\(\int \frac {\arctan (a x)^2}{(c+a^2 c x^2)^2} \, dx\) [294]

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

Optimal result

Integrand size = 19, antiderivative size = 100 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {x}{4 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)}{4 a c^2}+\frac {\arctan (a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5012, 5050, 205, 211} \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {x \arctan (a x)^2}{2 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)}{2 a c^2 \left (a^2 x^2+1\right )}-\frac {x}{4 c^2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^3}{6 a c^2}-\frac {\arctan (a x)}{4 a c^2} \]

[In]

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

[Out]

-1/4*x/(c^2*(1 + a^2*x^2)) - ArcTan[a*x]/(4*a*c^2) + ArcTan[a*x]/(2*a*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x]^2)/(
2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^3/(6*a*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 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 {x \arctan (a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a c^2}-a \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx \\ & = \frac {\arctan (a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a c^2}-\frac {1}{2} \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx \\ & = -\frac {x}{4 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a c^2}-\frac {\int \frac {1}{c+a^2 c x^2} \, dx}{4 c} \\ & = -\frac {x}{4 c^2 \left (1+a^2 x^2\right )}-\frac {\arctan (a x)}{4 a c^2}+\frac {\arctan (a x)}{2 a c^2 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{6 a c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.65 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {-3 a x+\left (3-3 a^2 x^2\right ) \arctan (a x)+6 a x \arctan (a x)^2+2 \left (1+a^2 x^2\right ) \arctan (a x)^3}{12 c^2 \left (a+a^3 x^2\right )} \]

[In]

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

[Out]

(-3*a*x + (3 - 3*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*c^2*(a + a^3*
x^2))

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.75

method result size
parallelrisch \(\frac {2 \arctan \left (a x \right )^{3} x^{2} a^{2}-3 a^{2} \arctan \left (a x \right ) x^{2}+6 a \arctan \left (a x \right )^{2} x +2 \arctan \left (a x \right )^{3}-3 a x +3 \arctan \left (a x \right )}{12 c^{2} \left (a^{2} x^{2}+1\right ) a}\) \(75\)
derivativedivides \(\frac {\frac {a x \arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{2 c^{2}}-\frac {\frac {\arctan \left (a x \right )^{3}}{3}-\frac {\arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )}+\frac {a x}{4 a^{2} x^{2}+4}+\frac {\arctan \left (a x \right )}{4}}{c^{2}}}{a}\) \(93\)
default \(\frac {\frac {a x \arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{2 c^{2}}-\frac {\frac {\arctan \left (a x \right )^{3}}{3}-\frac {\arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )}+\frac {a x}{4 a^{2} x^{2}+4}+\frac {\arctan \left (a x \right )}{4}}{c^{2}}}{a}\) \(93\)
parts \(\frac {x \arctan \left (a x \right )^{2}}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{2 a \,c^{2}}-\frac {\frac {\arctan \left (a x \right )^{3}}{3 a}+\frac {-\frac {\arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )}+\frac {a x}{4 a^{2} x^{2}+4}+\frac {\arctan \left (a x \right )}{4}}{a}}{c^{2}}\) \(99\)
risch \(\frac {i \ln \left (i a x +1\right )^{3}}{48 c^{2} a}-\frac {i \left (a^{2} x^{2} \ln \left (-i a x +1\right )+\ln \left (-i a x +1\right )-2 i a x \right ) \ln \left (i a x +1\right )^{2}}{16 c^{2} \left (a^{2} x^{2}+1\right ) a}+\frac {i \left (a^{2} x^{2} \ln \left (-i a x +1\right )^{2}+\ln \left (-i a x +1\right )^{2}-4 i a x \ln \left (-i a x +1\right )-4\right ) \ln \left (i a x +1\right )}{16 c^{2} \left (a x +i\right ) \left (a x -i\right ) a}-\frac {i \left (a^{2} x^{2} \ln \left (-i a x +1\right )^{3}+6 a^{2} x^{2} \ln \left (-i a x +1\right )-6 \ln \left (i a x +1\right ) a^{2} x^{2}-6 i a x \ln \left (-i a x +1\right )^{2}-12 i a x +\ln \left (-i a x +1\right )^{3}-6 \ln \left (-i a x +1\right )-6 \ln \left (i a x +1\right )\right )}{48 c^{2} \left (a x +i\right ) \left (a x -i\right ) a}\) \(280\)

[In]

int(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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.67 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {6 \, a x \arctan \left (a x\right )^{2} + 2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 3 \, a x - 3 \, {\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{12 \, {\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \]

[In]

integrate(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^3*c^2*x^2
+ a*c^2)

Sympy [F]

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

[In]

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

[Out]

Integral(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 = 146, normalized size of antiderivative = 1.46 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {1}{2} \, {\left (\frac {x}{a^{2} c^{2} x^{2} + c^{2}} + \frac {\arctan \left (a x\right )}{a c^{2}}\right )} \arctan \left (a x\right )^{2} + \frac {{\left (2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 3 \, a x - 3 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a^{2}}{12 \, {\left (a^{5} c^{2} x^{2} + a^{3} c^{2}\right )}} - \frac {{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 1\right )} a \arctan \left (a x\right )}{2 \, {\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}} \]

[In]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\mathrm {atan}\left (a\,x\right )}{2\,\left (a^3\,c^2\,x^2+a\,c^2\right )}-\frac {x}{2\,\left (2\,a^2\,c^2\,x^2+2\,c^2\right )}+\frac {x\,{\mathrm {atan}\left (a\,x\right )}^2}{2\,\left (a^2\,c^2\,x^2+c^2\right )}-\frac {\mathrm {atan}\left (a\,x\right )}{4\,a\,c^2}+\frac {{\mathrm {atan}\left (a\,x\right )}^3}{6\,a\,c^2} \]

[In]

int(atan(a*x)^2/(c + a^2*c*x^2)^2,x)

[Out]

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