\(\int \frac {x^2 \arctan (a x)}{(c+a^2 c x^2)^2} \, dx\) [185]

Optimal result

Integrand size = 20, antiderivative size = 64 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {1}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^2}{4 a^3 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5054, 5004} \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\arctan (a x)^2}{4 a^3 c^2}-\frac {x \arctan (a x)}{2 a^2 c^2 \left (a^2 x^2+1\right )}-\frac {1}{4 a^3 c^2 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {-1-2 a x \arctan (a x)+\left (1+a^2 x^2\right ) \arctan (a x)^2}{4 a^3 c^2 \left (1+a^2 x^2\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\(\frac {x^{2} \arctan \left (a x \right )^{2} a^{2}+a^{2} x^{2}-2 x \arctan \left (a x \right ) a +\arctan \left (a x \right )^{2}}{4 c^{2} \left (a^{2} x^{2}+1\right ) a^{3}}\)	\(55\)
derivativedivides	\(\frac {-\frac {a x \arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2 c^{2}}-\frac {\frac {\arctan \left (a x \right )^{2}}{2}+\frac {1}{2 a^{2} x^{2}+2}}{2 c^{2}}}{a^{3}}\)	\(66\)
default	\(\frac {-\frac {a x \arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2 c^{2}}-\frac {\frac {\arctan \left (a x \right )^{2}}{2}+\frac {1}{2 a^{2} x^{2}+2}}{2 c^{2}}}{a^{3}}\)	\(66\)
parts	\(-\frac {x \arctan \left (a x \right )}{2 a^{2} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2 a^{3} c^{2}}-\frac {\frac {\arctan \left (a x \right )^{2}}{2 a^{3}}+\frac {1}{2 a^{3} \left (a^{2} x^{2}+1\right )}}{2 c^{2}}\)	\(73\)
risch	\(-\frac {\ln \left (i a x +1\right )^{2}}{16 a^{3} c^{2}}+\frac {\left (a^{2} x^{2} \ln \left (-i a x +1\right )+\ln \left (-i a x +1\right )+2 i a x \right ) \ln \left (i a x +1\right )}{8 a^{3} c^{2} \left (a^{2} x^{2}+1\right )}-\frac {a^{2} x^{2} \ln \left (-i a x +1\right )^{2}+\ln \left (-i a x +1\right )^{2}+4 i a x \ln \left (-i a x +1\right )+4}{16 a^{3} c^{2} \left (a x +i\right ) \left (a x -i\right )}\)	\(142\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.77 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {2 \, a x \arctan \left (a x\right ) - {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 1}{4 \, {\left (a^{5} c^{2} x^{2} + a^{3} c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {1}{2} \, {\left (\frac {x}{a^{4} c^{2} x^{2} + a^{2} c^{2}} - \frac {\arctan \left (a x\right )}{a^{3} c^{2}}\right )} \arctan \left (a x\right ) - \frac {{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 1\right )} a}{4 \, {\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2-2\,a\,x\,\mathrm {atan}\left (a\,x\right )+{\mathrm {atan}\left (a\,x\right )}^2-1}{4\,a^3\,c^2\,\left (a^2\,x^2+1\right )} \]

[In]

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

[Out]

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