\(\int \frac {\arctan (a x)}{(c+a^2 c x^2)^3} \, dx\) [195]

Optimal result

Integrand size = 17, antiderivative size = 105 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3}{16 a c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)}{8 c^3 \left (1+a^2 x^2\right )}+\frac {3 \arctan (a x)^2}{16 a c^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {5016, 5012, 267} \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3 x \arctan (a x)}{8 c^3 \left (a^2 x^2+1\right )}+\frac {x \arctan (a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3}{16 a c^3 \left (a^2 x^2+1\right )}+\frac {1}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \arctan (a x)^2}{16 a c^3} \]

[In]

Int[ArcTan[a*x]/(c + a^2*c*x^2)^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {4+3 a^2 x^2+2 a x \left (5+3 a^2 x^2\right ) \arctan (a x)+3 \left (1+a^2 x^2\right )^2 \arctan (a x)^2}{16 a c^3 \left (1+a^2 x^2\right )^2} \]

[In]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(\frac {3 a^{4} \arctan \left (a x \right )^{2} x^{4}-4 a^{4} x^{4}+6 \arctan \left (a x \right ) x^{3} a^{3}+6 x^{2} \arctan \left (a x \right )^{2} a^{2}-5 a^{2} x^{2}+10 x \arctan \left (a x \right ) a +3 \arctan \left (a x \right )^{2}}{16 c^{3} \left (a^{2} x^{2}+1\right )^{2} a}\)	\(93\)
derivativedivides	\(\frac {\frac {a x \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}-\frac {3}{2 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{2}}{8 c^{3}}}{a}\)	\(101\)
default	\(\frac {\frac {a x \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}-\frac {3}{2 \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{2}}{8 c^{3}}}{a}\)	\(101\)
parts	\(\frac {x \arctan \left (a x \right )}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{2}}{8 a \,c^{3}}-\frac {\frac {-\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}-\frac {3}{2 \left (a^{2} x^{2}+1\right )}}{a}+\frac {3 \arctan \left (a x \right )^{2}}{2 a}}{8 c^{3}}\)	\(106\)
risch	\(-\frac {3 \ln \left (i a x +1\right )^{2}}{64 c^{3} a}+\frac {\left (3 x^{4} \ln \left (-i a x +1\right ) a^{4}+6 a^{2} x^{2} \ln \left (-i a x +1\right )-6 i a^{3} x^{3}+3 \ln \left (-i a x +1\right )-10 i a x \right ) \ln \left (i a x +1\right )}{32 c^{3} \left (a^{2} x^{2}+1\right )^{2} a}-\frac {3 a^{4} x^{4} \ln \left (-i a x +1\right )^{2}+6 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-12 i a^{3} x^{3} \ln \left (-i a x +1\right )-12 a^{2} x^{2}+3 \ln \left (-i a x +1\right )^{2}-20 i a x \ln \left (-i a x +1\right )-16}{64 \left (a x +i\right )^{2} a \,c^{3} \left (a x -i\right )^{2}}\)	\(216\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{2} x^{2} + 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 2 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \arctan \left (a x\right ) + 4}{16 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\text {Exception raised: RecursionError} \]

[In]

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

[Out]

Exception raised: RecursionError >> maximum recursion depth exceeded in comparison

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.23 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{8} \, {\left (\frac {3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{3} x^{4} + 2 \, a^{2} c^{3} x^{2} + c^{3}} + \frac {3 \, \arctan \left (a x\right )}{a c^{3}}\right )} \arctan \left (a x\right ) + \frac {{\left (3 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a}{16 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

[In]

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

[Out]

1/8*((3*a^2*x^3 + 5*x)/(a^4*c^3*x^4 + 2*a^2*c^3*x^2 + c^3) + 3*arctan(a*x)/(a*c^3))*arctan(a*x) + 1/16*(3*a^2*
x^2 - 3*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 4)*a/(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int \frac {\arctan (a x)}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3\,a^4\,x^4\,{\mathrm {atan}\left (a\,x\right )}^2+6\,a^3\,x^3\,\mathrm {atan}\left (a\,x\right )+6\,a^2\,x^2\,{\mathrm {atan}\left (a\,x\right )}^2+3\,a^2\,x^2+10\,a\,x\,\mathrm {atan}\left (a\,x\right )+3\,{\mathrm {atan}\left (a\,x\right )}^2+4}{16\,a\,c^3\,{\left (a^2\,x^2+1\right )}^2} \]

[In]

int(atan(a*x)/(c + a^2*c*x^2)^3,x)

[Out]

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