3.1.60 \(\int x \cos (a+b x) \text {Si}(a+b x) \, dx\) [60]

Optimal. Leaf size=108 \[ -\frac {x}{2 b}-\frac {a \text {CosIntegral}(2 a+2 b x)}{2 b^2}+\frac {a \log (a+b x)}{2 b^2}+\frac {\cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {\cos (a+b x) \text {Si}(a+b x)}{b^2}+\frac {x \sin (a+b x) \text {Si}(a+b x)}{b}-\frac {\text {Si}(2 a+2 b x)}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6654, 6874, 2715, 8, 3393, 3383, 6646, 4491, 12, 3380} \begin {gather*} -\frac {a \text {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}+\frac {\text {Si}(a+b x) \cos (a+b x)}{b^2}+\frac {a \log (a+b x)}{2 b^2}+\frac {\sin (a+b x) \cos (a+b x)}{2 b^2}+\frac {x \text {Si}(a+b x) \sin (a+b x)}{b}-\frac {x}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x]*SinIntegral[a + b*x],x]

[Out]

-1/2*x/b - (a*CosIntegral[2*a + 2*b*x])/(2*b^2) + (a*Log[a + b*x])/(2*b^2) + (Cos[a + b*x]*Sin[a + b*x])/(2*b^
2) + (Cos[a + b*x]*SinIntegral[a + b*x])/b^2 + (x*Sin[a + b*x]*SinIntegral[a + b*x])/b - SinIntegral[2*a + 2*b
*x]/(2*b^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 3383

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 6646

Int[Sin[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(-Cos[a + b*x])*(SinIntegral[c
+ d*x]/b), x] + Dist[d/b, Int[Cos[a + b*x]*(Sin[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]

Rule 6654

Int[Cos[(a_.) + (b_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(e +
f*x)^m*Sin[a + b*x]*(SinIntegral[c + d*x]/b), x] + (-Dist[d/b, Int[(e + f*x)^m*Sin[a + b*x]*(Sin[c + d*x]/(c
+ d*x)), x], x] - Dist[f*(m/b), Int[(e + f*x)^(m - 1)*Sin[a + b*x]*SinIntegral[c + d*x], x], x]) /; FreeQ[{a,
b, c, d, e, f}, x] && IGtQ[m, 0]

Rule 6874

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

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 74, normalized size = 0.69 \begin {gather*} \frac {-2 b x-2 a \text {CosIntegral}(2 (a+b x))+2 a \log (a+b x)+\sin (2 (a+b x))+4 (\cos (a+b x)+b x \sin (a+b x)) \text {Si}(a+b x)-2 \text {Si}(2 (a+b x))}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x]*SinIntegral[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.57, size = 94, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x +a \right ) \left (-a \sin \left (b x +a \right )+\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a \ln \left (b x +a \right )}{2}-\frac {a \cosineIntegral \left (2 b x +2 a \right )}{2}-\frac {\sinIntegral \left (2 b x +2 a \right )}{2}+\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}-\frac {b x}{2}-\frac {a}{2}}{b^{2}}\)	\(94\)
default	\(\frac {\sinIntegral \left (b x +a \right ) \left (-a \sin \left (b x +a \right )+\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a \ln \left (b x +a \right )}{2}-\frac {a \cosineIntegral \left (2 b x +2 a \right )}{2}-\frac {\sinIntegral \left (2 b x +2 a \right )}{2}+\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}-\frac {b x}{2}-\frac {a}{2}}{b^{2}}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*cos(b*x + a)*sin_integral(b*x + a), x)

