\(\int \frac {x^2 \arctan (a x)^2}{(c+a^2 c x^2)^2} \, dx\) [292]

Optimal result

Integrand size = 22, antiderivative size = 106 \[ \int \frac {x^2 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {x}{4 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)}{4 a^3 c^2}-\frac {\arctan (a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a^3 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 106, 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 = {5056, 5050, 205, 211} \[ \int \frac {x^2 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\arctan (a x)^3}{6 a^3 c^2}+\frac {\arctan (a x)}{4 a^3 c^2}-\frac {x \arctan (a x)^2}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {x}{4 a^2 c^2 \left (a^2 x^2+1\right )}-\frac {\arctan (a x)}{2 a^3 c^2 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 5050

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] - Dist[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]

Rule 5056

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(a + b*ArcTan
[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (Dist[b*(p/(2*c)), Int[x*((a + b*ArcTan[c*x])^(p - 1)/(d + e*x^2)^2
), x], x] - Simp[x*((a + b*ArcTan[c*x])^p/(2*c^2*d*(d + e*x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c
^2*d] && GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {x \arctan (a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a^3 c^2}+\frac {\int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{a} \\ & = -\frac {\arctan (a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a^3 c^2}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 a^2} \\ & = \frac {x}{4 a^2 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a^3 c^2}+\frac {\int \frac {1}{c+a^2 c x^2} \, dx}{4 a^2 c} \\ & = \frac {x}{4 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)}{4 a^3 c^2}-\frac {\arctan (a x)}{2 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a^3 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int \frac {x^2 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3 a x+3 \left (-1+a^2 x^2\right ) \arctan (a x)-6 a x \arctan (a x)^2+2 \left (1+a^2 x^2\right ) \arctan (a x)^3}{12 a^3 c^2 \left (1+a^2 x^2\right )} \]

[In]

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

[Out]

(3*a*x + 3*(-1 + a^2*x^2)*ArcTan[a*x] - 6*a*x*ArcTan[a*x]^2 + 2*(1 + a^2*x^2)*ArcTan[a*x]^3)/(12*a^3*c^2*(1 +
a^2*x^2))

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {2 \arctan \left (a x \right )^{3} x^{2} a^{2}+3 a^{2} \arctan \left (a x \right ) x^{2}-6 a \arctan \left (a x \right )^{2} x +2 \arctan \left (a x \right )^{3}+3 a x -3 \arctan \left (a x \right )}{12 c^{2} \left (a^{2} x^{2}+1\right ) a^{3}}\)	\(75\)
derivativedivides	\(\frac {-\frac {a x \arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{2 c^{2}}-\frac {\frac {\arctan \left (a x \right )^{3}}{3}+\frac {\arctan \left (a x \right )}{2 a^{2} x^{2}+2}-\frac {a x}{4 \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{4}}{c^{2}}}{a^{3}}\)	\(93\)
default	\(\frac {-\frac {a x \arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{2 c^{2}}-\frac {\frac {\arctan \left (a x \right )^{3}}{3}+\frac {\arctan \left (a x \right )}{2 a^{2} x^{2}+2}-\frac {a x}{4 \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{4}}{c^{2}}}{a^{3}}\)	\(93\)
parts	\(-\frac {x \arctan \left (a x \right )^{2}}{2 a^{2} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{2 a^{3} c^{2}}-\frac {\frac {\arctan \left (a x \right )^{3}}{3 a^{3}}-\frac {-\frac {\arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )}+\frac {a x}{4 a^{2} x^{2}+4}+\frac {\arctan \left (a x \right )}{4}}{a^{3}}}{c^{2}}\)	\(103\)
risch	\(\frac {i \ln \left (i a x +1\right )^{3}}{48 c^{2} a^{3}}-\frac {i \left (a^{2} x^{2} \ln \left (-i a x +1\right )+\ln \left (-i a x +1\right )+2 i a x \right ) \ln \left (i a x +1\right )^{2}}{16 a^{3} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {i \left (a^{2} x^{2} \ln \left (-i a x +1\right )^{2}+\ln \left (-i a x +1\right )^{2}+4 i a x \ln \left (-i a x +1\right )+4\right ) \ln \left (i a x +1\right )}{16 a^{3} c^{2} \left (a x +i\right ) \left (a x -i\right )}+\frac {i \left (-a^{2} x^{2} \ln \left (-i a x +1\right )^{3}-6 i a x \ln \left (-i a x +1\right )^{2}+6 \ln \left (i a x -1\right ) a^{2} x^{2}-6 \ln \left (-i a x -1\right ) a^{2} x^{2}-12 i a x -\ln \left (-i a x +1\right )^{3}+6 \ln \left (i a x -1\right )-6 \ln \left (-i a x -1\right )-12 \ln \left (-i a x +1\right )\right )}{48 a^{3} c^{2} \left (a x +i\right ) \left (a x -i\right )}\)	\(293\)

[In]

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

[Out]

1/12*(2*arctan(a*x)^3*x^2*a^2+3*a^2*arctan(a*x)*x^2-6*a*arctan(a*x)^2*x+2*arctan(a*x)^3+3*a*x-3*arctan(a*x))/c
^2/(a^2*x^2+1)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.65 \[ \int \frac {x^2 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {6 \, a x \arctan \left (a x\right )^{2} - 2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 3 \, a x - 3 \, {\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{12 \, {\left (a^{5} c^{2} x^{2} + a^{3} c^{2}\right )}} \]

[In]

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

[Out]

-1/12*(6*a*x*arctan(a*x)^2 - 2*(a^2*x^2 + 1)*arctan(a*x)^3 - 3*a*x - 3*(a^2*x^2 - 1)*arctan(a*x))/(a^5*c^2*x^2
+ a^3*c^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.42 \[ \int \frac {x^2 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {1}{2} \, {\left (\frac {x}{a^{4} c^{2} x^{2} + a^{2} c^{2}} - \frac {\arctan \left (a x\right )}{a^{3} c^{2}}\right )} \arctan \left (a x\right )^{2} + \frac {{\left (2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} + 3 \, a x + 3 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a^{2}}{12 \, {\left (a^{7} c^{2} x^{2} + a^{5} c^{2}\right )}} - \frac {{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 1\right )} a \arctan \left (a x\right )}{2 \, {\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )}} \]

[In]

integrate(x^2*arctan(a*x)^2/(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)^2 + 1/12*(2*(a^2*x^2 + 1)*arctan(a*x)^3 +
3*a*x + 3*(a^2*x^2 + 1)*arctan(a*x))*a^2/(a^7*c^2*x^2 + a^5*c^2) - 1/2*((a^2*x^2 + 1)*arctan(a*x)^2 + 1)*a*ar
ctan(a*x)/(a^6*c^2*x^2 + a^4*c^2)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {x}{2\,\left (2\,a^4\,c^2\,x^2+2\,a^2\,c^2\right )}+\frac {\mathrm {atan}\left (a\,x\right )}{4\,a^3\,c^2}+\frac {{\mathrm {atan}\left (a\,x\right )}^3}{6\,a^3\,c^2}-\frac {\mathrm {atan}\left (a\,x\right )}{2\,a^5\,c^2\,\left (\frac {1}{a^2}+x^2\right )}-\frac {x\,{\mathrm {atan}\left (a\,x\right )}^2}{2\,a^4\,c^2\,\left (\frac {1}{a^2}+x^2\right )} \]

[In]

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

[Out]

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