\(\int \frac {x^2}{(c+a^2 c x^2)^2 \arctan (a x)^{5/2}} \, dx\) [1059]

Optimal result

Integrand size = 24, antiderivative size = 180 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}+\frac {16 \sqrt {\arctan (a x)}}{3 a^3 c^2}-\frac {32 \sqrt {\arctan (a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {8 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a^3 c^2} \] Output:

-2/3*x^2/a/c^2/(a^2*x^2+1)/arctan(a*x)^(3/2)-8/3*x/a^2/c^2/(a^2*x^2+1)/arc 
tan(a*x)^(1/2)+16/3*arctan(a*x)^(1/2)/a^3/c^2-32/3*arctan(a*x)^(1/2)/a^3/c 
^2/(a^2*x^2+1)+16/3*(-a^2*x^2+1)*arctan(a*x)^(1/2)/a^3/c^2/(a^2*x^2+1)+8/3 
*Pi^(1/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))/a^3/c^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.90 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\frac {-2 a x (a x+4 \arctan (a x))+4 \sqrt {\pi } \left (1+a^2 x^2\right ) \arctan (a x)^{3/2} \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\sqrt {2} \left (1+a^2 x^2\right ) (-i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )+\sqrt {2} \left (1+a^2 x^2\right ) (i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )}{3 a^3 c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}} \] Input:

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

Output:

(-2*a*x*(a*x + 4*ArcTan[a*x]) + 4*Sqrt[Pi]*(1 + a^2*x^2)*ArcTan[a*x]^(3/2) 
*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + Sqrt[2]*(1 + a^2*x^2)*((-I)*Ar 
cTan[a*x])^(3/2)*Gamma[1/2, (-2*I)*ArcTan[a*x]] + Sqrt[2]*(1 + a^2*x^2)*(I 
*ArcTan[a*x])^(3/2)*Gamma[1/2, (2*I)*ArcTan[a*x]])/(3*a^3*c^2*(1 + a^2*x^2 
)*ArcTan[a*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5477, 27, 5467, 5465, 5439, 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.

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

\(\Big \downarrow \) 5477

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5467

\(\displaystyle \frac {4 \left (16 \int \frac {x \sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^2}dx-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 a c^2}-\frac {2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\)

\(\Big \downarrow \) 5465

\(\displaystyle \frac {4 \left (16 \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx}{4 a}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 a c^2}-\frac {2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\)

\(\Big \downarrow \) 5439

\(\displaystyle \frac {4 \left (16 \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 a c^2}-\frac {2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \left (16 \left (\frac {\int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^2}{\sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 a c^2}-\frac {2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {4 \left (16 \left (\frac {\int \left (\frac {\cos (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}+\frac {1}{2 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 a c^2}-\frac {2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 \left (16 \left (\frac {\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\sqrt {\arctan (a x)}}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )-\frac {2 x}{a \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}+\frac {4 \left (1-a^2 x^2\right ) \sqrt {\arctan (a x)}}{a^2 \left (a^2 x^2+1\right )}\right )}{3 a c^2}-\frac {2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}\)

Input:

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

Output:

(-2*x^2)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) + (4*((-2*x)/(a*(1 + a^ 
2*x^2)*Sqrt[ArcTan[a*x]]) + (4*(1 - a^2*x^2)*Sqrt[ArcTan[a*x]])/(a^2*(1 + 
a^2*x^2)) + 16*(-1/2*Sqrt[ArcTan[a*x]]/(a^2*(1 + a^2*x^2)) + (Sqrt[ArcTan[ 
a*x]] + (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/2)/(4*a^2))))/ 
(3*a*c^2)
 

Defintions of rubi rules used

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 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 5439

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

rule 5465

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

rule 5467

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, 
 x_Symbol] :> Simp[x*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x^2 
))), x] + (-Simp[(1 - c^2*x^2)*((a + b*ArcTan[c*x])^(p + 2)/(b^2*e*(p + 1)* 
(p + 2)*(d + e*x^2))), x] - Simp[4/(b^2*(p + 1)*(p + 2))   Int[x*((a + b*Ar 
cTan[c*x])^(p + 2)/(d + e*x^2)^2), x], x]) /; FreeQ[{a, b, c, d, e}, x] && 
EqQ[e, c^2*d] && LtQ[p, -1] && NeQ[p, -2]
 

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 
]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.34

method	result	size
default	\(-\frac {-8 \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+4 \arctan \left (a x \right ) \sin \left (2 \arctan \left (a x \right )\right )-\cos \left (2 \arctan \left (a x \right )\right )+1}{3 a^{3} c^{2} \arctan \left (a x \right )^{\frac {3}{2}}}\)	\(62\)

Input:

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

Output:

-1/3/a^3/c^2*(-8*Pi^(1/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a* 
x)^(3/2)+4*arctan(a*x)*sin(2*arctan(a*x))-cos(2*arctan(a*x))+1)/arctan(a*x 
)^(3/2)
 

Fricas [F(-2)]

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

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

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

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

Output:

Integral(x**2/(a**4*x**4*atan(a*x)**(5/2) + 2*a**2*x**2*atan(a*x)**(5/2) + 
 atan(a*x)**(5/2)), x)/c**2
 

Maxima [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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