\(\int \frac {x^2 \sqrt {\arctan (a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\) [751]

Optimal result

Integrand size = 26, antiderivative size = 163 \[ \int \frac {x^2 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {x^3 \sqrt {\arctan (a x)}}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{12 a^3 c^2 \sqrt {c+a^2 c x^2}} \] Output:

1/3*x^3*arctan(a*x)^(1/2)/c/(a^2*c*x^2+c)^(3/2)-1/8*2^(1/2)*Pi^(1/2)*(a^2* 
x^2+1)^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))/a^3/c^2/(a^2*c*x 
^2+c)^(1/2)+1/72*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)*FresnelS(6^(1/2)/Pi^(1 
/2)*arctan(a*x)^(1/2))/a^3/c^2/(a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.82 \[ \int \frac {x^2 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {24 a^3 x^3 \sqrt {\arctan (a x)}-9 \sqrt {2 \pi } \left (1+a^2 x^2\right )^{3/2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\sqrt {6 \pi } \left (1+a^2 x^2\right )^{3/2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{72 a^3 c^2 \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2}} \] Input:

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

Output:

(24*a^3*x^3*Sqrt[ArcTan[a*x]] - 9*Sqrt[2*Pi]*(1 + a^2*x^2)^(3/2)*FresnelS[ 
Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] + Sqrt[6*Pi]*(1 + a^2*x^2)^(3/2)*FresnelS[Sq 
rt[6/Pi]*Sqrt[ArcTan[a*x]]])/(72*a^3*c^2*(1 + a^2*x^2)*Sqrt[c + a^2*c*x^2] 
)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5479, 5506, 5505, 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.

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

\(\Big \downarrow \) 5479

\(\displaystyle \frac {x^3 \sqrt {\arctan (a x)}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {1}{6} a \int \frac {x^3}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx\)

\(\Big \downarrow \) 5506

\(\displaystyle \frac {x^3 \sqrt {\arctan (a x)}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {a \sqrt {a^2 x^2+1} \int \frac {x^3}{\left (a^2 x^2+1\right )^{5/2} \sqrt {\arctan (a x)}}dx}{6 c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 5505

\(\displaystyle \frac {x^3 \sqrt {\arctan (a x)}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \int \frac {a^3 x^3}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{6 a^3 c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {x^3 \sqrt {\arctan (a x)}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \int \frac {\sin (\arctan (a x))^3}{\sqrt {\arctan (a x)}}d\arctan (a x)}{6 a^3 c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {x^3 \sqrt {\arctan (a x)}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \int \left (\frac {3 a x}{4 \sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}-\frac {\sin (3 \arctan (a x))}{4 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{6 a^3 c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^3 \sqrt {\arctan (a x)}}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {\sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{6 a^3 c^2 \sqrt {a^2 c x^2+c}}\)

Input:

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

Output:

(x^3*Sqrt[ArcTan[a*x]])/(3*c*(c + a^2*c*x^2)^(3/2)) - (Sqrt[1 + a^2*x^2]*( 
(3*Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/2 - (Sqrt[Pi/6]*Fres 
nelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/2))/(6*a^3*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 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 5479

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + 
 b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Simp[b*c*(p/(f*(m + 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 + 3, 0] && GtQ[p, 0] 
&& NeQ[m, -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 [F]

\[\int \frac {x^{2} \sqrt {\arctan \left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}d x\]

Input:

int(x^2*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)
 

Output:

int(x^2*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)
 

Fricas [F(-2)]

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

integrate(x^2*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^2*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^2*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((x^2*atan(a*x)^(1/2))/(c + a^2*c*x^2)^(5/2),x)
 

Output:

int((x^2*atan(a*x)^(1/2))/(c + a^2*c*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^2 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (2 \sqrt {a^{2} x^{2}+1}\, \sqrt {\mathit {atan} \left (a x \right )}\, x^{3}-\left (\int \frac {\sqrt {a^{2} x^{2}+1}\, \sqrt {\mathit {atan} \left (a x \right )}\, x^{3}}{\mathit {atan} \left (a x \right ) a^{6} x^{6}+3 \mathit {atan} \left (a x \right ) a^{4} x^{4}+3 \mathit {atan} \left (a x \right ) a^{2} x^{2}+\mathit {atan} \left (a x \right )}d x \right ) a^{5} x^{4}-2 \left (\int \frac {\sqrt {a^{2} x^{2}+1}\, \sqrt {\mathit {atan} \left (a x \right )}\, x^{3}}{\mathit {atan} \left (a x \right ) a^{6} x^{6}+3 \mathit {atan} \left (a x \right ) a^{4} x^{4}+3 \mathit {atan} \left (a x \right ) a^{2} x^{2}+\mathit {atan} \left (a x \right )}d x \right ) a^{3} x^{2}-\left (\int \frac {\sqrt {a^{2} x^{2}+1}\, \sqrt {\mathit {atan} \left (a x \right )}\, x^{3}}{\mathit {atan} \left (a x \right ) a^{6} x^{6}+3 \mathit {atan} \left (a x \right ) a^{4} x^{4}+3 \mathit {atan} \left (a x \right ) a^{2} x^{2}+\mathit {atan} \left (a x \right )}d x \right ) a \right )}{6 c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )} \] Input:

int(x^2*atan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(c)*(2*sqrt(a**2*x**2 + 1)*sqrt(atan(a*x))*x**3 - int((sqrt(a**2*x**2 
 + 1)*sqrt(atan(a*x))*x**3)/(atan(a*x)*a**6*x**6 + 3*atan(a*x)*a**4*x**4 + 
 3*atan(a*x)*a**2*x**2 + atan(a*x)),x)*a**5*x**4 - 2*int((sqrt(a**2*x**2 + 
 1)*sqrt(atan(a*x))*x**3)/(atan(a*x)*a**6*x**6 + 3*atan(a*x)*a**4*x**4 + 3 
*atan(a*x)*a**2*x**2 + atan(a*x)),x)*a**3*x**2 - int((sqrt(a**2*x**2 + 1)* 
sqrt(atan(a*x))*x**3)/(atan(a*x)*a**6*x**6 + 3*atan(a*x)*a**4*x**4 + 3*ata 
n(a*x)*a**2*x**2 + atan(a*x)),x)*a))/(6*c**3*(a**4*x**4 + 2*a**2*x**2 + 1) 
)