\(\int \frac {\arctan (a x)^2}{(c+a^2 c x^2)^3} \, dx\) [302]

Optimal result

Integrand size = 19, antiderivative size = 169 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac {15 \arctan (a x)}{64 a c^3}+\frac {\arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{8 a c^3} \]

[Out]

-1/32*x/c^3/(a^2*x^2+1)^2-15/64*x/c^3/(a^2*x^2+1)-15/64*arctan(a*x)/a/c^3+1/8*arctan(a*x)/a/c^3/(a^2*x^2+1)^2+
3/8*arctan(a*x)/a/c^3/(a^2*x^2+1)+1/4*x*arctan(a*x)^2/c^3/(a^2*x^2+1)^2+3/8*x*arctan(a*x)^2/c^3/(a^2*x^2+1)+1/
8*arctan(a*x)^3/a/c^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5020, 5012, 5050, 205, 211} \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {3 x \arctan (a x)^2}{8 c^3 \left (a^2 x^2+1\right )}+\frac {x \arctan (a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \arctan (a x)}{8 a c^3 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}-\frac {15 x}{64 c^3 \left (a^2 x^2+1\right )}-\frac {x}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac {\arctan (a x)^3}{8 a c^3}-\frac {15 \arctan (a x)}{64 a c^3} \]

[In]

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

[Out]

-1/32*x/(c^3*(1 + a^2*x^2)^2) - (15*x)/(64*c^3*(1 + a^2*x^2)) - (15*ArcTan[a*x])/(64*a*c^3) + ArcTan[a*x]/(8*a
*c^3*(1 + a^2*x^2)^2) + (3*ArcTan[a*x])/(8*a*c^3*(1 + a^2*x^2)) + (x*ArcTan[a*x]^2)/(4*c^3*(1 + a^2*x^2)^2) +
(3*x*ArcTan[a*x]^2)/(8*c^3*(1 + a^2*x^2)) + ArcTan[a*x]^3/(8*a*c^3)

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 5020

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[b*p*(d + e*x^2)^(q +
1)*((a + b*ArcTan[c*x])^(p - 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])^p, x], x] - Dist[b^2*p*((p - 1)/(4*(q + 1)^2)), Int[(d + e*x^2)^q*(a + b*ArcTan[c*x])^(
p - 2), x], x] - Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(2*d*(q + 1))), x]) /; FreeQ[{a, b, c, d, e
}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && GtQ[p, 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)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {x \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {1}{8} \int \frac {1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac {3 \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c} \\ & = -\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {\arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {x \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{8 a c^3}-\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{32 c}-\frac {(3 a) \int \frac {x \arctan (a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c} \\ & = -\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {3 x}{64 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{8 a c^3}-\frac {3 \int \frac {1}{c+a^2 c x^2} \, dx}{64 c^2}-\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{8 c} \\ & = -\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac {3 \arctan (a x)}{64 a c^3}+\frac {\arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{8 a c^3}-\frac {3 \int \frac {1}{c+a^2 c x^2} \, dx}{16 c^2} \\ & = -\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac {15 \arctan (a x)}{64 a c^3}+\frac {\arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \arctan (a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac {x \arctan (a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \arctan (a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^3}{8 a c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.58 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {-a x \left (17+15 a^2 x^2\right )+\left (17-6 a^2 x^2-15 a^4 x^4\right ) \arctan (a x)+8 a x \left (5+3 a^2 x^2\right ) \arctan (a x)^2+8 \left (1+a^2 x^2\right )^2 \arctan (a x)^3}{64 a c^3 \left (1+a^2 x^2\right )^2} \]

[In]

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

[Out]

(-(a*x*(17 + 15*a^2*x^2)) + (17 - 6*a^2*x^2 - 15*a^4*x^4)*ArcTan[a*x] + 8*a*x*(5 + 3*a^2*x^2)*ArcTan[a*x]^2 +
8*(1 + a^2*x^2)^2*ArcTan[a*x]^3)/(64*a*c^3*(1 + a^2*x^2)^2)

