\(\int \frac {x \arctan (a x)^2}{(c+a^2 c x^2)^2} \, dx\) [293]

Optimal result

Integrand size = 20, antiderivative size = 91 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1}{4 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^2}{4 a^2 c^2}-\frac {\arctan (a x)^2}{2 a^2 c^2 \left (1+a^2 x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5050, 5012, 267} \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {\arctan (a x)^2}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {x \arctan (a x)}{2 a c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^2}{4 a^2 c^2}+\frac {1}{4 a^2 c^2 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5012

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

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]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.52 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1+2 a x \arctan (a x)+\left (-1+a^2 x^2\right ) \arctan (a x)^2}{4 a^2 c^2 \left (1+a^2 x^2\right )} \]

[In]

Integrate[(x*ArcTan[a*x]^2)/(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^2*c^2*(1 + a^2*x^2))

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(\frac {x^{2} \arctan \left (a x \right )^{2} a^{2}-a^{2} x^{2}+2 x \arctan \left (a x \right ) a -\arctan \left (a x \right )^{2}}{4 c^{2} \left (a^{2} x^{2}+1\right ) a^{2}}\)	\(58\)
derivativedivides	\(\frac {-\frac {\arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\frac {x \arctan \left (a x \right ) a}{2 a^{2} x^{2}+2}+\frac {\arctan \left (a x \right )^{2}}{4}+\frac {1}{4 a^{2} x^{2}+4}}{c^{2}}}{a^{2}}\)	\(73\)
default	\(\frac {-\frac {\arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\frac {x \arctan \left (a x \right ) a}{2 a^{2} x^{2}+2}+\frac {\arctan \left (a x \right )^{2}}{4}+\frac {1}{4 a^{2} x^{2}+4}}{c^{2}}}{a^{2}}\)	\(73\)
parts	\(-\frac {\arctan \left (a x \right )^{2}}{2 a^{2} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\frac {x \arctan \left (a x \right ) a}{2 a^{2} x^{2}+2}+\frac {\arctan \left (a x \right )^{2}}{4}+\frac {1}{4 a^{2} x^{2}+4}}{c^{2} a^{2}}\)	\(75\)
risch	\(-\frac {\left (a^{2} x^{2}-1\right ) \ln \left (i a x +1\right )^{2}}{16 a^{2} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (-\ln \left (-i a x +1\right )+a^{2} x^{2} \ln \left (-i a x +1\right )-2 i a x \right ) \ln \left (i a x +1\right )}{8 \left (a x +i\right ) a^{2} c^{2} \left (a x -i\right )}-\frac {-4+a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-\ln \left (-i a x +1\right )^{2}-4 i a x \ln \left (-i a x +1\right )}{16 \left (a x +i\right ) a^{2} c^{2} \left (a x -i\right )}\)	\(171\)

[In]

int(x*arctan(a*x)^2/(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^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.53 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {2 \, a x \arctan \left (a x\right ) + {\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )^{2} + 1}{4 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}} \]

[In]

integrate(x*arctan(a*x)^2/(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^4*c^2*x^2 + a^2*c^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {{\left (\frac {x}{a^{2} c x^{2} + c} + \frac {\arctan \left (a x\right )}{a c}\right )} \arctan \left (a x\right )}{2 \, a c} - \frac {{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 1}{4 \, {\left (a^{4} c x^{2} + a^{2} c\right )} c} - \frac {\arctan \left (a x\right )^{2}}{2 \, {\left (a^{2} c x^{2} + c\right )} a^{2} c} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.55 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+2\,a\,x\,\mathrm {atan}\left (a\,x\right )-{\mathrm {atan}\left (a\,x\right )}^2+1}{4\,a^2\,c^2\,\left (a^2\,x^2+1\right )} \]

[In]

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

[Out]

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