3.20 \(\int x \text {Si}(a+b x) \, dx\)

Optimal. Leaf size=71 \[ -\frac {a^2 \text {Si}(a+b x)}{2 b^2}-\frac {\sin (a+b x)}{2 b^2}-\frac {a \cos (a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {Si}(a+b x)+\frac {x \cos (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6503, 6742, 2638, 3296, 2637, 3299} \[ -\frac {a^2 \text {Si}(a+b x)}{2 b^2}-\frac {\sin (a+b x)}{2 b^2}-\frac {a \cos (a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {Si}(a+b x)+\frac {x \cos (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*SinIntegral[a + b*x],x]

[Out]

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

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

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Mathematica [A]  time = 0.12, size = 50, normalized size = 0.70 \[ \frac {\left (b^2 x^2-a^2\right ) \text {Si}(a+b x)-\sin (a+b x)+(b x-a) \cos (a+b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*SinIntegral[a + b*x],x]

[Out]

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

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fricas [F]  time = 2.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {Si}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*sin_integral(b*x + a), x)

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giac [C]  time = 0.77, size = 191, normalized size = 2.69 \[ \frac {1}{2} \, x^{2} \operatorname {Si}\left (b x + a\right ) - \frac {{\left (a^{2} \Im \left (\operatorname {Ci}\left (b x + a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - a^{2} \Im \left (\operatorname {Ci}\left (-b x - a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a^{2} \operatorname {Si}\left (b x + a\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + a^{2} \Im \left (\operatorname {Ci}\left (b x + a\right ) \right ) - a^{2} \Im \left (\operatorname {Ci}\left (-b x - a\right ) \right ) + 2 \, a^{2} \operatorname {Si}\left (b x + a\right ) - 2 \, a \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, b x + 2 \, a + 4 \, \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )} b}{4 \, {\left (b^{3} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*sin_integral(b*x + a) - 1/4*(a^2*imag_part(cos_integral(b*x + a))*tan(1/2*b*x + 1/2*a)^2 - a^2*imag_pa
rt(cos_integral(-b*x - a))*tan(1/2*b*x + 1/2*a)^2 + 2*a^2*sin_integral(b*x + a)*tan(1/2*b*x + 1/2*a)^2 + 2*b*x
*tan(1/2*b*x + 1/2*a)^2 + a^2*imag_part(cos_integral(b*x + a)) - a^2*imag_part(cos_integral(-b*x - a)) + 2*a^2
*sin_integral(b*x + a) - 2*a*tan(1/2*b*x + 1/2*a)^2 - 2*b*x + 2*a + 4*tan(1/2*b*x + 1/2*a))*b/(b^3*tan(1/2*b*x
+ 1/2*a)^2 + b^3)

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maple [A]  time = 0.02, size = 61, normalized size = 0.86 \[ \frac {\Si \left (b x +a \right ) \left (\frac {\left (b x +a \right )^{2}}{2}-a \left (b x +a \right )\right )-\frac {\sin \left (b x +a \right )}{2}+\frac {\left (b x +a \right ) \cos \left (b x +a \right )}{2}-a \cos \left (b x +a \right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Si(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \frac {x^2\,\mathrm {sinint}\left (a+b\,x\right )}{2}-\frac {\sin \left (a+b\,x\right )+a\,\cos \left (a+b\,x\right )+a^2\,\mathrm {sinint}\left (a+b\,x\right )-b\,x\,\cos \left (a+b\,x\right )}{2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinint(a + b*x),x)

[Out]

(x^2*sinint(a + b*x))/2 - (sin(a + b*x) + a*cos(a + b*x) + a^2*sinint(a + b*x) - b*x*cos(a + b*x))/(2*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*Si(a + b*x), x)

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