\(\int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx\) [124]

Optimal result

Integrand size = 14, antiderivative size = 109 \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\frac {x}{2 b}-\frac {x \cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b}-\frac {a \operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {a \log (a+b x)}{2 b^2}+\frac {\cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6655, 6874, 2715, 8, 3393, 3383, 6647, 4491, 12, 3380} \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=-\frac {a \operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}+\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b^2}-\frac {\text {Si}(2 a+2 b x)}{2 b^2}-\frac {a \log (a+b x)}{2 b^2}+\frac {\sin (a+b x) \cos (a+b x)}{2 b^2}-\frac {x \operatorname {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {x}{2 b} \]

[In]

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

[Out]

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

Rule 6874

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.70 \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\frac {2 b x-2 a \operatorname {CosIntegral}(2 (a+b x))-2 a \log (a+b x)+\operatorname {CosIntegral}(a+b x) (-4 b x \cos (a+b x)+4 \sin (a+b x))+\sin (2 (a+b x))-2 \text {Si}(2 (a+b x))}{4 b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (a \cos \left (b x +a \right )+\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )-a \left (\frac {\ln \left (b x +a \right )}{2}+\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}\right )-\frac {\operatorname {Si}\left (2 b x +2 a \right )}{2}+\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}}{b^{2}}\)	\(96\)
default	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (a \cos \left (b x +a \right )+\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )-a \left (\frac {\ln \left (b x +a \right )}{2}+\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}\right )-\frac {\operatorname {Si}\left (2 b x +2 a \right )}{2}+\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}}{b^{2}}\)	\(96\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (99) = 198\).

Time = 0.28 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.51 \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=-\frac {2 \, \pi b^{2} x \cos \left (b x + a\right ) \operatorname {C}\left (b x + a\right ) - 2 \, \pi b \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + \sqrt {b^{2}} {\left ({\left (\pi a + 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) - \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left ({\left (\pi a - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (\pi a + 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (\pi a - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right )}{2 \, \pi b^{3}} \]

[In]

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

[Out]

-1/2*(2*pi*b^2*x*cos(b*x + a)*fresnel_cos(b*x + a) - 2*pi*b*fresnel_cos(b*x + a)*sin(b*x + a) - 2*b*cos(b*x +
a)*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) + sqrt(b^2)*((pi*a + 1)*cos(1/2/pi) - pi*sin(1/2/pi))*fresnel_c
os((pi*b*x + pi*a + 1)*sqrt(b^2)/(pi*b)) + sqrt(b^2)*((pi*a - 1)*cos(1/2/pi) + pi*sin(1/2/pi))*fresnel_cos((pi
*b*x + pi*a - 1)*sqrt(b^2)/(pi*b)) + sqrt(b^2)*(pi*cos(1/2/pi) + (pi*a + 1)*sin(1/2/pi))*fresnel_sin((pi*b*x +
pi*a + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*(pi*cos(1/2/pi) - (pi*a - 1)*sin(1/2/pi))*fresnel_sin((pi*b*x + pi*a
- 1)*sqrt(b^2)/(pi*b)))/(pi*b^3)

Sympy [F]

\[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int x \sin {\left (a + b x \right )} \operatorname {Ci}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

Integral(x*sin(a + b*x)*Ci(a + b*x), x)

Maxima [F]

\[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int { x \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int { x \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx=\int x\,\mathrm {cosint}\left (a+b\,x\right )\,\sin \left (a+b\,x\right ) \,d x \]

[In]

int(x*cosint(a + b*x)*sin(a + b*x),x)

[Out]

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