\(\int \frac {x^3}{(c+a^2 c x^2)^{5/2} \arctan (a x)^2} \, dx\) [588]

Optimal result

Integrand size = 24, antiderivative size = 118 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=-\frac {x^3}{a c \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)}+\frac {3 \sqrt {1+a^2 x^2} \operatorname {CosIntegral}(\arctan (a x))}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {3 \sqrt {1+a^2 x^2} \operatorname {CosIntegral}(3 \arctan (a x))}{4 a^4 c^2 \sqrt {c+a^2 c x^2}} \] Output:

-x^3/a/c/(a^2*c*x^2+c)^(3/2)/arctan(a*x)+3/4*(a^2*x^2+1)^(1/2)*Ci(arctan(a 
*x))/a^4/c^2/(a^2*c*x^2+c)^(1/2)-3/4*(a^2*x^2+1)^(1/2)*Ci(3*arctan(a*x))/a 
^4/c^2/(a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\frac {-\frac {4 a^3 c x^3}{\left (1+a^2 x^2\right ) \arctan (a x)}+3 c \sqrt {1+a^2 x^2} (\operatorname {CosIntegral}(\arctan (a x))-\operatorname {CosIntegral}(3 \arctan (a x)))}{4 a^4 c^3 \sqrt {c+a^2 c x^2}} \] Input:

Integrate[x^3/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]^2),x]
 

Output:

((-4*a^3*c*x^3)/((1 + a^2*x^2)*ArcTan[a*x]) + 3*c*Sqrt[1 + a^2*x^2]*(CosIn 
tegral[ArcTan[a*x]] - CosIntegral[3*ArcTan[a*x]]))/(4*a^4*c^3*Sqrt[c + a^2 
*c*x^2])
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5477, 5506, 5505, 4906, 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.

\(\displaystyle \int \frac {x^3}{\arctan (a x)^2 \left (a^2 c x^2+c\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 5477

\(\displaystyle \frac {3 \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2} \arctan (a x)}dx}{a}-\frac {x^3}{a c \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 5506

\(\displaystyle \frac {3 \sqrt {a^2 x^2+1} \int \frac {x^2}{\left (a^2 x^2+1\right )^{5/2} \arctan (a x)}dx}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {x^3}{a c \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 5505

\(\displaystyle \frac {3 \sqrt {a^2 x^2+1} \int \frac {a^2 x^2}{\left (a^2 x^2+1\right )^{3/2} \arctan (a x)}d\arctan (a x)}{a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {x^3}{a c \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {3 \sqrt {a^2 x^2+1} \int \left (\frac {1}{4 \sqrt {a^2 x^2+1} \arctan (a x)}-\frac {\cos (3 \arctan (a x))}{4 \arctan (a x)}\right )d\arctan (a x)}{a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {x^3}{a c \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \sqrt {a^2 x^2+1} \left (\frac {1}{4} \operatorname {CosIntegral}(\arctan (a x))-\frac {1}{4} \operatorname {CosIntegral}(3 \arctan (a x))\right )}{a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {x^3}{a c \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}}\)

Input:

Int[x^3/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]^2),x]
 

Output:

-(x^3/(a*c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])) + (3*Sqrt[1 + a^2*x^2]*(Cos 
Integral[ArcTan[a*x]]/4 - CosIntegral[3*ArcTan[a*x]]/4))/(a^4*c^2*Sqrt[c + 
 a^2*c*x^2])
 

Defintions of rubi rules used

rule 2009

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

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG 
tQ[p, 0]
 

rule 5477

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_. 
)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*ArcT 
an[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[f*(m/(b*c*(p + 1)))   Int[(f*x 
)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, e, f, m, q}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1 
]
 

rule 5505

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(q_), x_Symbol] :> Simp[d^q/c^(m + 1)   Subst[Int[(a + b*x)^p*(Sin[x]^m/ 
Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p 
}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q 
] || GtQ[d, 0])
 

rule 5506

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(q_), x_Symbol] :> Simp[d^(q + 1/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]) 
  Int[x^m*(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, 
d, e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(I 
ntegerQ[q] || GtQ[d, 0])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 9.81 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.62

