\(\int \frac {x^3}{(c+a^2 c x^2)^3 \arctan (a x)^{3/2}} \, dx\) [998]

Optimal result

Integrand size = 24, antiderivative size = 96 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{3/2}} \, dx=-\frac {2 x^3}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}-\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^4 c^3}+\frac {\sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{a^4 c^3} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.54 \[ \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{3/2}} \, dx=\frac {-2 \sqrt {2 \pi } \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+16 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {-\frac {32 a^3 x^3}{\left (1+a^2 x^2\right )^2}+3 i \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-4 i \arctan (a x)\right )-3 i \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},4 i \arctan (a x)\right )}{\sqrt {\arctan (a x)}}}{16 a^4 c^3} \] Input:

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

Output:

(-2*Sqrt[2*Pi]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] + 16*Sqrt[Pi]*Fres 
nelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + ((-32*a^3*x^3)/(1 + a^2*x^2)^2 + (3 
*I)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-4*I)*ArcTan[a*x]] - (3*I)*Sqrt[I*A 
rcTan[a*x]]*Gamma[1/2, (4*I)*ArcTan[a*x]])/Sqrt[ArcTan[a*x]])/(16*a^4*c^3)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.71, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5503, 27, 5505, 3042, 3793, 2009, 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.

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

\(\Big \downarrow \) 5503

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5505

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

\(\displaystyle -\frac {2 \int \left (-\frac {\cos (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}+\frac {\cos (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}+\frac {3}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^4 c^3}+\frac {6 \int \frac {a^2 x^2}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a^4 c^3}-\frac {2 x^3}{a c^3 \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {6 \int \frac {a^2 x^2}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{a^4 c^3}-\frac {2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}\right )}{a^4 c^3}-\frac {2 x^3}{a c^3 \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {6 \int \left (\frac {1}{8 \sqrt {\arctan (a x)}}-\frac {\cos (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^4 c^3}-\frac {2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}\right )}{a^4 c^3}-\frac {2 x^3}{a c^3 \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {6 \left (\frac {1}{4} \sqrt {\arctan (a x)}-\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^4 c^3}-\frac {2 \left (\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}\right )}{a^4 c^3}-\frac {2 x^3}{a c^3 \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}\)

Input:

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

Output:

(-2*x^3)/(a*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) + (6*(Sqrt[ArcTan[a*x]] 
/4 - (Sqrt[Pi/2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/8))/(a^4*c^3) - 
 (2*((3*Sqrt[ArcTan[a*x]])/4 + (Sqrt[Pi/2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcT 
an[a*x]]])/8 - (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/2))/(a^ 
4*c^3)
 

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 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 5503

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

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])
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )-4 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )+2 \sin \left (2 \arctan \left (a x \right )\right )-\sin \left (4 \arctan \left (a x \right )\right )}{4 a^{4} c^{3} \sqrt {\arctan \left (a x \right )}}\)	\(86\)

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**3/(a**6*x**6*atan(a*x)**(3/2) + 3*a**4*x**4*atan(a*x)**(3/2) + 
 3*a**2*x**2*atan(a*x)**(3/2) + atan(a*x)**(3/2)), x)/c**3
 

Maxima [F(-2)]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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