x + a\right )}{6 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*sin_integral(b*x + a) - 1/6*(a^3*(I*Ei(I*b*x + I*a) - I*Ei(-I*b*x - I*a)) - 2*((b*x + a)^2 - 3*(b*x +
a)*a + 3*a^2 - 2)*cos(b*x + a) + 2*(2*b*x - a)*sin(b*x + a))/b^3

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Fricas [A]
time = 0.34, size = 64, normalized size = 0.54 \begin {gather*} \frac {{\left (b^{2} x^{2} - a b x + a^{2} - 2\right )} \cos \left (b x + a\right ) - {\left (2 \, b x - a\right )} \sin \left (b x + a\right ) + {\left (b^{3} x^{3} + a^{3}\right )} \operatorname {Si}\left (b x + a\right )}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {Si}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*Si(a + b*x), x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.42, size = 252, normalized size = 2.14 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {Si}\left (b x + a\right ) - \frac {{\left (2 \, b^{2} x^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - a^{3} \Im \left ( \operatorname {Ci}\left (b x + a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + a^{3} \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^{3} \operatorname {Si}\left (b x + a\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, a b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, b^{2} x^{2} - a^{3} \Im \left ( \operatorname {Ci}\left (b x + a\right ) \right ) + a^{3} \Im \left ( \operatorname {Ci}\left (-b x - a\right ) \right ) - 2 \, a^{3} \operatorname {Si}\left (b x + a\right ) + 2 \, a^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a b x + 8 \, b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 2 \, a^{2} - 4 \, a \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 4 \, \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 4\right )} b}{6 \, {\left (b^{4} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*sin_integral(b*x + a) - 1/6*(2*b^2*x^2*tan(1/2*b*x + 1/2*a)^2 - a^3*imag_part(cos_integral(b*x + a))*t
an(1/2*b*x + 1/2*a)^2 + a^3*imag_part(cos_integral(-b*x - a))*tan(1/2*b*x + 1/2*a)^2 - 2*a^3*sin_integral(b*x
+ a)*tan(1/2*b*x + 1/2*a)^2 - 2*a*b*x*tan(1/2*b*x + 1/2*a)^2 - 2*b^2*x^2 - a^3*imag_part(cos_integral(b*x + a)
) + a^3*imag_part(cos_integral(-b*x - a)) - 2*a^3*sin_integral(b*x + a) + 2*a^2*tan(1/2*b*x + 1/2*a)^2 + 2*a*b
*x + 8*b*x*tan(1/2*b*x + 1/2*a) - 2*a^2 - 4*a*tan(1/2*b*x + 1/2*a) - 4*tan(1/2*b*x + 1/2*a)^2 + 4)*b/(b^4*tan(
1/2*b*x + 1/2*a)^2 + b^4)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\mathrm {sinint}\left (a+b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sinint(a + b*x),x)

[Out]

int(x^2*sinint(a + b*x), x)

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