\(\int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx\) [112]

Optimal result

Integrand size = 12, antiderivative size = 111 \[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=\frac {x^2}{4 b}+\frac {\cos ^2(b x)}{4 b^3}+\frac {2 \cos (b x) \operatorname {CosIntegral}(b x)}{b^3}-\frac {x^2 \cos (b x) \operatorname {CosIntegral}(b x)}{b}-\frac {\operatorname {CosIntegral}(2 b x)}{b^3}-\frac {\log (x)}{b^3}+\frac {x \cos (b x) \sin (b x)}{2 b^2}+\frac {2 x \operatorname {CosIntegral}(b x) \sin (b x)}{b^2}-\frac {\sin ^2(b x)}{b^3} \] Output:

1/4*x^2/b+1/4*cos(b*x)^2/b^3+2*cos(b*x)*Ci(b*x)/b^3-x^2*cos(b*x)*Ci(b*x)/b 
-Ci(2*b*x)/b^3-ln(x)/b^3+1/2*x*cos(b*x)*sin(b*x)/b^2+2*x*Ci(b*x)*sin(b*x)/ 
b^2-sin(b*x)^2/b^3
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.65 \[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=\frac {2 b^2 x^2+5 \cos (2 b x)-8 \operatorname {CosIntegral}(2 b x)-8 \log (x)-8 \operatorname {CosIntegral}(b x) \left (\left (-2+b^2 x^2\right ) \cos (b x)-2 b x \sin (b x)\right )+2 b x \sin (2 b x)}{8 b^3} \] Input:

Integrate[x^2*CosIntegral[b*x]*Sin[b*x],x]
 

Output:

(2*b^2*x^2 + 5*Cos[2*b*x] - 8*CosIntegral[2*b*x] - 8*Log[x] - 8*CosIntegra 
l[b*x]*((-2 + b^2*x^2)*Cos[b*x] - 2*b*x*Sin[b*x]) + 2*b*x*Sin[2*b*x])/(8*b 
^3)
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {7074, 27, 3042, 3791, 15, 7068, 27, 3042, 3044, 15, 7072, 27, 3042, 3793, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx\)

\(\Big \downarrow \) 7074

\(\displaystyle \frac {2 \int x \cos (b x) \operatorname {CosIntegral}(b x)dx}{b}+\int \frac {x \cos ^2(b x)}{b}dx-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \int x \cos (b x) \operatorname {CosIntegral}(b x)dx}{b}+\frac {\int x \cos ^2(b x)dx}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int x \cos (b x) \operatorname {CosIntegral}(b x)dx}{b}+\frac {\int x \sin \left (b x+\frac {\pi }{2}\right )^2dx}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 3791

\(\displaystyle \frac {\frac {\int xdx}{2}+\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}}{b}+\frac {2 \int x \cos (b x) \operatorname {CosIntegral}(b x)dx}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 \int x \cos (b x) \operatorname {CosIntegral}(b x)dx}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 7068

\(\displaystyle \frac {2 \left (-\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\int \frac {\cos (b x) \sin (b x)}{b}dx+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\frac {\int \cos (b x) \sin (b x)dx}{b}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (-\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\frac {\int \cos (b x) \sin (b x)dx}{b}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 3044

\(\displaystyle \frac {2 \left (-\frac {\int \sin (b x)d\sin (b x)}{b^2}-\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 \left (-\frac {\int \operatorname {CosIntegral}(b x) \sin (b x)dx}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 7072

\(\displaystyle \frac {2 \left (-\frac {\int \frac {\cos ^2(b x)}{b x}dx-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-\frac {\frac {\int \frac {\cos ^2(b x)}{x}dx}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (-\frac {\frac {\int \frac {\sin \left (b x+\frac {\pi }{2}\right )^2}{x}dx}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {2 \left (-\frac {\frac {\int \left (\frac {\cos (2 b x)}{2 x}+\frac {1}{2 x}\right )dx}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \operatorname {CosIntegral}(b x) \sin (b x)}{b}-\frac {\frac {\frac {\operatorname {CosIntegral}(2 b x)}{2}+\frac {\log (x)}{2}}{b}-\frac {\operatorname {CosIntegral}(b x) \cos (b x)}{b}}{b}\right )}{b}+\frac {\frac {\cos ^2(b x)}{4 b^2}+\frac {x \sin (b x) \cos (b x)}{2 b}+\frac {x^2}{4}}{b}-\frac {x^2 \operatorname {CosIntegral}(b x) \cos (b x)}{b}\)

Input:

Int[x^2*CosIntegral[b*x]*Sin[b*x],x]
 

Output:

-((x^2*Cos[b*x]*CosIntegral[b*x])/b) + (x^2/4 + Cos[b*x]^2/(4*b^2) + (x*Co 
s[b*x]*Sin[b*x])/(2*b))/b + (2*(-((-((Cos[b*x]*CosIntegral[b*x])/b) + (Cos 
Integral[2*b*x]/2 + Log[x]/2)/b)/b) + (x*CosIntegral[b*x]*Sin[b*x])/b - Si 
n[b*x]^2/(2*b^2)))/b
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3791

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

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[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 7068

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] + (-Simp[d/b   Int[(e + f*x)^m*Sin[a + b*x]*(Cos[c + d*x]/(c + d* 
x)), x], x] - Simp[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 7072

