\(\int \frac {x^2 \arctan (a x)^3}{(c+a^2 c x^2)^2} \, dx\) [397]

Optimal result

Integrand size = 22, antiderivative size = 135 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3}{8 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {3 x \arctan (a x)}{4 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^2}{8 a^3 c^2}-\frac {3 \arctan (a x)^2}{4 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^3}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^4}{8 a^3 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 135, 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, 5012, 267} \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\arctan (a x)^4}{8 a^3 c^2}+\frac {3 \arctan (a x)^2}{8 a^3 c^2}-\frac {x \arctan (a x)^3}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {3 x \arctan (a x)}{4 a^2 c^2 \left (a^2 x^2+1\right )}-\frac {3 \arctan (a x)^2}{4 a^3 c^2 \left (a^2 x^2+1\right )}+\frac {3}{8 a^3 c^2 \left (a^2 x^2+1\right )} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.55 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3+6 a x \arctan (a x)+3 \left (-1+a^2 x^2\right ) \arctan (a x)^2-4 a x \arctan (a x)^3+\left (1+a^2 x^2\right ) \arctan (a x)^4}{8 a^3 c^2 \left (1+a^2 x^2\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(\frac {\arctan \left (a x \right )^{4} x^{2} a^{2}+3 x^{2} \arctan \left (a x \right )^{2} a^{2}-4 \arctan \left (a x \right )^{3} a x -3 a^{2} x^{2}+\arctan \left (a x \right )^{4}+6 x \arctan \left (a x \right ) a -3 \arctan \left (a x \right )^{2}}{8 c^{2} \left (a^{2} x^{2}+1\right ) a^{3}}\)	\(88\)
derivativedivides	\(\frac {-\frac {\arctan \left (a x \right )^{3} a x}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{4}}{2 c^{2}}-\frac {3 \left (\frac {\arctan \left (a x \right )^{4}}{4}+\frac {\arctan \left (a x \right )^{2}}{2 a^{2} x^{2}+2}-\frac {\arctan \left (a x \right ) a x}{2 \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )^{2}}{4}-\frac {1}{4 \left (a^{2} x^{2}+1\right )}\right )}{2 c^{2}}}{a^{3}}\)	\(114\)
default	\(\frac {-\frac {\arctan \left (a x \right )^{3} a x}{2 c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{4}}{2 c^{2}}-\frac {3 \left (\frac {\arctan \left (a x \right )^{4}}{4}+\frac {\arctan \left (a x \right )^{2}}{2 a^{2} x^{2}+2}-\frac {\arctan \left (a x \right ) a x}{2 \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )^{2}}{4}-\frac {1}{4 \left (a^{2} x^{2}+1\right )}\right )}{2 c^{2}}}{a^{3}}\)	\(114\)
parts	\(-\frac {x \arctan \left (a x \right )^{3}}{2 a^{2} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{4}}{2 a^{3} c^{2}}-\frac {3 \left (\frac {\arctan \left (a x \right )^{4}}{4 a^{3}}-\frac {-\frac {\arctan \left (a x \right )^{2}}{2 \left (a^{2} x^{2}+1\right )}+\frac {x \arctan \left (a x \right ) a}{2 a^{2} x^{2}+2}+\frac {\arctan \left (a x \right )^{2}}{4}+\frac {1}{4 a^{2} x^{2}+4}}{a^{3}}\right )}{2 c^{2}}\)	\(124\)
risch	\(\frac {\ln \left (i a x +1\right )^{4}}{128 c^{2} a^{3}}-\frac {\left (a^{2} x^{2} \ln \left (-i a x +1\right )+\ln \left (-i a x +1\right )+2 i a x \right ) \ln \left (i a x +1\right )^{3}}{32 a^{3} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {3 \left (a^{2} x^{2} \ln \left (-i a x +1\right )^{2}+4 i a x \ln \left (-i a x +1\right )-2 a^{2} x^{2}+\ln \left (-i a x +1\right )^{2}+2\right ) \ln \left (i a x +1\right )^{2}}{64 a^{3} c^{2} \left (a x +i\right ) \left (a x -i\right )}-\frac {\left (a^{2} x^{2} \ln \left (-i a x +1\right )^{3}-6 a^{2} x^{2} \ln \left (-i a x +1\right )+\ln \left (-i a x +1\right )^{3}+6 i a x \ln \left (-i a x +1\right )^{2}+6 \ln \left (-i a x +1\right )+12 i a x \right ) \ln \left (i a x +1\right )}{32 a^{3} c^{2} \left (a x +i\right ) \left (a x -i\right )}+\frac {a^{2} x^{2} \ln \left (-i a x +1\right )^{4}-12 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}+\ln \left (-i a x +1\right )^{4}+8 i a x \ln \left (-i a x +1\right )^{3}+12 \ln \left (-i a x +1\right )^{2}+48 i a x \ln \left (-i a x +1\right )+48}{128 a^{3} c^{2} \left (a x +i\right ) \left (a x -i\right )}\)	\(379\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.56 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {4 \, a x \arctan \left (a x\right )^{3} - {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{4} - 6 \, a x \arctan \left (a x\right ) - 3 \, {\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )^{2} - 3}{8 \, {\left (a^{5} c^{2} x^{2} + a^{3} c^{2}\right )}} \]