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {8 a^{4} \arctan \left (a x \right )^{3} x^{4}-15 \arctan \left (a x \right ) a^{4} x^{4}+24 a^{3} \arctan \left (a x \right )^{2} x^{3}+16 \arctan \left (a x \right )^{3} x^{2} a^{2}-15 a^{3} x^{3}-6 a^{2} \arctan \left (a x \right ) x^{2}+40 a \arctan \left (a x \right )^{2} x +8 \arctan \left (a x \right )^{3}-17 a x +17 \arctan \left (a x \right )}{64 c^{3} \left (a^{2} x^{2}+1\right )^{2} a}\)	\(123\)
derivativedivides	\(\frac {\frac {a x \arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x \arctan \left (a x \right )^{2}}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{3}}{8 c^{3}}-\frac {-\frac {3 \arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {15}{8} a^{3} x^{3}+\frac {17}{8} a x}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {15 \arctan \left (a x \right )}{16}+\arctan \left (a x \right )^{3}}{4 c^{3}}}{a}\)	\(143\)
default	\(\frac {\frac {a x \arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 a x \arctan \left (a x \right )^{2}}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{3}}{8 c^{3}}-\frac {-\frac {3 \arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {15}{8} a^{3} x^{3}+\frac {17}{8} a x}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {15 \arctan \left (a x \right )}{16}+\arctan \left (a x \right )^{3}}{4 c^{3}}}{a}\)	\(143\)
parts	\(\frac {x \arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (a x \right )^{2}}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )^{3}}{8 a \,c^{3}}-\frac {\frac {-\frac {3 \arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {\frac {15}{8} a^{3} x^{3}+\frac {17}{8} a x}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {15 \arctan \left (a x \right )}{16}}{a}+\frac {\arctan \left (a x \right )^{3}}{a}}{4 c^{3}}\)	\(149\)
risch	\(\frac {i \ln \left (i a x +1\right )^{3}}{64 c^{3} a}-\frac {i \left (3 x^{4} \ln \left (-i a x +1\right ) a^{4}+6 a^{2} x^{2} \ln \left (-i a x +1\right )-6 i a^{3} x^{3}+3 \ln \left (-i a x +1\right )-10 i a x \right ) \ln \left (i a x +1\right )^{2}}{64 c^{3} \left (a^{2} x^{2}+1\right )^{2} a}+\frac {i \left (3 a^{4} x^{4} \ln \left (-i a x +1\right )^{2}+6 a^{2} x^{2} \ln \left (-i a x +1\right )^{2}-12 i x^{3} \ln \left (-i a x +1\right ) a^{3}-12 a^{2} x^{2}+3 \ln \left (-i a x +1\right )^{2}-20 i a x \ln \left (-i a x +1\right )-16\right ) \ln \left (i a x +1\right )}{64 \left (a x +i\right )^{2} a \,c^{3} \left (a x -i\right )^{2}}-\frac {i \left (2 a^{4} x^{4} \ln \left (-i a x +1\right )^{3}+15 \ln \left (i a x -1\right ) a^{4} x^{4}-15 \ln \left (-i a x -1\right ) a^{4} x^{4}+4 a^{2} x^{2} \ln \left (-i a x +1\right )^{3}-12 i a^{3} x^{3} \ln \left (-i a x +1\right )^{2}+30 \ln \left (i a x -1\right ) a^{2} x^{2}-30 \ln \left (-i a x -1\right ) a^{2} x^{2}-24 a^{2} x^{2} \ln \left (-i a x +1\right )-30 i a^{3} x^{3}+2 \ln \left (-i a x +1\right )^{3}-20 i a x \ln \left (-i a x +1\right )^{2}+15 \ln \left (i a x -1\right )-15 \ln \left (-i a x -1\right )-32 \ln \left (-i a x +1\right )-34 i a x \right )}{128 \left (a x +i\right )^{2} a \,c^{3} \left (a x -i\right )^{2}}\)	\(461\)

