\(\int \frac {x \arctan (a x)^2}{(c+a^2 c x^2)^3} \, dx\) [301]

Optimal result

Integrand size = 20, antiderivative size = 138 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{32 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {3}{32 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)}{16 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^2}{32 a^2 c^3}-\frac {\arctan (a x)^2}{4 a^2 c^3 \left (1+a^2 x^2\right )^2} \]

[Out]

1/32/a^2/c^3/(a^2*x^2+1)^2+3/32/a^2/c^3/(a^2*x^2+1)+1/8*x*arctan(a*x)/a/c^3/(a^2*x^2+1)^2+3/16*x*arctan(a*x)/a
/c^3/(a^2*x^2+1)+3/32*arctan(a*x)^2/a^2/c^3-1/4*arctan(a*x)^2/a^2/c^3/(a^2*x^2+1)^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5050, 5016, 5012, 267} \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {\arctan (a x)^2}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x \arctan (a x)}{16 a c^3 \left (a^2 x^2+1\right )}+\frac {x \arctan (a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \arctan (a x)^2}{32 a^2 c^3}+\frac {3}{32 a^2 c^3 \left (a^2 x^2+1\right )}+\frac {1}{32 a^2 c^3 \left (a^2 x^2+1\right )^2} \]

[In]

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

[Out]

1/(32*a^2*c^3*(1 + a^2*x^2)^2) + 3/(32*a^2*c^3*(1 + a^2*x^2)) + (x*ArcTan[a*x])/(8*a*c^3*(1 + a^2*x^2)^2) + (3
*x*ArcTan[a*x])/(16*a*c^3*(1 + a^2*x^2)) + (3*ArcTan[a*x]^2)/(32*a^2*c^3) - ArcTan[a*x]^2/(4*a^2*c^3*(1 + a^2*
x^2)^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 5016

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[b*((d + e*x^2)^(q + 1)/(4
*c*d*(q + 1)^2)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x] - Si
mp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d
] && LtQ[q, -1] && NeQ[q, -3/2]

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}{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 )^3} \, dx}{2 a} \\ & = \frac {1}{32 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {x \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}-\frac {\arctan (a x)^2}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{8 a c} \\ & = \frac {1}{32 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {x \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)}{16 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^2}{32 a^2 c^3}-\frac {\arctan (a x)^2}{4 a^2 c^3 \left (1+a^2 x^2\right )^2}-\frac {3 \int \frac {x}{\left (c+a^2 c x^2\right )^2} \, dx}{16 c} \\ & = \frac {1}{32 a^2 c^3 \left (1+a^2 x^2\right )^2}+\frac {3}{32 a^2 c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)}{16 a c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^2}{32 a^2 c^3}-\frac {\arctan (a x)^2}{4 a^2 c^3 \left (1+a^2 x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.51 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {4+3 a^2 x^2+2 a x \left (5+3 a^2 x^2\right ) \arctan (a x)+\left (-5+6 a^2 x^2+3 a^4 x^4\right ) \arctan (a x)^2}{32 c^3 \left (a+a^3 x^2\right )^2} \]

[In]

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

[Out]

