\(\int \frac {x^3 \arctan (a x)^2}{(c+a^2 c x^2)^3} \, dx\) [299]

Optimal result

Integrand size = 22, antiderivative size = 140 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {x^4}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {3}{32 a^4 c^3 \left (1+a^2 x^2\right )}+\frac {x^3 \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)}{16 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 \arctan (a x)^2}{32 a^4 c^3}+\frac {x^4 \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2} \] Output:

-1/32*x^4/c^3/(a^2*x^2+1)^2+3/32/a^4/c^3/(a^2*x^2+1)+1/8*x^3*arctan(a*x)/a 
/c^3/(a^2*x^2+1)^2+3/16*x*arctan(a*x)/a^3/c^3/(a^2*x^2+1)-3/32*arctan(a*x) 
^2/a^4/c^3+1/4*x^4*arctan(a*x)^2/c^3/(a^2*x^2+1)^2
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.53 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {4+5 a^2 x^2+2 a x \left (3+5 a^2 x^2\right ) \arctan (a x)+\left (-3-6 a^2 x^2+5 a^4 x^4\right ) \arctan (a x)^2}{32 a^4 c^3 \left (1+a^2 x^2\right )^2} \] Input:

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

Output:

(4 + 5*a^2*x^2 + 2*a*x*(3 + 5*a^2*x^2)*ArcTan[a*x] + (-3 - 6*a^2*x^2 + 5*a 
^4*x^4)*ArcTan[a*x]^2)/(32*a^4*c^3*(1 + a^2*x^2)^2)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5479, 27, 5473, 5469, 5419}

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.

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

\(\Big \downarrow \) 5479

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5473

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

\(\Big \downarrow \) 5469

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

\(\Big \downarrow \) 5419

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

Input:

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

Output:

(x^4*ArcTan[a*x]^2)/(4*c^3*(1 + a^2*x^2)^2) - (a*(x^4/(16*a*(1 + a^2*x^2)^ 
2) - (x^3*ArcTan[a*x])/(4*a^2*(1 + a^2*x^2)^2) + (3*(-1/4*1/(a^3*(1 + a^2* 
x^2)) - (x*ArcTan[a*x])/(2*a^2*(1 + a^2*x^2)) + ArcTan[a*x]^2/(4*a^3)))/(4 
*a^2)))/(2*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 5419

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

rule 5469

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x 
_Symbol] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (Simp 
[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*c^2*d*(q + 1))), x] - Simp[1 
/(2*c^2*d*(q + 1))   Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x]) / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]
 

