\(\int x^3 \operatorname {CosIntegral}(a+b x) \, dx\) [86]

Optimal result

Integrand size = 10, antiderivative size = 184 \[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx=\frac {3 \cos (a+b x)}{2 b^4}-\frac {a^2 \cos (a+b x)}{4 b^4}+\frac {a x \cos (a+b x)}{2 b^3}-\frac {3 x^2 \cos (a+b x)}{4 b^2}-\frac {a^4 \operatorname {CosIntegral}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \operatorname {CosIntegral}(a+b x)-\frac {a \sin (a+b x)}{2 b^4}+\frac {a^3 \sin (a+b x)}{4 b^4}+\frac {3 x \sin (a+b x)}{2 b^3}-\frac {a^2 x \sin (a+b x)}{4 b^3}+\frac {a x^2 \sin (a+b x)}{4 b^2}-\frac {x^3 \sin (a+b x)}{4 b} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.52 \[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx=\frac {-\left (\left (-6+a^2-2 a b x+3 b^2 x^2\right ) \cos (a+b x)\right )+\left (-a^4+b^4 x^4\right ) \operatorname {CosIntegral}(a+b x)+\left (-2 a+a^3+6 b x-a^2 b x+a b^2 x^2-b^3 x^3\right ) \sin (a+b x)}{4 b^4} \] Input:

Integrate[x^3*CosIntegral[a + b*x],x]
 

Output:

(-((-6 + a^2 - 2*a*b*x + 3*b^2*x^2)*Cos[a + b*x]) + (-a^4 + b^4*x^4)*CosIn 
tegral[a + b*x] + (-2*a + a^3 + 6*b*x - a^2*b*x + a*b^2*x^2 - b^3*x^3)*Sin 
[a + b*x])/(4*b^4)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {7058, 7293, 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.

\(\displaystyle \int x^3 \operatorname {CosIntegral}(a+b x) \, dx\)

\(\Big \downarrow \) 7058

\(\displaystyle \frac {1}{4} x^4 \operatorname {CosIntegral}(a+b x)-\frac {1}{4} b \int \frac {x^4 \cos (a+b x)}{a+b x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{4} x^4 \operatorname {CosIntegral}(a+b x)-\frac {1}{4} b \int \left (\frac {\cos (a+b x) a^4}{b^4 (a+b x)}-\frac {\cos (a+b x) a^3}{b^4}+\frac {x \cos (a+b x) a^2}{b^3}-\frac {x^2 \cos (a+b x) a}{b^2}+\frac {x^3 \cos (a+b x)}{b}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} x^4 \operatorname {CosIntegral}(a+b x)-\frac {1}{4} b \left (\frac {a^4 \operatorname {CosIntegral}(a+b x)}{b^5}-\frac {a^3 \sin (a+b x)}{b^5}+\frac {a^2 \cos (a+b x)}{b^5}+\frac {a^2 x \sin (a+b x)}{b^4}+\frac {2 a \sin (a+b x)}{b^5}-\frac {6 \cos (a+b x)}{b^5}-\frac {6 x \sin (a+b x)}{b^4}-\frac {2 a x \cos (a+b x)}{b^4}-\frac {a x^2 \sin (a+b x)}{b^3}+\frac {3 x^2 \cos (a+b x)}{b^3}+\frac {x^3 \sin (a+b x)}{b^2}\right )\)

Input:

Int[x^3*CosIntegral[a + b*x],x]
 

Output:

(x^4*CosIntegral[a + b*x])/4 - (b*((-6*Cos[a + b*x])/b^5 + (a^2*Cos[a + b* 
x])/b^5 - (2*a*x*Cos[a + b*x])/b^4 + (3*x^2*Cos[a + b*x])/b^3 + (a^4*CosIn 
tegral[a + b*x])/b^5 + (2*a*Sin[a + b*x])/b^5 - (a^3*Sin[a + b*x])/b^5 - ( 
6*x*Sin[a + b*x])/b^4 + (a^2*x*Sin[a + b*x])/b^4 - (a*x^2*Sin[a + b*x])/b^ 
3 + (x^3*Sin[a + b*x])/b^2))/4
 