method	result	size
default	\(-\frac {\left (3 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (-i \arctan \left (a x \right )\right ) a^{4} x^{4}-3 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (3 i \arctan \left (a x \right )\right ) a^{4} x^{4}-3 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (-3 i \arctan \left (a x \right )\right ) a^{4} x^{4}+3 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (i \arctan \left (a x \right )\right ) a^{4} x^{4}+8 \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+6 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (-i \arctan \left (a x \right )\right ) a^{2} x^{2}-6 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (3 i \arctan \left (a x \right )\right ) a^{2} x^{2}-6 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (-3 i \arctan \left (a x \right )\right ) a^{2} x^{2}+6 \arctan \left (a x \right ) \operatorname {expIntegral}_{1}\left (i \arctan \left (a x \right )\right ) a^{2} x^{2}+3 \,\operatorname {expIntegral}_{1}\left (-i \arctan \left (a x \right )\right ) \arctan \left (a x \right )-3 \,\operatorname {expIntegral}_{1}\left (3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right )-3 \,\operatorname {expIntegral}_{1}\left (-3 i \arctan \left (a x \right )\right ) \arctan \left (a x \right )+3 \,\operatorname {expIntegral}_{1}\left (i \arctan \left (a x \right )\right ) \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 \sqrt {a^{2} x^{2}+1}\, \arctan \left (a x \right ) a^{4} c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(309\)

Input:

int(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/8*(3*arctan(a*x)*Ei(1,-I*arctan(a*x))*a^4*x^4-3*arctan(a*x)*Ei(1,3*I*ar 
ctan(a*x))*a^4*x^4-3*arctan(a*x)*Ei(1,-3*I*arctan(a*x))*a^4*x^4+3*arctan(a 
*x)*Ei(1,I*arctan(a*x))*a^4*x^4+8*(a^2*x^2+1)^(1/2)*a^3*x^3+6*arctan(a*x)* 
Ei(1,-I*arctan(a*x))*a^2*x^2-6*arctan(a*x)*Ei(1,3*I*arctan(a*x))*a^2*x^2-6 
*arctan(a*x)*Ei(1,-3*I*arctan(a*x))*a^2*x^2+6*arctan(a*x)*Ei(1,I*arctan(a* 
x))*a^2*x^2+3*Ei(1,-I*arctan(a*x))*arctan(a*x)-3*Ei(1,3*I*arctan(a*x))*arc 
tan(a*x)-3*Ei(1,-3*I*arctan(a*x))*arctan(a*x)+3*Ei(1,I*arctan(a*x))*arctan 
(a*x))/(a^2*x^2+1)^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/arctan(a*x)/a^4/c^3/(a^ 
4*x^4+2*a^2*x^2+1)
 

Fricas [F]

\[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\int { \frac {x^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )^{2}} \,d x } \] Input:

integrate(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^2,x, algorithm="fricas")
 

Output:

integral(sqrt(a^2*c*x^2 + c)*x^3/((a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3 
*x^2 + c^3)*arctan(a*x)^2), x)
 

Sympy [F]

\[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\int \frac {x^{3}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}^{2}{\left (a x \right )}}\, dx \] Input:

integrate(x**3/(a**2*c*x**2+c)**(5/2)/atan(a*x)**2,x)
 

Output:

Integral(x**3/((c*(a**2*x**2 + 1))**(5/2)*atan(a*x)**2), x)
 

Maxima [F]

\[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\int { \frac {x^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )^{2}} \,d x } \] Input:

integrate(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^2,x, algorithm="maxima")
 

Output:

integrate(x^3/((a^2*c*x^2 + c)^(5/2)*arctan(a*x)^2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^3/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^2,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\int \frac {x^3}{{\mathrm {atan}\left (a\,x\right )}^2\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(x^3/(atan(a*x)^2*(c + a^2*c*x^2)^(5/2)),x)
 

Output:

int(x^3/(atan(a*x)^2*(c + a^2*c*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^2} \, dx=\frac {\int \frac {x^{3}}{\sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right )^{2} a^{4} x^{4}+2 \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right )^{2} a^{2} x^{2}+\sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right )^{2}}d x}{\sqrt {c}\, c^{2}} \] Input:

int(x^3/(a^2*c*x^2+c)^(5/2)/atan(a*x)^2,x)
 

Output:

int(x**3/(sqrt(a**2*x**2 + 1)*atan(a*x)**2*a**4*x**4 + 2*sqrt(a**2*x**2 + 
1)*atan(a*x)**2*a**2*x**2 + sqrt(a**2*x**2 + 1)*atan(a*x)**2),x)/(sqrt(c)* 
c**2)