rule 5473

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66

method	result	size
parallelrisch	\(\frac {5 a^{4} \arctan \left (a x \right )^{2} x^{4}-4 a^{4} x^{4}+10 \arctan \left (a x \right ) x^{3} a^{3}-6 \arctan \left (a x \right )^{2} x^{2} a^{2}-3 a^{2} x^{2}+6 \arctan \left (a x \right ) a x -3 \arctan \left (a x \right )^{2}}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2} a^{4}}\)	\(93\)
derivativedivides	\(\frac {-\frac {\arctan \left (a x \right )^{2}}{2 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {-\frac {5 \arctan \left (a x \right ) a^{3} x^{3}}{8 \left (a^{2} x^{2}+1\right )^{2}}-\frac {3 \arctan \left (a x \right ) a x}{8 \left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \arctan \left (a x \right )^{2}}{16}-\frac {5}{16 \left (a^{2} x^{2}+1\right )}+\frac {1}{16 \left (a^{2} x^{2}+1\right )^{2}}}{2 c^{3}}}{a^{4}}\)	\(132\)
default	\(\frac {-\frac {\arctan \left (a x \right )^{2}}{2 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {-\frac {5 \arctan \left (a x \right ) a^{3} x^{3}}{8 \left (a^{2} x^{2}+1\right )^{2}}-\frac {3 \arctan \left (a x \right ) a x}{8 \left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \arctan \left (a x \right )^{2}}{16}-\frac {5}{16 \left (a^{2} x^{2}+1\right )}+\frac {1}{16 \left (a^{2} x^{2}+1\right )^{2}}}{2 c^{3}}}{a^{4}}\)	\(132\)
parts	\(-\frac {\arctan \left (a x \right )^{2}}{2 c^{3} a^{4} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{4 c^{3} a^{4} \left (a^{2} x^{2}+1\right )^{2}}-\frac {-\frac {5 \arctan \left (a x \right ) x^{3}}{8 a \left (a^{2} x^{2}+1\right )^{2}}-\frac {3 \arctan \left (a x \right ) x}{8 a^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \arctan \left (a x \right )^{2}}{8 a^{4}}+\frac {-\frac {5}{2 \left (a^{2} x^{2}+1\right )}+\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {5 \arctan \left (a x \right )^{2}}{2}}{8 a^{4}}}{2 c^{3}}\)	\(153\)
risch	\(-\frac {\left (5 a^{4} x^{4}-6 a^{2} x^{2}-3\right ) \ln \left (i a x +1\right )^{2}}{128 a^{4} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\left (-6 a^{2} x^{2} \ln \left (-i a x +1\right )-3 \ln \left (-i a x +1\right )+5 x^{4} \ln \left (-i a x +1\right ) a^{4}-10 i a^{3} x^{3}-6 i a x \right ) \ln \left (i a x +1\right )}{64 a^{4} \left (a x +i\right )^{2} \left (a x -i\right )^{2} c^{3}}-\frac {5 a^{4} x^{4} \ln \left (-i a x +1\right )^{2}-6 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-3 \ln \left (-i a x +1\right )^{2}-20 i x^{3} \ln \left (-i a x +1\right ) a^{3}-12 i a x \ln \left (-i a x +1\right )-20 a^{2} x^{2}-16}{128 a^{4} \left (a x +i\right )^{2} \left (a x -i\right )^{2} c^{3}}\)	\(250\)
orering	\(-\frac {\left (a^{2} x^{2}+1\right ) \left (12 a^{6} x^{6}-13 a^{4} x^{4}+9 a^{2} x^{2}+15\right ) \arctan \left (a x \right )^{2}}{16 a^{4} \left (a^{2} c \,x^{2}+c \right )^{3}}-\frac {\left (a^{2} x^{2}+1\right )^{2} \left (16 a^{4} x^{4}-5 a^{2} x^{2}-12\right ) \left (\frac {3 x^{2} \arctan \left (a x \right )^{2}}{\left (a^{2} c \,x^{2}+c \right )^{3}}+\frac {2 x^{3} \arctan \left (a x \right ) a}{\left (a^{2} c \,x^{2}+c \right )^{3} \left (a^{2} x^{2}+1\right )}-\frac {6 x^{4} \arctan \left (a x \right )^{2} c \,a^{2}}{\left (a^{2} c \,x^{2}+c \right )^{4}}\right )}{32 x^{2} a^{4}}-\frac {\left (4 a^{2} x^{2}+3\right ) \left (a^{2} x^{2}+1\right )^{3} \left (\frac {6 x \arctan \left (a x \right )^{2}}{\left (a^{2} c \,x^{2}+c \right )^{3}}+\frac {12 x^{2} \arctan \left (a x \right ) a}{\left (a^{2} c \,x^{2}+c \right )^{3} \left (a^{2} x^{2}+1\right )}-\frac {42 x^{3} \arctan \left (a x \right )^{2} c \,a^{2}}{\left (a^{2} c \,x^{2}+c \right )^{4}}+\frac {2 x^{3} a^{2}}{\left (a^{2} x^{2}+1\right )^{2} \left (a^{2} c \,x^{2}+c \right )^{3}}-\frac {24 x^{4} \arctan \left (a x \right ) a^{3} c}{\left (a^{2} c \,x^{2}+c \right )^{4} \left (a^{2} x^{2}+1\right )}-\frac {4 x^{4} \arctan \left (a x \right ) a^{3}}{\left (a^{2} c \,x^{2}+c \right )^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {48 x^{5} \arctan \left (a x \right )^{2} c^{2} a^{4}}{\left (a^{2} c \,x^{2}+c \right )^{5}}\right )}{64 x \,a^{4}}\)	\(423\)