[In]

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

[Out]

1/64*(8*a^4*arctan(a*x)^3*x^4-15*arctan(a*x)*a^4*x^4+24*a^3*arctan(a*x)^2*x^3+16*arctan(a*x)^3*x^2*a^2-15*a^3*
x^3-6*a^2*arctan(a*x)*x^2+40*a*arctan(a*x)^2*x+8*arctan(a*x)^3-17*a*x+17*arctan(a*x))/c^3/(a^2*x^2+1)^2/a

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.67 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {15 \, a^{3} x^{3} - 8 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 8 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \arctan \left (a x\right )^{2} + 17 \, a x + {\left (15 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 17\right )} \arctan \left (a x\right )}{64 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]

[In]

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

[Out]

-1/64*(15*a^3*x^3 - 8*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^3 - 8*(3*a^3*x^3 + 5*a*x)*arctan(a*x)^2 + 17*a*x +
(15*a^4*x^4 + 6*a^2*x^2 - 17)*arctan(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^2 + a*c^3)

Sympy [F]

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

[In]

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

[Out]

Integral(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 = 0.34 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.37 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {1}{8} \, {\left (\frac {3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{3} x^{4} + 2 \, a^{2} c^{3} x^{2} + c^{3}} + \frac {3 \, \arctan \left (a x\right )}{a c^{3}}\right )} \arctan \left (a x\right )^{2} - \frac {{\left (15 \, a^{3} x^{3} - 8 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} + 17 \, a x + 15 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a^{2}}{64 \, {\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} + \frac {{\left (3 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a \arctan \left (a x\right )}{8 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]

[In]

integrate(arctan(a*x)^2/(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)^2 - 1/64*(15*a
^3*x^3 - 8*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^3 + 17*a*x + 15*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x))*a^2/(a
^7*c^3*x^4 + 2*a^5*c^3*x^2 + a^3*c^3) + 1/8*(3*a^2*x^2 - 3*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 4)*a*arct
an(a*x)/(a^6*c^3*x^4 + 2*a^4*c^3*x^2 + a^2*c^3)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.93 \[ \int \frac {\arctan (a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (a\,x\right )\,\left (\frac {1}{2\,a^3\,c^3}+\frac {3\,x^2}{8\,a\,c^3}\right )}{\frac {1}{a^2}+2\,x^2+a^2\,x^4}-\frac {15\,\mathrm {atan}\left (a\,x\right )}{64\,a\,c^3}-\frac {\frac {15\,a^2\,x^3}{8}+\frac {17\,x}{8}}{8\,a^4\,c^3\,x^4+16\,a^2\,c^3\,x^2+8\,c^3}+\frac {{\mathrm {atan}\left (a\,x\right )}^2\,\left (\frac {3\,x^3}{8\,c^3}+\frac {5\,x}{8\,a^2\,c^3}\right )}{\frac {1}{a^2}+2\,x^2+a^2\,x^4}+\frac {{\mathrm {atan}\left (a\,x\right )}^3}{8\,a\,c^3} \]

[In]

int(atan(a*x)^2/(c + a^2*c*x^2)^3,x)

[Out]

(atan(a*x)*(1/(2*a^3*c^3) + (3*x^2)/(8*a*c^3)))/(1/a^2 + 2*x^2 + a^2*x^4) - (15*atan(a*x))/(64*a*c^3) - ((17*x
)/8 + (15*a^2*x^3)/8)/(8*c^3 + 16*a^2*c^3*x^2 + 8*a^4*c^3*x^4) + (atan(a*x)^2*((3*x^3)/(8*c^3) + (5*x)/(8*a^2*
c^3)))/(1/a^2 + 2*x^2 + a^2*x^4) + atan(a*x)^3/(8*a*c^3)