\(\int \frac {x \arctan (a x)^{3/2}}{(c+a^2 c x^2)^{5/2}} \, dx\) [832]

Optimal result

Integrand size = 24, antiderivative size = 248 \[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {3 x \sqrt {\arctan (a x)}}{8 a c^2 \sqrt {c+a^2 c x^2}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{8 a^2 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{24 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \sin (3 \arctan (a x))}{24 a^2 c^2 \sqrt {c+a^2 c x^2}} \] Output:

3/8*x*arctan(a*x)^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)-1/3*arctan(a*x)^(3/2)/a^ 
2/c/(a^2*c*x^2+c)^(3/2)-3/16*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^2/c^2/(a^2*c*x^2+c)^(1/2)-1/144*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^2/c^2/(a^2*c*x^2+c)^(1/2)+1/24*(a^2*x^2+1)^(1/2)*arctan(a*x)^(1/2)*sin( 
3*arctan(a*x))/a^2/c^2/(a^2*c*x^2+c)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.83 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.05 \[ \int \frac {x \arctan (a x)^{3/2}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {48 \left (3 a x+2 a^3 x^3-2 \arctan (a x)\right ) \arctan (a x)-4 \sqrt {6 \pi } \left (1+a^2 x^2\right )^{3/2} \sqrt {\arctan (a x)} \left (3 \sqrt {3} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )+3 \left (1+a^2 x^2\right )^{3/2} \left (3 \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+3 \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )+\sqrt {3} \left (\sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+\sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )\right )\right )}{288 a^2 c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}} \] Input:

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

Output:

(48*(3*a*x + 2*a^3*x^3 - 2*ArcTan[a*x])*ArcTan[a*x] - 4*Sqrt[6*Pi]*(1 + a^ 
2*x^2)^(3/2)*Sqrt[ArcTan[a*x]]*(3*Sqrt[3]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[ 
a*x]]] - FresnelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]]) + 3*(1 + a^2*x^2)^(3/2)*( 
3*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] + 3*Sqrt[I*ArcTan[a* 
x]]*Gamma[1/2, I*ArcTan[a*x]] + Sqrt[3]*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, 
 (-3*I)*ArcTan[a*x]] + Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])) 
)/(288*a^2*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5465, 5440, 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.

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

\(\Big \downarrow \) 5465

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

\(\Big \downarrow \) 5440

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

\(\Big \downarrow \) 5439

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2 x^2+1} \left (\frac {3 a x \sqrt {\arctan (a x)}}{4 \sqrt {a^2 x^2+1}}-\frac {3}{4} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{12} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{12} \sqrt {\arctan (a x)} \sin (3 \arctan (a x))\right )}{2 a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x)^{3/2}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}\)

Input:

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

Output:

-1/3*ArcTan[a*x]^(3/2)/(a^2*c*(c + a^2*c*x^2)^(3/2)) + (Sqrt[1 + a^2*x^2]* 
((3*a*x*Sqrt[ArcTan[a*x]])/(4*Sqrt[1 + a^2*x^2]) - (3*Sqrt[Pi/2]*FresnelS[ 
Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/4 - (Sqrt[Pi/6]*FresnelS[Sqrt[6/Pi]*Sqrt[Ar 
cTan[a*x]]])/12 + (Sqrt[ArcTan[a*x]]*Sin[3*ArcTan[a*x]])/12))/(2*a^2*c^2*S 
qrt[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 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 5440

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_ 
Symbol] :> Simp[d^(q + 1/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2])   Int[(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] && 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 [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

integrate(x*arctan(a*x)^(3/2)/(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]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x*arctan(a*x)^(3/2)/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*( - 4*sqrt(a**2*x**2 + 1)*sqrt(atan(a*x))*atan(a*x) + 4*sqrt(a**2 
*x**2 + 1)*sqrt(atan(a*x))*a**3*x**3 + 6*sqrt(a**2*x**2 + 1)*sqrt(atan(a*x 
))*a*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*atan(a*x)*a**2*x**2 + atan(a*x)),x)*a**8* 
x**4 - 4*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**6*x* 
*2 - 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**4 - 3* 
int((sqrt(a**2*x**2 + 1)*sqrt(atan(a*x))*x)/(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**6*x**4 - 6*int(( 
sqrt(a**2*x**2 + 1)*sqrt(atan(a*x))*x)/(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**4*x**2 - 3*int((sqrt( 
a**2*x**2 + 1)*sqrt(atan(a*x))*x)/(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**2))/(12*a**2*c**3*(a**4*x* 
*4 + 2*a**2*x**2 + 1))