Input:

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

Output:

1/32*(5*a^4*arctan(a*x)^2*x^4-4*a^4*x^4+10*arctan(a*x)*x^3*a^3-6*arctan(a* 
x)^2*x^2*a^2-3*a^2*x^2+6*arctan(a*x)*a*x-3*arctan(a*x)^2)/c^3/(a^2*x^2+1)^ 
2/a^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {5 \, a^{2} x^{2} + {\left (5 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (5 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right ) + 4}{32 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \] Input:

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

Output:

1/32*(5*a^2*x^2 + (5*a^4*x^4 - 6*a^2*x^2 - 3)*arctan(a*x)^2 + 2*(5*a^3*x^3 
 + 3*a*x)*arctan(a*x) + 4)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)
 

Sympy [F]

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

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

Output:

Integral(x**3*atan(a*x)**2/(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1), x) 
/c**3
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{16} \, a {\left (\frac {5 \, a^{2} x^{3} + 3 \, x}{a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}} + \frac {5 \, \arctan \left (a x\right )}{a^{5} c^{3}}\right )} \arctan \left (a x\right ) + \frac {{\left (5 \, a^{2} x^{2} - 5 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a^{2}}{32 \, {\left (a^{10} c^{3} x^{4} + 2 \, a^{8} c^{3} x^{2} + a^{6} c^{3}\right )}} - \frac {{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2}}{4 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \] Input:

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

Output:

1/16*a*((5*a^2*x^3 + 3*x)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3) + 5*arct 
an(a*x)/(a^5*c^3))*arctan(a*x) + 1/32*(5*a^2*x^2 - 5*(a^4*x^4 + 2*a^2*x^2 
+ 1)*arctan(a*x)^2 + 4)*a^2/(a^10*c^3*x^4 + 2*a^8*c^3*x^2 + a^6*c^3) - 1/4 
*(2*a^2*x^2 + 1)*arctan(a*x)^2/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.61 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {5\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+10\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )-6\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+5\,a^2\,x^2+6\,a\,x\,\mathrm {atan}\left (a\,x\right )-3\,{\mathrm {atan}\left (a\,x\right )}^2+4}{32\,a^4\,c^3\,{\left (a^2\,x^2+1\right )}^2} \] Input:

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

Output:

(5*a^2*x^2 - 3*atan(a*x)^2 + 10*a^3*x^3*atan(a*x) + 6*a*x*atan(a*x) - 6*a^ 
2*x^2*atan(a*x)^2 + 5*a^4*x^4*atan(a*x)^2 + 4)/(32*a^4*c^3*(a^2*x^2 + 1)^2 
)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {10 \mathit {atan} \left (a x \right )^{2} a^{4} x^{4}-12 \mathit {atan} \left (a x \right )^{2} a^{2} x^{2}-6 \mathit {atan} \left (a x \right )^{2}+20 \mathit {atan} \left (a x \right ) a^{3} x^{3}+12 \mathit {atan} \left (a x \right ) a x -5 a^{4} x^{4}+3}{64 a^{4} c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )} \] Input:

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

Output:

(10*atan(a*x)**2*a**4*x**4 - 12*atan(a*x)**2*a**2*x**2 - 6*atan(a*x)**2 + 
20*atan(a*x)*a**3*x**3 + 12*atan(a*x)*a*x - 5*a**4*x**4 + 3)/(64*a**4*c**3 
*(a**4*x**4 + 2*a**2*x**2 + 1))