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

rule 7074

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] + (Simp[d/b   Int[(e + f*x)^m*Cos[a + b*x]*(Cos[c + d*x]/(c + 
d*x)), x], x] + Simp[f*(m/b)   Int[(e + f*x)^(m - 1)*Cos[a + b*x]*CosIntegr 
al[c + d*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 4.60 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\operatorname {Ci}\left (b x \right ) \left (-b^{2} x^{2} \cos \left (b x \right )+2 \cos \left (b x \right )+2 b x \sin \left (b x \right )\right )+b x \left (\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )-\frac {b^{2} x^{2}}{4}-\frac {\sin \left (b x \right )^{2}}{4}+\cos \left (b x \right )^{2}-\ln \left (b x \right )-\operatorname {Ci}\left (2 b x \right )}{b^{3}}\)	\(91\)
default	\(\frac {\operatorname {Ci}\left (b x \right ) \left (-b^{2} x^{2} \cos \left (b x \right )+2 \cos \left (b x \right )+2 b x \sin \left (b x \right )\right )+b x \left (\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )-\frac {b^{2} x^{2}}{4}-\frac {\sin \left (b x \right )^{2}}{4}+\cos \left (b x \right )^{2}-\ln \left (b x \right )-\operatorname {Ci}\left (2 b x \right )}{b^{3}}\)	\(91\)

Input:

int(x^2*Ci(b*x)*sin(b*x),x,method=_RETURNVERBOSE)
 

Output:

1/b^3*(Ci(b*x)*(-b^2*x^2*cos(b*x)+2*cos(b*x)+2*b*x*sin(b*x))+b*x*(1/2*sin( 
b*x)*cos(b*x)+1/2*b*x)-1/4*b^2*x^2-1/4*sin(b*x)^2+cos(b*x)^2-ln(b*x)-Ci(2* 
b*x))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (105) = 210\).

Time = 0.11 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.67 \[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=-\frac {2 \, {\left (\pi ^{2} b^{3} x^{2} - 2 \, \pi ^{2} b\right )} \cos \left (b x\right ) \operatorname {C}\left (b x\right ) + \sqrt {b^{2}} {\left ({\left (2 \, \pi ^{2} - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) + \sqrt {b^{2}} {\left ({\left (2 \, \pi ^{2} - 1\right )} \cos \left (\frac {1}{2 \, \pi }\right ) + \pi \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {C}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (2 \, \pi ^{2} - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x + 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - \sqrt {b^{2}} {\left (\pi \cos \left (\frac {1}{2 \, \pi }\right ) - {\left (2 \, \pi ^{2} - 1\right )} \sin \left (\frac {1}{2 \, \pi }\right )\right )} \operatorname {S}\left (\frac {{\left (\pi b x - 1\right )} \sqrt {b^{2}}}{\pi b}\right ) - 2 \, {\left (\pi b^{2} x \cos \left (b x\right ) - 2 \, \pi b \sin \left (b x\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, {\left (2 \, \pi ^{2} b^{2} x \operatorname {C}\left (b x\right ) - b \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )} \sin \left (b x\right )}{2 \, \pi ^{2} b^{4}} \] Input:

integrate(x^2*fresnel_cos(b*x)*sin(b*x),x, algorithm="fricas")
 

Output:

-1/2*(2*(pi^2*b^3*x^2 - 2*pi^2*b)*cos(b*x)*fresnel_cos(b*x) + sqrt(b^2)*(( 
2*pi^2 - 1)*cos(1/2/pi) + pi*sin(1/2/pi))*fresnel_cos((pi*b*x + 1)*sqrt(b^ 
2)/(pi*b)) + sqrt(b^2)*((2*pi^2 - 1)*cos(1/2/pi) + pi*sin(1/2/pi))*fresnel 
_cos((pi*b*x - 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*(pi*cos(1/2/pi) - (2*pi^2 
- 1)*sin(1/2/pi))*fresnel_sin((pi*b*x + 1)*sqrt(b^2)/(pi*b)) - sqrt(b^2)*( 
pi*cos(1/2/pi) - (2*pi^2 - 1)*sin(1/2/pi))*fresnel_sin((pi*b*x - 1)*sqrt(b 
^2)/(pi*b)) - 2*(pi*b^2*x*cos(b*x) - 2*pi*b*sin(b*x))*sin(1/2*pi*b^2*x^2) 
- 2*(2*pi^2*b^2*x*fresnel_cos(b*x) - b*cos(1/2*pi*b^2*x^2))*sin(b*x))/(pi^ 
2*b^4)
 

Sympy [F]

\[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=\int x^{2} \sin {\left (b x \right )} \operatorname {Ci}{\left (b x \right )}\, dx \] Input:

integrate(x**2*Ci(b*x)*sin(b*x),x)
 

Output:

Integral(x**2*sin(b*x)*Ci(b*x), x)
 

Maxima [F]

\[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=\int { x^{2} \operatorname {C}\left (b x\right ) \sin \left (b x\right ) \,d x } \] Input:

integrate(x^2*fresnel_cos(b*x)*sin(b*x),x, algorithm="maxima")
 

Output:

integrate(x^2*fresnel_cos(b*x)*sin(b*x), x)
 

Giac [F]

\[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=\int { x^{2} \operatorname {C}\left (b x\right ) \sin \left (b x\right ) \,d x } \] Input:

integrate(x^2*fresnel_cos(b*x)*sin(b*x),x, algorithm="giac")
 

Output:

integrate(x^2*fresnel_cos(b*x)*sin(b*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=\int x^2\,\mathrm {cosint}\left (b\,x\right )\,\sin \left (b\,x\right ) \,d x \] Input:

int(x^2*cosint(b*x)*sin(b*x),x)
 

Output:

int(x^2*cosint(b*x)*sin(b*x), x)
 

Reduce [F]

\[ \int x^2 \operatorname {CosIntegral}(b x) \sin (b x) \, dx=\int \mathit {ci} \left (b x \right ) \sin \left (b x \right ) x^{2}d x \] Input:

int(x^2*Ci(b*x)*sin(b*x),x)
 

Output:

int(ci(b*x)*sin(b*x)*x**2,x)