(4 + 3*a^2*x^2 + 2*a*x*(5 + 3*a^2*x^2)*ArcTan[a*x] + (-5 + 6*a^2*x^2 + 3*a^4*x^4)*ArcTan[a*x]^2)/(32*c^3*(a +
a^3*x^2)^2)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {3 a^{4} \arctan \left (a x \right )^{2} x^{4}-4 a^{4} x^{4}+6 \arctan \left (a x \right ) x^{3} a^{3}+6 x^{2} \arctan \left (a x \right )^{2} a^{2}-5 a^{2} x^{2}+10 x \arctan \left (a x \right ) a -5 \arctan \left (a x \right )^{2}}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2} a^{2}}\)	\(93\)
derivativedivides	\(\frac {-\frac {\arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {\arctan \left (a x \right ) a x}{4 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \arctan \left (a x \right ) a x}{8 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{16}+\frac {1}{16 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3}{16 \left (a^{2} x^{2}+1\right )}}{2 c^{3}}}{a^{2}}\)	\(106\)
default	\(\frac {-\frac {\arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {\arctan \left (a x \right ) a x}{4 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \arctan \left (a x \right ) a x}{8 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{16}+\frac {1}{16 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3}{16 \left (a^{2} x^{2}+1\right )}}{2 c^{3}}}{a^{2}}\)	\(106\)
parts	\(-\frac {\arctan \left (a x \right )^{2}}{4 a^{2} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {\arctan \left (a x \right ) a x}{4 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \arctan \left (a x \right ) a x}{8 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{16}+\frac {1}{16 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3}{16 \left (a^{2} x^{2}+1\right )}}{2 c^{3} a^{2}}\)	\(108\)
risch	\(-\frac {\left (3 a^{4} x^{4}+6 a^{2} x^{2}-5\right ) \ln \left (i a x +1\right )^{2}}{128 a^{2} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\left (-5 \ln \left (-i a x +1\right )+3 x^{4} \ln \left (-i a x +1\right ) a^{4}+6 a^{2} x^{2} \ln \left (-i a x +1\right )-6 i a^{3} x^{3}-10 i a x \right ) \ln \left (i a x +1\right )}{64 \left (a x +i\right )^{2} c^{3} \left (a x -i\right )^{2} a^{2}}-\frac {3 a^{4} x^{4} \ln \left (-i a x +1\right )^{2}+6 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-5 \ln \left (-i a x +1\right )^{2}-12 i x^{3} \ln \left (-i a x +1\right ) a^{3}-20 i a x \ln \left (-i a x +1\right )-12 a^{2} x^{2}-16}{128 \left (a x +i\right )^{2} c^{3} \left (a x -i\right )^{2} a^{2}}\)	\(250\)

[In]

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

[Out]

1/32*(3*a^4*arctan(a*x)^2*x^4-4*a^4*x^4+6*arctan(a*x)*x^3*a^3+6*x^2*arctan(a*x)^2*a^2-5*a^2*x^2+10*x*arctan(a*
x)*a-5*arctan(a*x)^2)/c^3/(a^2*x^2+1)^2/a^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.63 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{2} x^{2} + {\left (3 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 5\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \arctan \left (a x\right ) + 4}{32 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

[In]

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

[Out]

1/32*(3*a^2*x^2 + (3*a^4*x^4 + 6*a^2*x^2 - 5)*arctan(a*x)^2 + 2*(3*a^3*x^3 + 5*a*x)*arctan(a*x) + 4)/(a^6*c^3*
x^4 + 2*a^4*c^3*x^2 + a^2*c^3)

Sympy [F]

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

[In]

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

[Out]

Integral(x*atan(a*x)**2/(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 = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.18 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {{\left (\frac {3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}} + \frac {3 \, \arctan \left (a x\right )}{a c^{2}}\right )} \arctan \left (a x\right )}{16 \, a c} + \frac {3 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4}{32 \, {\left (a^{6} c^{2} x^{4} + 2 \, a^{4} c^{2} x^{2} + a^{2} c^{2}\right )} c} - \frac {\arctan \left (a x\right )^{2}}{4 \, {\left (a^{2} c x^{2} + c\right )}^{2} a^{2} c} \]

[In]

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

[Out]

1/16*((3*a^2*x^3 + 5*x)/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2) + 3*arctan(a*x)/(a*c^2))*arctan(a*x)/(a*c) + 1/32*
(3*a^2*x^2 - 3*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 4)/((a^6*c^2*x^4 + 2*a^4*c^2*x^2 + a^2*c^2)*c) - 1/4*
arctan(a*x)^2/((a^2*c*x^2 + c)^2*a^2*c)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62 \[ \int \frac {x \arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+6\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )+6\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+3\,a^2\,x^2+10\,a\,x\,\mathrm {atan}\left (a\,x\right )-5\,{\mathrm {atan}\left (a\,x\right )}^2+4}{32\,a^2\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

[In]

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

[Out]

(3*a^2*x^2 - 5*atan(a*x)^2 + 6*a^3*x^3*atan(a*x) + 10*a*x*atan(a*x) + 6*a^2*x^2*atan(a*x)^2 + 3*a^4*x^4*atan(a
*x)^2 + 4)/(32*a^2*c^3*(a^2*x^2 + 1)^2)