\(\int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx\) [128]

Optimal result

Integrand size = 14, antiderivative size = 96 \[ \int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx=\frac {\cos (2 a+2 b x)}{4 b^2}+\frac {\cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^2}-\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {\log (a+b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}+\frac {a \text {Si}(2 a+2 b x)}{2 b^2} \]

[Out]

-1/2*Ci(2*b*x+2*a)/b^2+Ci(b*x+a)*cos(b*x+a)/b^2+1/4*cos(2*b*x+2*a)/b^2-1/2*ln(b*x+a)/b^2+1/2*a*Si(2*b*x+2*a)/b
^2+x*Ci(b*x+a)*sin(b*x+a)/b

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6649, 4669, 6873, 6874, 2718, 3380, 6653, 3393, 3383} \[ \int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx=-\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}+\frac {\operatorname {CosIntegral}(a+b x) \cos (a+b x)}{b^2}+\frac {a \text {Si}(2 a+2 b x)}{2 b^2}-\frac {\log (a+b x)}{2 b^2}+\frac {\cos (2 a+2 b x)}{4 b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b} \]

[In]

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

[Out]

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 4669

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

Rule 6649

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

Rule 6653

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

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\int \operatorname {CosIntegral}(a+b x) \sin (a+b x) \, dx}{b}-\int \frac {x \cos (a+b x) \sin (a+b x)}{a+b x} \, dx \\ & = \frac {\cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {1}{2} \int \frac {x \sin (2 (a+b x))}{a+b x} \, dx-\frac {\int \frac {\cos ^2(a+b x)}{a+b x} \, dx}{b} \\ & = \frac {\cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {1}{2} \int \frac {x \sin (2 a+2 b x)}{a+b x} \, dx-\frac {\int \left (\frac {1}{2 (a+b x)}+\frac {\cos (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b} \\ & = \frac {\cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^2}-\frac {\log (a+b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {1}{2} \int \left (\frac {\sin (2 a+2 b x)}{b}+\frac {a \sin (2 a+2 b x)}{b (-a-b x)}\right ) \, dx-\frac {\int \frac {\cos (2 a+2 b x)}{a+b x} \, dx}{2 b} \\ & = \frac {\cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^2}-\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {\log (a+b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\int \sin (2 a+2 b x) \, dx}{2 b}-\frac {a \int \frac {\sin (2 a+2 b x)}{-a-b x} \, dx}{2 b} \\ & = \frac {\cos (2 a+2 b x)}{4 b^2}+\frac {\cos (a+b x) \operatorname {CosIntegral}(a+b x)}{b^2}-\frac {\operatorname {CosIntegral}(2 a+2 b x)}{2 b^2}-\frac {\log (a+b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}+\frac {a \text {Si}(2 a+2 b x)}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.72 \[ \int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx=\frac {\cos (2 (a+b x))-2 \operatorname {CosIntegral}(2 (a+b x))-2 \log (a+b x)+4 \operatorname {CosIntegral}(a+b x) (\cos (a+b x)+b x \sin (a+b x))+2 a \text {Si}(2 (a+b x))}{4 b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (-a \sin \left (b x +a \right )+\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a \,\operatorname {Si}\left (2 b x +2 a \right )}{2}-\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}+\frac {\cos \left (b x +a \right )^{2}}{2}}{b^{2}}\)	\(82\)
default	\(\frac {\operatorname {Ci}\left (b x +a \right ) \left (-a \sin \left (b x +a \right )+\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {a \,\operatorname {Si}\left (2 b x +2 a \right )}{2}-\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Ci}\left (2 b x +2 a \right )}{2}+\frac {\cos \left (b x +a \right )^{2}}{2}}{b^{2}}\)	\(82\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (88) = 176\).

Time = 0.29 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.88 \[ \int x \cos (a+b x) \operatorname {CosIntegral}(a+b x) \, dx=\frac {2 \, \pi b^{2} x \operatorname {C}\left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, \pi b \cos \left (b x + a\right ) \operatorname {C}\left (b x + a\right ) - 2 \, b \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) \sin \left (b x + a\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) + {\left (\pi a + 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (\pi a - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left ({\left (\pi a + 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) - \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left ({\left (\pi a - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + \pi a - 1\right )} \sqrt {b^{2}}}{\pi b}\right )}{2 \, \pi b^{3}} \]

[In]

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

[Out]

1/2*(2*pi*b^2*x*fresnel_cos(b*x + a)*sin(b*x + a) + 2*pi*b*cos(b*x + a)*fresnel_cos(b*x + a) - 2*b*sin(1/2*pi*
b^2*x^2 + pi*a*b*x + 1/2*pi*a^2)*sin(b*x + a) - sqrt(b^2)*(pi*cos(1/2/pi) + (pi*a + 1)*sin(1/2/pi))*fresnel_co
s((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_cos((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_sin((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_sin((pi*b*x + pi*a -
1)*sqrt(b^2)/(pi*b)))/(pi*b^3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(x*cosint(a + b*x)*cos(a + b*x),x)

[Out]

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