\(\int \frac {\arctan (a x)^{3/2}}{(c+a^2 c x^2)^2} \, dx\) [786]

Optimal result

Integrand size = 21, antiderivative size = 124 \[ \int \frac {\arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {3 \sqrt {\arctan (a x)}}{16 a c^2}+\frac {3 \sqrt {\arctan (a x)}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}-\frac {3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{32 a c^2} \] Output:

-3/16*arctan(a*x)^(1/2)/a/c^2+3/8*arctan(a*x)^(1/2)/a/c^2/(a^2*x^2+1)+1/2* 
x*arctan(a*x)^(3/2)/c^2/(a^2*x^2+1)+1/5*arctan(a*x)^(5/2)/a/c^2-3/32*Pi^(1 
/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))/a/c^2
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \frac {\arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\frac {2 \sqrt {\arctan (a x)} \left (15-15 a^2 x^2+40 a x \arctan (a x)+16 \left (1+a^2 x^2\right ) \arctan (a x)^2\right )}{1+a^2 x^2}-15 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{160 a c^2} \] Input:

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

Output:

((2*Sqrt[ArcTan[a*x]]*(15 - 15*a^2*x^2 + 40*a*x*ArcTan[a*x] + 16*(1 + a^2* 
x^2)*ArcTan[a*x]^2))/(1 + a^2*x^2) - 15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a 
*x]])/Sqrt[Pi]])/(160*a*c^2)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5427, 27, 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.

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

\(\Big \downarrow \) 5427

\(\displaystyle -\frac {3}{4} a \int \frac {x \sqrt {\arctan (a x)}}{c^2 \left (a^2 x^2+1\right )^2}dx+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 a \int \frac {x \sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^2}dx}{4 c^2}+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}\)

\(\Big \downarrow \) 5465

\(\displaystyle -\frac {3 a \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx}{4 a}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 c^2}+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}\)

\(\Big \downarrow \) 5439

\(\displaystyle -\frac {3 a \left (\frac {\int \frac {1}{\left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 c^2}+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 a \left (\frac {\int \frac {\sin \left (\arctan (a x)+\frac {\pi }{2}\right )^2}{\sqrt {\arctan (a x)}}d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 c^2}+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}\)

\(\Big \downarrow \) 3793

\(\displaystyle -\frac {3 a \left (\frac {\int \left (\frac {\cos (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}+\frac {1}{2 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 c^2}+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 a \left (\frac {\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\sqrt {\arctan (a x)}}{4 a^2}-\frac {\sqrt {\arctan (a x)}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 c^2}+\frac {x \arctan (a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{5/2}}{5 a c^2}\)

Input:

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

Output:

(x*ArcTan[a*x]^(3/2))/(2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(5*a*c^2) 
- (3*a*(-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)))/(4*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 5427

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

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]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {32 \arctan \left (a x \right )^{3}+40 \arctan \left (a x \right )^{2} \sin \left (2 \arctan \left (a x \right )\right )+30 \arctan \left (a x \right ) \cos \left (2 \arctan \left (a x \right )\right )-15 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )}{160 a \,c^{2} \sqrt {\arctan \left (a x \right )}}\)	\(75\)

Input:

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

Output:

1/160/a/c^2/arctan(a*x)^(1/2)*(32*arctan(a*x)^3+40*arctan(a*x)^2*sin(2*arc 
tan(a*x))+30*arctan(a*x)*cos(2*arctan(a*x))-15*arctan(a*x)^(1/2)*Pi^(1/2)* 
FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2)))
 

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {16 \sqrt {\mathit {atan} \left (a x \right )}\, \mathit {atan} \left (a x \right )^{2} a^{2} x^{2}+16 \sqrt {\mathit {atan} \left (a x \right )}\, \mathit {atan} \left (a x \right )^{2}+40 \sqrt {\mathit {atan} \left (a x \right )}\, \mathit {atan} \left (a x \right ) a x -30 \sqrt {\mathit {atan} \left (a x \right )}\, a^{2} x^{2}+15 \left (\int \frac {\sqrt {\mathit {atan} \left (a x \right )}\, x^{2}}{\mathit {atan} \left (a x \right ) a^{4} x^{4}+2 \mathit {atan} \left (a x \right ) a^{2} x^{2}+\mathit {atan} \left (a x \right )}d x \right ) a^{5} x^{2}+15 \left (\int \frac {\sqrt {\mathit {atan} \left (a x \right )}\, x^{2}}{\mathit {atan} \left (a x \right ) a^{4} x^{4}+2 \mathit {atan} \left (a x \right ) a^{2} x^{2}+\mathit {atan} \left (a x \right )}d x \right ) a^{3}}{80 a \,c^{2} \left (a^{2} x^{2}+1\right )} \] Input:

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

Output:

(16*sqrt(atan(a*x))*atan(a*x)**2*a**2*x**2 + 16*sqrt(atan(a*x))*atan(a*x)* 
*2 + 40*sqrt(atan(a*x))*atan(a*x)*a*x - 30*sqrt(atan(a*x))*a**2*x**2 + 15* 
int((sqrt(atan(a*x))*x**2)/(atan(a*x)*a**4*x**4 + 2*atan(a*x)*a**2*x**2 + 
atan(a*x)),x)*a**5*x**2 + 15*int((sqrt(atan(a*x))*x**2)/(atan(a*x)*a**4*x* 
*4 + 2*atan(a*x)*a**2*x**2 + atan(a*x)),x)*a**3)/(80*a*c**2*(a**2*x**2 + 1 
))