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

Optimal. Leaf size=111 \[ \frac{1}{16 a^3 c^3 \left (a^2 x^2+1\right )}-\frac{1}{16 a^3 c^3 \left (a^2 x^2+1\right )^2}+\frac{x \tan ^{-1}(a x)}{8 a^2 c^3 \left (a^2 x^2+1\right )}-\frac{x \tan ^{-1}(a x)}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^2}{16 a^3 c^3} \]

[Out]

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

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Rubi [A]  time = 0.0748615, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4934, 4892, 261} \[ \frac{1}{16 a^3 c^3 \left (a^2 x^2+1\right )}-\frac{1}{16 a^3 c^3 \left (a^2 x^2+1\right )^2}+\frac{x \tan ^{-1}(a x)}{8 a^2 c^3 \left (a^2 x^2+1\right )}-\frac{x \tan ^{-1}(a x)}{4 a^2 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^2}{16 a^3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4934

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q
+ 1))/(4*c^3*d*(q + 1)^2), x] + (-Dist[1/(2*c^2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x],
x] + Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(2*c^2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]

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

Mathematica [A]  time = 0.0526421, size = 64, normalized size = 0.58 \[ \frac{a^2 x^2+2 a x \left (a^2 x^2-1\right ) \tan ^{-1}(a x)+\left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2}{16 a^3 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 101, normalized size = 0.9 \begin{align*}{\frac{\arctan \left ( ax \right ){x}^{3}}{8\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{x\arctan \left ( ax \right ) }{8\,{a}^{2}{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{16\,{a}^{3}{c}^{3}}}-{\frac{1}{16\,{a}^{3}{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{1}{16\,{a}^{3}{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.70585, size = 174, normalized size = 1.57 \begin{align*} \frac{1}{8} \,{\left (\frac{a^{2} x^{3} - x}{a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}} + \frac{\arctan \left (a x\right )}{a^{3} c^{3}}\right )} \arctan \left (a x\right ) + \frac{{\left (a^{2} x^{2} -{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2}\right )} a}{16 \,{\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.59638, size = 176, normalized size = 1.59 \begin{align*} \frac{a^{2} x^{2} +{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 2 \,{\left (a^{3} x^{3} - a x\right )} \arctan \left (a x\right )}{16 \,{\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(a^2*x^2 + (a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 2*(a^3*x^3 - a*x)*arctan(a*x))/(a^7*c^3*x^4 + 2*a^5*
c^3*x^2 + a^3*c^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{2} \operatorname{atan}{\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**2*atan(a*x)/(a**2*c*x**2+c)**3,x)

[Out]

Integral(x**2*atan(a*x)/(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^{2} \arctan \left (a x\right )}{{\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^2*arctan(a*x)/(a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

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