3.1.50 \(\int x \cos (b x) \text {Si}(b x) \, dx\) [50]

Optimal. Leaf size=61 \[ -\frac {x}{2 b}+\frac {\cos (b x) \sin (b x)}{2 b^2}+\frac {\cos (b x) \text {Si}(b x)}{b^2}+\frac {x \sin (b x) \text {Si}(b x)}{b}-\frac {\text {Si}(2 b x)}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6654, 12, 2715, 8, 6646, 4491, 3380} \begin {gather*} -\frac {\text {Si}(2 b x)}{2 b^2}+\frac {\text {Si}(b x) \cos (b x)}{b^2}+\frac {\sin (b x) \cos (b x)}{2 b^2}+\frac {x \text {Si}(b x) \sin (b x)}{b}-\frac {x}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x/b + (Cos[b*x]*Sin[b*x])/(2*b^2) + (Cos[b*x]*SinIntegral[b*x])/b^2 + (x*Sin[b*x]*SinIntegral[b*x])/b - S
inIntegral[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 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]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 42, normalized size = 0.69 \begin {gather*} \frac {-2 b x+\sin (2 b x)+4 (\cos (b x)+b x \sin (b x)) \text {Si}(b x)-2 \text {Si}(2 b x)}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.45, size = 44, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x \right ) \left (\cos \left (b x \right )+b x \sin \left (b x \right )\right )-\frac {\sinIntegral \left (2 b x \right )}{2}+\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}-\frac {b x}{2}}{b^{2}}\)	\(44\)
default	\(\frac {\sinIntegral \left (b x \right ) \left (\cos \left (b x \right )+b x \sin \left (b x \right )\right )-\frac {\sinIntegral \left (2 b x \right )}{2}+\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}-\frac {b x}{2}}{b^{2}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*sin_integral(b*x),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 43, normalized size = 0.70 \begin {gather*} -\frac {b x - {\left (2 \, b x \operatorname {Si}\left (b x\right ) + \cos \left (b x\right )\right )} \sin \left (b x\right ) - 2 \, \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) + \operatorname {Si}\left (2 \, b x\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*cos(b*x)*Si(b*x), x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.41, size = 124, normalized size = 2.03 \begin {gather*} {\left (\frac {x \sin \left (b x\right )}{b} + \frac {\cos \left (b x\right )}{b^{2}}\right )} \operatorname {Si}\left (b x\right ) - \frac {2 \, b x \tan \left (b x\right )^{2} + \Im \left ( \operatorname {Ci}\left (2 \, b x\right ) \right ) \tan \left (b x\right )^{2} - \Im \left ( \operatorname {Ci}\left (-2 \, b x\right ) \right ) \tan \left (b x\right )^{2} + 2 \, \operatorname {Si}\left (2 \, b x\right ) \tan \left (b x\right )^{2} + 2 \, b x + \Im \left ( \operatorname {Ci}\left (2 \, b x\right ) \right ) - \Im \left ( \operatorname {Ci}\left (-2 \, b x\right ) \right ) + 2 \, \operatorname {Si}\left (2 \, b x\right ) - 2 \, \tan \left (b x\right )}{4 \, {\left (b^{2} \tan \left (b x\right )^{2} + b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*sin(b*x)/b + cos(b*x)/b^2)*sin_integral(b*x) - 1/4*(2*b*x*tan(b*x)^2 + imag_part(cos_integral(2*b*x))*tan(b
*x)^2 - imag_part(cos_integral(-2*b*x))*tan(b*x)^2 + 2*sin_integral(2*b*x)*tan(b*x)^2 + 2*b*x + imag_part(cos_
integral(2*b*x)) - imag_part(cos_integral(-2*b*x)) + 2*sin_integral(2*b*x) - 2*tan(b*x))/(b^2*tan(b*x)^2 + b^2
)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,\mathrm {sinint}\left (b\,x\right )\,\cos \left (b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*sinint(b*x)*cos(b*x), x)

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