\(\int \frac {x^2 \arctan (a x)}{(c+a^2 c x^2)^3} \, dx\) [193]

Optimal result

Integrand size = 20, antiderivative size = 111 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {1}{16 a^3 c^3 \left (1+a^2 x^2\right )^2}+\frac {1}{16 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {x \arctan (a x)}{8 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^2}{16 a^3 c^3} \]

[Out]

-1/16/a^3/c^3/(a^2*x^2+1)^2+1/16/a^3/c^3/(a^2*x^2+1)-1/4*x*arctan(a*x)/a^2/c^3/(a^2*x^2+1)^2+1/8*x*arctan(a*x)
/a^2/c^3/(a^2*x^2+1)+1/16*arctan(a*x)^2/a^3/c^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 111, 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 = {5054, 5012, 267} \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\arctan (a x)^2}{16 a^3 c^3}+\frac {x \arctan (a x)}{8 a^2 c^3 \left (a^2 x^2+1\right )}-\frac {x \arctan (a x)}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac {1}{16 a^3 c^3 \left (a^2 x^2+1\right )}-\frac {1}{16 a^3 c^3 \left (a^2 x^2+1\right )^2} \]

[In]

Int[(x^2*ArcTan[a*x])/(c + a^2*c*x^2)^3,x]

[Out]

-1/16*1/(a^3*c^3*(1 + a^2*x^2)^2) + 1/(16*a^3*c^3*(1 + a^2*x^2)) - (x*ArcTan[a*x])/(4*a^2*c^3*(1 + a^2*x^2)^2)
+ (x*ArcTan[a*x])/(8*a^2*c^3*(1 + a^2*x^2)) + ArcTan[a*x]^2/(16*a^3*c^3)

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 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}{16 a^3 c^3 \left (1+a^2 x^2\right )^2}-\frac {x \arctan (a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {\int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{4 a^2 c} \\ & = -\frac {1}{16 a^3 c^3 \left (1+a^2 x^2\right )^2}-\frac {x \arctan (a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {x \arctan (a x)}{8 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^2}{16 a^3 c^3}-\frac {\int \frac {x}{\left (c+a^2 c x^2\right )^2} \, dx}{8 a c} \\ & = -\frac {1}{16 a^3 c^3 \left (1+a^2 x^2\right )^2}+\frac {1}{16 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {x \arctan (a x)}{8 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^2}{16 a^3 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.58 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {a^2 x^2+2 a x \left (-1+a^2 x^2\right ) \arctan (a x)+\left (1+a^2 x^2\right )^2 \arctan (a x)^2}{16 a^3 c^3 \left (1+a^2 x^2\right )^2} \]

[In]

Integrate[(x^2*ArcTan[a*x])/(c + a^2*c*x^2)^3,x]

[Out]

(a^2*x^2 + 2*a*x*(-1 + a^2*x^2)*ArcTan[a*x] + (1 + a^2*x^2)^2*ArcTan[a*x]^2)/(16*a^3*c^3*(1 + a^2*x^2)^2)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {a^{4} \arctan \left (a x \right )^{2} x^{4}+2 \arctan \left (a x \right ) x^{3} a^{3}+2 x^{2} \arctan \left (a x \right )^{2} a^{2}+a^{2} x^{2}-2 x \arctan \left (a x \right ) a +\arctan \left (a x \right )^{2}}{16 c^{3} \left (a^{2} x^{2}+1\right )^{2} a^{3}}\)	\(81\)
derivativedivides	\(\frac {\frac {\arctan \left (a x \right ) a^{3} x^{3}}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {\frac {\arctan \left (a x \right )^{2}}{2}-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}}{8 c^{3}}}{a^{3}}\)	\(105\)
default	\(\frac {\frac {\arctan \left (a x \right ) a^{3} x^{3}}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {\frac {\arctan \left (a x \right )^{2}}{2}-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}}{8 c^{3}}}{a^{3}}\)	\(105\)
parts	\(\frac {\arctan \left (a x \right ) x^{3}}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {x \arctan \left (a x \right )}{8 a^{2} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\arctan \left (a x \right )^{2}}{8 a^{3} c^{3}}-\frac {\frac {\arctan \left (a x \right )^{2}}{2 a^{3}}+\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )}+\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}}{a^{3}}}{8 c^{3}}\)	\(111\)
risch	\(-\frac {\ln \left (i a x +1\right )^{2}}{64 a^{3} c^{3}}+\frac {\left (x^{4} \ln \left (-i a x +1\right ) a^{4}+2 a^{2} x^{2} \ln \left (-i a x +1\right )-2 i a^{3} x^{3}+\ln \left (-i a x +1\right )+2 i a x \right ) \ln \left (i a x +1\right )}{32 a^{3} c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {a^{4} x^{4} \ln \left (-i a x +1\right )^{2}+2 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-4 i a^{3} x^{3} \ln \left (-i a x +1\right )-4 a^{2} x^{2}+\ln \left (-i a x +1\right )^{2}+4 i a x \ln \left (-i a x +1\right )}{64 \left (a x +i\right )^{2} c^{3} \left (a x -i\right )^{2} a^{3}}\)	\(209\)

[In]

int(x^2*arctan(a*x)/(a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/16*(a^4*arctan(a*x)^2*x^4+2*arctan(a*x)*x^3*a^3+2*x^2*arctan(a*x)^2*a^2+a^2*x^2-2*x*arctan(a*x)*a+arctan(a*x
)^2)/c^3/(a^2*x^2+1)^2/a^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {a^{2} x^{2} + {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (a^{3} x^{3} - a x\right )} \arctan \left (a x\right )}{16 \, {\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} \]

[In]

integrate(x^2*arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

1/16*(a^2*x^2 + (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 2*(a^3*x^3 - a*x)*arctan(a*x))/(a^7*c^3*x^4 + 2*a^5*
c^3*x^2 + a^3*c^3)

Sympy [F]

\[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\int \frac {x^{2} \operatorname {atan}{\left (a x \right )}}{a^{6} x^{6} + 3 a^{4} x^{4} + 3 a^{2} x^{2} + 1}\, dx}{c^{3}} \]

[In]

integrate(x**2*atan(a*x)/(a**2*c*x**2+c)**3,x)

[Out]

Integral(x**2*atan(a*x)/(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1), x)/c**3

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{8} \, {\left (\frac {a^{2} x^{3} - x}{a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}} + \frac {\arctan \left (a x\right )}{a^{3} c^{3}}\right )} \arctan \left (a x\right ) + \frac {{\left (a^{2} x^{2} - {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2}\right )} a}{16 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

[In]

integrate(x^2*arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

1/8*((a^2*x^3 - x)/(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3) + arctan(a*x)/(a^3*c^3))*arctan(a*x) + 1/16*(a^2*x^
2 - (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2)*a/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

Giac [F]

\[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\int { \frac {x^{2} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{3}} \,d x } \]

[In]

integrate(x^2*arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.72 \[ \int \frac {x^2 \arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+2\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )+2\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+a^2\,x^2-2\,a\,x\,\mathrm {atan}\left (a\,x\right )+{\mathrm {atan}\left (a\,x\right )}^2}{16\,a^3\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

[In]

int((x^2*atan(a*x))/(c + a^2*c*x^2)^3,x)

[Out]

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