3.2.11 \(\int x \text {CosIntegral}(b x) \sin (b x) \, dx\) [111]

Optimal. Leaf size=62 \[ \frac {x}{2 b}-\frac {x \cos (b x) \text {CosIntegral}(b x)}{b}+\frac {\cos (b x) \sin (b x)}{2 b^2}+\frac {\text {CosIntegral}(b x) \sin (b x)}{b^2}-\frac {\text {Si}(2 b x)}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 62, 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 = {6655, 12, 2715, 8, 6647, 4491, 3380} \begin {gather*} \frac {\text {CosIntegral}(b x) \sin (b x)}{b^2}-\frac {\text {Si}(2 b x)}{2 b^2}+\frac {\sin (b x) \cos (b x)}{2 b^2}-\frac {x \text {CosIntegral}(b x) \cos (b x)}{b}+\frac {x}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*CosIntegral[b*x]*Sin[b*x],x]

[Out]

x/(2*b) - (x*Cos[b*x]*CosIntegral[b*x])/b + (Cos[b*x]*Sin[b*x])/(2*b^2) + (CosIntegral[b*x]*Sin[b*x])/b^2 - Si
nIntegral[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 6647

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

Rule 6655

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 44, normalized size = 0.71 \begin {gather*} \frac {2 b x+\text {CosIntegral}(b x) (-4 b x \cos (b x)+4 \sin (b x))+\sin (2 b x)-2 \text {Si}(2 b x)}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*CosIntegral[b*x]*Sin[b*x],x]

[Out]

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

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Maple [A]
time = 0.41, size = 45, normalized size = 0.73

method	result	size
derivativedivides	\(\frac {\cosineIntegral \left (b x \right ) \left (\sin \left (b x \right )-b x \cos \left (b x \right )\right )-\frac {\sinIntegral \left (2 b x \right )}{2}+\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}}{b^{2}}\)	\(45\)
default	\(\frac {\cosineIntegral \left (b x \right ) \left (\sin \left (b x \right )-b x \cos \left (b x \right )\right )-\frac {\sinIntegral \left (2 b x \right )}{2}+\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}}{b^{2}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*Ci(b*x)*sin(b*x),x,method=_RETURNVERBOSE)

[Out]

1/b^2*(Ci(b*x)*(sin(b*x)-b*x*cos(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*fresnel_cos(b*x)*sin(b*x),x, algorithm="maxima")

[Out]

integrate(x*fresnel_cos(b*x)*sin(b*x), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (56) = 112\).
time = 0.41, size = 219, normalized size = 3.53 \begin {gather*} -\frac {2 \, \pi b^{2} x \cos \left (b x\right ) \operatorname {C}\left (b x\right ) - 2 \, \pi b \operatorname {C}\left (b x\right ) \sin \left (b x\right ) - 2 \, b \cos \left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - \sqrt {b^{2}} {\left (\pi \sin \left (\frac {1}{2 \, \pi }\right ) - \cos \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left (\pi \sin \left (\frac {1}{2 \, \pi }\right ) - \cos \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right )}{2 \, \pi b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnel_cos(b*x)*sin(b*x),x, algorithm="fricas")

[Out]

-1/2*(2*pi*b^2*x*cos(b*x)*fresnel_cos(b*x) - 2*pi*b*fresnel_cos(b*x)*sin(b*x) - 2*b*cos(b*x)*sin(1/2*pi*b^2*x^
2) - sqrt(b^2)*(pi*sin(1/2/pi) - cos(1/2/pi))*fresnel_cos((pi*b*x + 1)*sqrt(b^2)/(pi*b)) + sqrt(b^2)*(pi*sin(1
/2/pi) - cos(1/2/pi))*fresnel_cos((pi*b*x - 1)*sqrt(b^2)/(pi*b)) + sqrt(b^2)*(pi*cos(1/2/pi) + sin(1/2/pi))*fr
esnel_sin((pi*b*x + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*(pi*cos(1/2/pi) + sin(1/2/pi))*fresnel_sin((pi*b*x - 1)*s
qrt(b^2)/(pi*b)))/(pi*b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Ci(b*x)*sin(b*x),x)

[Out]

Integral(x*sin(b*x)*Ci(b*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnel_cos(b*x)*sin(b*x),x, algorithm="giac")

[Out]

integrate(x*fresnel_cos(b*x)*sin(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosint(b*x)*sin(b*x),x)

[Out]

int(x*cosint(b*x)*sin(b*x), x)

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