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Fricas [A]
time = 0.35, size = 91, normalized size = 0.84 \begin {gather*} -\frac {2 \, b x + a \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) + a \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) - 2 \, a \log \left (b x + a\right ) - 2 \, {\left (2 \, b x \operatorname {Si}\left (b x + a\right ) + \cos \left (b x + a\right )\right )} \sin \left (b x + a\right ) - 4 \, \cos \left (b x + a\right ) \operatorname {Si}\left (b x + a\right ) + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b*x + a*cos_integral(2*b*x + 2*a) + a*cos_integral(-2*b*x - 2*a) - 2*a*log(b*x + a) - 2*(2*b*x*sin_int
egral(b*x + a) + cos(b*x + a))*sin(b*x + a) - 4*cos(b*x + a)*sin_integral(b*x + a) + 2*sin_integral(2*b*x + 2*
a))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*cos(a + b*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 = 528, normalized size = 4.89 \begin {gather*} {\left (\frac {x \sin \left (b x + a\right )}{b} + \frac {\cos \left (b x + a\right )}{b^{2}}\right )} \operatorname {Si}\left (b x + a\right ) - \frac {2 \, b x \tan \left (b x\right )^{2} \tan \left (a\right )^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} \tan \left (a\right )^{2} + a \Re \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \left (a\right )^{2} + a \Re \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \left (a\right )^{2} + \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \left (a\right )^{2} - \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} \tan \left (a\right )^{2} + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} \tan \left (a\right )^{2} + 2 \, b x \tan \left (b x\right )^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (b x\right )^{2} + a \Re \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + a \Re \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + 2 \, b x \tan \left (a\right )^{2} - 2 \, a \log \left ({\left | b x + a \right |}\right ) \tan \left (a\right )^{2} + a \Re \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (a\right )^{2} + a \Re \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (a\right )^{2} + \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (b x\right )^{2} - \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (b x\right )^{2} + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (b x\right )^{2} + 2 \, \tan \left (b x\right )^{2} \tan \left (a\right ) + \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) \tan \left (a\right )^{2} - \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) \tan \left (a\right )^{2} + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) \tan \left (a\right )^{2} + 2 \, \tan \left (b x\right ) \tan \left (a\right )^{2} + 2 \, b x - 2 \, a \log \left ({\left | b x + a \right |}\right ) + a \Re \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) + a \Re \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + \Im \left ( \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - \Im \left ( \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right ) - 2 \, \tan \left (b x\right ) - 2 \, \tan \left (a\right )}{4 \, {\left (b^{2} \tan \left (b x\right )^{2} \tan \left (a\right )^{2} + b^{2} \tan \left (b x\right )^{2} + b^{2} \tan \left (a\right )^{2} + b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*sin(b*x + a)/b + cos(b*x + a)/b^2)*sin_integral(b*x + a) - 1/4*(2*b*x*tan(b*x)^2*tan(a)^2 - 2*a*log(abs(b*x
+ a))*tan(b*x)^2*tan(a)^2 + a*real_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 + a*real_part(cos_inte
gral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 + imag_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2*tan(a)^2 - imag_part
(cos_integral(-2*b*x - 2*a))*tan(b*x)^2*tan(a)^2 + 2*sin_integral(2*b*x + 2*a)*tan(b*x)^2*tan(a)^2 + 2*b*x*tan
(b*x)^2 - 2*a*log(abs(b*x + a))*tan(b*x)^2 + a*real_part(cos_integral(2*b*x + 2*a))*tan(b*x)^2 + a*real_part(c
os_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*b*x*tan(a)^2 - 2*a*log(abs(b*x + a))*tan(a)^2 + a*real_part(cos_inte
gral(2*b*x + 2*a))*tan(a)^2 + a*real_part(cos_integral(-2*b*x - 2*a))*tan(a)^2 + imag_part(cos_integral(2*b*x
+ 2*a))*tan(b*x)^2 - imag_part(cos_integral(-2*b*x - 2*a))*tan(b*x)^2 + 2*sin_integral(2*b*x + 2*a)*tan(b*x)^2
+ 2*tan(b*x)^2*tan(a) + imag_part(cos_integral(2*b*x + 2*a))*tan(a)^2 - imag_part(cos_integral(-2*b*x - 2*a))
*tan(a)^2 + 2*sin_integral(2*b*x + 2*a)*tan(a)^2 + 2*tan(b*x)*tan(a)^2 + 2*b*x - 2*a*log(abs(b*x + a)) + a*rea
l_part(cos_integral(2*b*x + 2*a)) + a*real_part(cos_integral(-2*b*x - 2*a)) + imag_part(cos_integral(2*b*x + 2
*a)) - imag_part(cos_integral(-2*b*x - 2*a)) + 2*sin_integral(2*b*x + 2*a) - 2*tan(b*x) - 2*tan(a))/(b^2*tan(b
*x)^2*tan(a)^2 + b^2*tan(b*x)^2 + b^2*tan(a)^2 + b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinint(a + b*x)*cos(a + b*x),x)

[Out]

int(x*sinint(a + b*x)*cos(a + b*x), x)

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