3.1.20 \(\int x \text {Si}(a+b x) \, dx\) [20]

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

[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 = 0.14, 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 = {6638, 6874, 2718, 3377, 2717, 3380} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*SinIntegral[a + b*x],x]

[Out]

-1/2*(a*Cos[a + b*x])/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 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 \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 = 0.07, size = 50, normalized size = 0.70 \begin {gather*} \frac {(-a+b x) \cos (a+b x)-\sin (a+b x)+\left (-a^2+b^2 x^2\right ) \text {Si}(a+b x)}{2 b^2} \end {gather*}

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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Maple [A]
time = 0.32, size = 61, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x +a \right ) \left (-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{2}\right )-a \cos \left (b x +a \right )-\frac {\sin \left (b x +a \right )}{2}+\frac {\left (b x +a \right ) \cos \left (b x +a \right )}{2}}{b^{2}}\)	\(61\)
default	\(\frac {\sinIntegral \left (b x +a \right ) \left (-a \left (b x +a \right )+\frac {\left (b x +a \right )^{2}}{2}\right )-a \cos \left (b x +a \right )-\frac {\sin \left (b x +a \right )}{2}+\frac {\left (b x +a \right ) \cos \left (b x +a \right )}{2}}{b^{2}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.30, size = 68, normalized size = 0.96 \begin {gather*} \frac {1}{2} \, x^{2} \operatorname {Si}\left (b x + a\right ) - \frac {a^{2} {\left (-i \, {\rm Ei}\left (i \, b x + i \, a\right ) + i \, {\rm Ei}\left (-i \, b x - i \, a\right )\right )} - 2 \, {\left (b x - a\right )} \cos \left (b x + a\right ) + 2 \, \sin \left (b x + a\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*sin_integral(b*x + a) - 1/4*(a^2*(-I*Ei(I*b*x + I*a) + I*Ei(-I*b*x - I*a)) - 2*(b*x - a)*cos(b*x + a)
+ 2*sin(b*x + a))/b^2

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Fricas [A]
time = 0.35, size = 48, normalized size = 0.68 \begin {gather*} \frac {{\left (b x - a\right )} \cos \left (b x + a\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \operatorname {Si}\left (b x + a\right ) - \sin \left (b x + a\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.42, size = 191, normalized size = 2.69 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin_integral(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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} \end {gather*}

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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