3.301 \(\int \frac{x \tan ^{-1}(a x)^2}{(c+a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=138 \[ \frac{3}{32 a^2 c^3 \left (a^2 x^2+1\right )}+\frac{1}{32 a^2 c^3 \left (a^2 x^2+1\right )^2}-\frac{\tan ^{-1}(a x)^2}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x \tan ^{-1}(a x)}{16 a c^3 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}(a x)^2}{32 a^2 c^3} \]

[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)

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Rubi [A]  time = 0.0964198, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4930, 4896, 4892, 261} \[ \frac{3}{32 a^2 c^3 \left (a^2 x^2+1\right )}+\frac{1}{32 a^2 c^3 \left (a^2 x^2+1\right )^2}-\frac{\tan ^{-1}(a x)^2}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x \tan ^{-1}(a x)}{16 a c^3 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac{3 \tan ^{-1}(a x)^2}{32 a^2 c^3} \]

Antiderivative was successfully verified.

[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 4930

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 4896

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 4892

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 261

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]

Rubi steps

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

Mathematica [A]  time = 0.0385074, size = 71, normalized size = 0.51 \[ \frac{3 a^2 x^2+2 a x \left (3 a^2 x^2+5\right ) \tan ^{-1}(a x)+\left (3 a^4 x^4+6 a^2 x^2-5\right ) \tan ^{-1}(a x)^2+4}{32 c^3 \left (a^3 x^2+a\right )^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.041, size = 127, normalized size = 0.9 \begin{align*}{\frac{1}{32\,{c}^{3}{a}^{2} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3}{32\,{c}^{3}{a}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{x\arctan \left ( ax \right ) }{8\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,x\arctan \left ( ax \right ) }{16\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{3\, \left ( \arctan \left ( ax \right ) \right ) ^{2}}{32\,{c}^{3}{a}^{2}}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{c}^{3}{a}^{2} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [A]  time = 1.58952, size = 220, normalized size = 1.59 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.12659, size = 192, normalized size = 1.39 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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