\(\int \frac {\arctan (a x)}{(c+a^2 c x^2)^2} \, dx\) [187]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-2)]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 17, antiderivative size = 61 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1}{4 a c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^2}{4 a c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5012, 267} \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {x \arctan (a x)}{2 c^2 \left (a^2 x^2+1\right )}+\frac {1}{4 a c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^2}{4 a c^2} \]

[In]

Int[ArcTan[a*x]/(c + a^2*c*x^2)^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1+2 a x \arctan (a x)+\left (1+a^2 x^2\right ) \arctan (a x)^2}{4 c^2 \left (a+a^3 x^2\right )} \]

[In]

Integrate[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*c^2*(a + a^3*x^2))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92

method result size
parallelrisch \(\frac {x^{2} \arctan \left (a x \right )^{2} a^{2}-a^{2} x^{2}+2 x \arctan \left (a x \right ) a +\arctan \left (a x \right )^{2}}{4 c^{2} \left (a^{2} x^{2}+1\right ) a}\) \(56\)
derivativedivides \(\frac {\frac {a x \arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2 c^{2}}-\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2}}{2 c^{2}}}{a}\) \(66\)
default \(\frac {\frac {a x \arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2 c^{2}}-\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2}}{2 c^{2}}}{a}\) \(66\)
parts \(\frac {x \arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{2}}{2 a \,c^{2}}-\frac {\frac {\arctan \left (a x \right )^{2}}{2 a}-\frac {1}{2 a \left (a^{2} x^{2}+1\right )}}{2 c^{2}}\) \(70\)
risch \(-\frac {\ln \left (i a x +1\right )^{2}}{16 c^{2} a}+\frac {\left (a^{2} x^{2} \ln \left (-i a x +1\right )+\ln \left (-i a x +1\right )-2 i a x \right ) \ln \left (i a x +1\right )}{8 c^{2} \left (a^{2} x^{2}+1\right ) a}-\frac {a^{2} x^{2} \ln \left (-i a x +1\right )^{2}+\ln \left (-i a x +1\right )^{2}-4 i a x \ln \left (-i a x +1\right )-4}{16 c^{2} \left (a x +i\right ) \left (a x -i\right ) a}\) \(142\)

[In]

int(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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {2 \, a x \arctan \left (a x\right ) + {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 1}{4 \, {\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \]

[In]

integrate(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^3*c^2*x^2 + a*c^2)

Sympy [F(-2)]

Exception generated. \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\text {Exception raised: RecursionError} \]

[In]

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

[Out]

Exception raised: RecursionError >> maximum recursion depth exceeded while calling a Python object

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.28 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1}{2} \, {\left (\frac {x}{a^{2} c^{2} x^{2} + c^{2}} + \frac {\arctan \left (a x\right )}{a c^{2}}\right )} \arctan \left (a x\right ) - \frac {{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 1\right )} a}{4 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+2\,a\,x\,\mathrm {atan}\left (a\,x\right )+{\mathrm {atan}\left (a\,x\right )}^2+1}{4\,a\,c^2\,\left (a^2\,x^2+1\right )} \]

[In]

int(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*c^2*(a^2*x^2 + 1))