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

Optimal result

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

[Out]

-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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5064, 5058, 5054, 5004} \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {3 \arctan (a x)^2}{32 a^4 c^3}+\frac {x^4 \arctan (a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {x^3 \arctan (a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}-\frac {x^4}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac {3}{32 a^4 c^3 \left (a^2 x^2+1\right )}+\frac {3 x \arctan (a x)}{16 a^3 c^3 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

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

Rule 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> 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 5054

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] + (-Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x]
, x] + Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*c^2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]

Rule 5058

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] + (Dist[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] - Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(c^2*d*m)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -1]

Rule 5064

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {x^4 \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {1}{2} a \int \frac {x^4 \arctan (a x)}{\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 {x^3 \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {x^4 \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {3 \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{8 a c} \\ & = -\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 {x^4 \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {3 \int \frac {\arctan (a x)}{c+a^2 c x^2} \, dx}{16 a^3 c^2} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.53 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {4+5 a^2 x^2+2 a x \left (3+5 a^2 x^2\right ) \arctan (a x)+\left (-3-6 a^2 x^2+5 a^4 x^4\right ) \arctan (a x)^2}{32 a^4 c^3 \left (1+a^2 x^2\right )^2} \]

[In]

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

[Out]

(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)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66

method	result	size
parallelrisch	\(\frac {5 a^{4} \arctan \left (a x \right )^{2} x^{4}-4 a^{4} x^{4}+10 \arctan \left (a x \right ) x^{3} a^{3}-6 x^{2} \arctan \left (a x \right )^{2} a^{2}-3 a^{2} x^{2}+6 x \arctan \left (a x \right ) a -3 \arctan \left (a x \right )^{2}}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2} a^{4}}\)	\(93\)
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	\(\frac {\arctan \left (a x \right )^{2}}{4 c^{3} a^{4} \left (a^{2} x^{2}+1\right )^{2}}-\frac {\arctan \left (a x \right )^{2}}{2 c^{3} a^{4} \left (a^{2} x^{2}+1\right )}-\frac {-\frac {5 \arctan \left (a x \right ) x^{3}}{8 a \left (a^{2} x^{2}+1\right )^{2}}-\frac {3 \arctan \left (a x \right ) x}{8 a^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \arctan \left (a x \right )^{2}}{8 a^{4}}+\frac {-\frac {5}{2 \left (a^{2} x^{2}+1\right )}+\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {5 \arctan \left (a x \right )^{2}}{2}}{8 a^{4}}}{2 c^{3}}\)	\(153\)
risch	\(-\frac {\left (5 a^{4} x^{4}-6 a^{2} x^{2}-3\right ) \ln \left (i a x +1\right )^{2}}{128 a^{4} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\left (-6 a^{2} x^{2} \ln \left (-i a x +1\right )-3 \ln \left (-i a x +1\right )+5 x^{4} \ln \left (-i a x +1\right ) a^{4}-10 i a^{3} x^{3}-6 i a x \right ) \ln \left (i a x +1\right )}{64 a^{4} \left (a x +i\right )^{2} \left (a x -i\right )^{2} c^{3}}-\frac {5 a^{4} x^{4} \ln \left (-i a x +1\right )^{2}-6 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-3 \ln \left (-i a x +1\right )^{2}-20 i x^{3} \ln \left (-i a x +1\right ) a^{3}-12 i a x \ln \left (-i a x +1\right )-20 a^{2} x^{2}-16}{128 a^{4} \left (a x +i\right )^{2} \left (a x -i\right )^{2} c^{3}}\)	\(250\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {5 \, a^{2} x^{2} + {\left (5 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (5 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right ) + 4}{32 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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)

none

Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{16} \, a {\left (\frac {5 \, a^{2} x^{3} + 3 \, x}{a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}} + \frac {5 \, \arctan \left (a x\right )}{a^{5} c^{3}}\right )} \arctan \left (a x\right ) + \frac {{\left (5 \, a^{2} x^{2} - 5 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a^{2}}{32 \, {\left (a^{10} c^{3} x^{4} + 2 \, a^{8} c^{3} x^{2} + a^{6} c^{3}\right )}} - \frac {{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2}}{4 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

[In]

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

[Out]

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*arctan(a*x)/(a^5*c^3))*arctan(a*x) + 1/3
2*(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]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.61 \[ \int \frac {x^3 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {5\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+10\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )-6\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+5\,a^2\,x^2+6\,a\,x\,\mathrm {atan}\left (a\,x\right )-3\,{\mathrm {atan}\left (a\,x\right )}^2+4}{32\,a^4\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

[In]

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

[Out]

(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)