Defintions of rubi rules used

rule 2009

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

rule 7058

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

rule 7293

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

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.83

method	result	size
parts	\(\frac {x^{4} \operatorname {Ci}\left (b x +a \right )}{4}-\frac {a^{4} \operatorname {Ci}\left (b x +a \right )-4 a^{3} \sin \left (b x +a \right )+6 a^{2} \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )-4 a \left (\left (b x +a \right )^{2} \sin \left (b x +a \right )-2 \sin \left (b x +a \right )+2 \left (b x +a \right ) \cos \left (b x +a \right )\right )+\left (b x +a \right )^{3} \sin \left (b x +a \right )+3 \left (b x +a \right )^{2} \cos \left (b x +a \right )-6 \cos \left (b x +a \right )-6 \left (b x +a \right ) \sin \left (b x +a \right )}{4 b^{4}}\)	\(153\)
derivativedivides	\(\frac {\frac {\operatorname {Ci}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {Ci}\left (b x +a \right )}{4}+a^{3} \sin \left (b x +a \right )-\frac {3 a^{2} \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )}{2}+a \left (\left (b x +a \right )^{2} \sin \left (b x +a \right )-2 \sin \left (b x +a \right )+2 \left (b x +a \right ) \cos \left (b x +a \right )\right )-\frac {\left (b x +a \right )^{3} \sin \left (b x +a \right )}{4}-\frac {3 \left (b x +a \right )^{2} \cos \left (b x +a \right )}{4}+\frac {3 \cos \left (b x +a \right )}{2}+\frac {3 \left (b x +a \right ) \sin \left (b x +a \right )}{2}}{b^{4}}\)	\(154\)
default	\(\frac {\frac {\operatorname {Ci}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {Ci}\left (b x +a \right )}{4}+a^{3} \sin \left (b x +a \right )-\frac {3 a^{2} \left (\cos \left (b x +a \right )+\left (b x +a \right ) \sin \left (b x +a \right )\right )}{2}+a \left (\left (b x +a \right )^{2} \sin \left (b x +a \right )-2 \sin \left (b x +a \right )+2 \left (b x +a \right ) \cos \left (b x +a \right )\right )-\frac {\left (b x +a \right )^{3} \sin \left (b x +a \right )}{4}-\frac {3 \left (b x +a \right )^{2} \cos \left (b x +a \right )}{4}+\frac {3 \cos \left (b x +a \right )}{2}+\frac {3 \left (b x +a \right ) \sin \left (b x +a \right )}{2}}{b^{4}}\)	\(154\)
orering	\(-\frac {\left (-b^{6} x^{6}+a^{4} b^{2} x^{2}-18 b^{4} x^{4}-6 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}-6 a^{3} b x +12 a^{4}+72 b^{2} x^{2}+72 b x a -24 a^{2}\right ) \operatorname {Ci}\left (b x +a \right )}{4 b^{6} x^{2}}+\frac {\left (-4 b^{4} x^{4}-a \,b^{3} x^{3}+a^{2} b^{2} x^{2}-a^{3} b x +3 a^{4}+18 b^{2} x^{2}+16 b x a -6 a^{2}\right ) \left (3 x^{2} \operatorname {Ci}\left (b x +a \right )+\frac {x^{3} \cos \left (b x +a \right ) b}{b x +a}\right )}{2 b^{6} x^{4}}-\frac {\left (-b^{3} x^{3}+a \,b^{2} x^{2}-a^{2} b x +a^{3}+6 b x -2 a \right ) \left (b x +a \right ) \left (6 x \,\operatorname {Ci}\left (b x +a \right )+\frac {6 x^{2} \cos \left (b x +a \right ) b}{b x +a}-\frac {x^{3} b^{2} \sin \left (b x +a \right )}{b x +a}-\frac {x^{3} \cos \left (b x +a \right ) b^{2}}{\left (b x +a \right )^{2}}\right )}{4 b^{6} x^{3}}\)	\(307\)

Input:

int(x^3*Ci(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

1/4*x^4*Ci(b*x+a)-1/4/b^4*(a^4*Ci(b*x+a)-4*a^3*sin(b*x+a)+6*a^2*(cos(b*x+a 
)+(b*x+a)*sin(b*x+a))-4*a*((b*x+a)^2*sin(b*x+a)-2*sin(b*x+a)+2*(b*x+a)*cos 
(b*x+a))+(b*x+a)^3*sin(b*x+a)+3*(b*x+a)^2*cos(b*x+a)-6*cos(b*x+a)-6*(b*x+a 
)*sin(b*x+a))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.96 \[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx=\frac {\pi ^{2} b^{5} x^{4} \operatorname {C}\left (b x + a\right ) + 6 \, \pi a^{2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi ^{2} a^{4} - 3\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (3 \, b^{2} x - 5 \, a b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - {\left (\pi b^{4} x^{3} - \pi a b^{3} x^{2} + \pi a^{2} b^{2} x - \pi a^{3} b\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{4 \, \pi ^{2} b^{5}} \] Input:

integrate(x^3*fresnel_cos(b*x+a),x, algorithm="fricas")
 

Output:

1/4*(pi^2*b^5*x^4*fresnel_cos(b*x + a) + 6*pi*a^2*sqrt(b^2)*fresnel_sin(sq 
rt(b^2)*(b*x + a)/b) - (pi^2*a^4 - 3)*sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x 
 + a)/b) - (3*b^2*x - 5*a*b)*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - 
 (pi*b^4*x^3 - pi*a*b^3*x^2 + pi*a^2*b^2*x - pi*a^3*b)*sin(1/2*pi*b^2*x^2 
+ pi*a*b*x + 1/2*pi*a^2))/(pi^2*b^5)
 

Sympy [F]

\[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx=\int x^{3} \operatorname {Ci}{\left (a + b x \right )}\, dx \] Input:

integrate(x**3*Ci(b*x+a),x)
 

Output:

Integral(x**3*Ci(a + b*x), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.10 (sec) , antiderivative size = 502, normalized size of antiderivative = 2.73 \[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx =\text {Too large to display} \] Input:

integrate(x^3*fresnel_cos(b*x+a),x, algorithm="maxima")
 

Output:

1/4*x^4*fresnel_cos(b*x + a) + 1/32*(16*(-I*pi^2*e^(1/2*I*pi*b^2*x^2 + I*p 
i*a*b*x + 1/2*I*pi*a^2) + I*pi^2*e^(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I 
*pi*a^2))*a^4 + 32*(pi*gamma(2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a 
^2) + pi*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^2 + 16 
*((-I*pi^2*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + I*pi^2*e^(-1 
/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^3 + 2*(pi*gamma(2, 1/2*I*p 
i*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + pi*gamma(2, -1/2*I*pi*b^2*x^2 - I 
*pi*a*b*x - 1/2*I*pi*a^2))*a)*b*x + (((I - 1)*sqrt(2)*pi^(5/2)*(erf(sqrt(1 
/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2)) - 1) - (I + 1)*sqrt(2)*pi^(5 
/2)*(erf(sqrt(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)) - 1))*a^4 + 
12*(-(I + 1)*sqrt(2)*pi*gamma(3/2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*p 
i*a^2) + (I - 1)*sqrt(2)*pi*gamma(3/2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/ 
2*I*pi*a^2))*a^2 + (4*I - 4)*sqrt(2)*gamma(5/2, 1/2*I*pi*b^2*x^2 + I*pi*a* 
b*x + 1/2*I*pi*a^2) - (4*I + 4)*sqrt(2)*gamma(5/2, -1/2*I*pi*b^2*x^2 - I*p 
i*a*b*x - 1/2*I*pi*a^2))*sqrt(2*pi*b^2*x^2 + 4*pi*a*b*x + 2*pi*a^2))*b/(pi 
^3*b^6*x + pi^3*a*b^5)
 

Giac [F]

\[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx=\int { x^{3} \operatorname {C}\left (b x + a\right ) \,d x } \] Input:

integrate(x^3*fresnel_cos(b*x+a),x, algorithm="giac")
 

Output:

integrate(x^3*fresnel_cos(b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx=\int x^3\,\mathrm {cosint}\left (a+b\,x\right ) \,d x \] Input:

int(x^3*cosint(a + b*x),x)
 

Output:

int(x^3*cosint(a + b*x), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int x^3 \operatorname {CosIntegral}(a+b x) \, dx=\frac {-\mathit {ci} \left (b x +a \right ) a^{4}+\mathit {ci} \left (b x +a \right ) b^{4} x^{4}-\cos \left (b x +a \right ) a^{2}+2 \cos \left (b x +a \right ) a b x -3 \cos \left (b x +a \right ) b^{2} x^{2}+6 \cos \left (b x +a \right )+\sin \left (b x +a \right ) a^{3}-\sin \left (b x +a \right ) a^{2} b x +\sin \left (b x +a \right ) a \,b^{2} x^{2}-2 \sin \left (b x +a \right ) a -\sin \left (b x +a \right ) b^{3} x^{3}+6 \sin \left (b x +a \right ) b x}{4 b^{4}} \] Input:

int(x^3*Ci(b*x+a),x)
 

Output:

( - ci(a + b*x)*a**4 + ci(a + b*x)*b**4*x**4 - cos(a + b*x)*a**2 + 2*cos(a 
 + b*x)*a*b*x - 3*cos(a + b*x)*b**2*x**2 + 6*cos(a + b*x) + sin(a + b*x)*a 
**3 - sin(a + b*x)*a**2*b*x + sin(a + b*x)*a*b**2*x**2 - 2*sin(a + b*x)*a 
- sin(a + b*x)*b**3*x**3 + 6*sin(a + b*x)*b*x)/(4*b**4)