[In]

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

[Out]

-1/8*(4*a*x*arctan(a*x)^3 - (a^2*x^2 + 1)*arctan(a*x)^4 - 6*a*x*arctan(a*x) - 3*(a^2*x^2 - 1)*arctan(a*x)^2 -
3)/(a^5*c^2*x^2 + a^3*c^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.61 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=-\frac {1}{2} \, {\left (\frac {x}{a^{4} c^{2} x^{2} + a^{2} c^{2}} - \frac {\arctan \left (a x\right )}{a^{3} c^{2}}\right )} \arctan \left (a x\right )^{3} - \frac {3 \, {\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 1\right )} a \arctan \left (a x\right )^{2}}{4 \, {\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )}} - \frac {1}{8} \, {\left (\frac {{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{4} + 3 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 3\right )} a^{2}}{a^{8} c^{2} x^{2} + a^{6} c^{2}} - \frac {2 \, {\left (2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} + 3 \, a x + 3 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a \arctan \left (a x\right )}{a^{7} c^{2} x^{2} + a^{5} c^{2}}\right )} a \]

[In]

integrate(x^2*arctan(a*x)^3/(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)^3 - 3/4*((a^2*x^2 + 1)*arctan(a*x)^2 + 1)
*a*arctan(a*x)^2/(a^6*c^2*x^2 + a^4*c^2) - 1/8*(((a^2*x^2 + 1)*arctan(a*x)^4 + 3*(a^2*x^2 + 1)*arctan(a*x)^2 -
3)*a^2/(a^8*c^2*x^2 + a^6*c^2) - 2*(2*(a^2*x^2 + 1)*arctan(a*x)^3 + 3*a*x + 3*(a^2*x^2 + 1)*arctan(a*x))*a*ar
ctan(a*x)/(a^7*c^2*x^2 + a^5*c^2))*a

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \arctan (a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {3}{2\,a^2\,\left (4\,a^3\,c^2\,x^2+4\,a\,c^2\right )}+{\mathrm {atan}\left (a\,x\right )}^2\,\left (\frac {3}{8\,a^3\,c^2}-\frac {3}{4\,a^5\,c^2\,\left (\frac {1}{a^2}+x^2\right )}\right )+\frac {{\mathrm {atan}\left (a\,x\right )}^4}{8\,a^3\,c^2}+\frac {3\,x\,\mathrm {atan}\left (a\,x\right )}{4\,a^4\,c^2\,\left (\frac {1}{a^2}+x^2\right )}-\frac {x\,{\mathrm {atan}\left (a\,x\right )}^3}{2\,a^4\,c^2\,\left (\frac {1}{a^2}+x^2\right )} \]

[In]

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

[Out]

3/(2*a^2*(4*a*c^2 + 4*a^3*c^2*x^2)) + atan(a*x)^2*(3/(8*a^3*c^2) - 3/(4*a^5*c^2*(1/a^2 + x^2))) + atan(a*x)^4/
(8*a^3*c^2) + (3*x*atan(a*x))/(4*a^4*c^2*(1/a^2 + x^2)) - (x*atan(a*x)^3)/(2*a^4*c^2*(1/a^2 + x^2))