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

Optimal. Leaf size=140 \[ -\frac{x^4}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac{3}{32 a^4 c^3 \left (a^2 x^2+1\right )}+\frac{x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{x^3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x \tan ^{-1}(a x)}{16 a^3 c^3 \left (a^2 x^2+1\right )}-\frac{3 \tan ^{-1}(a x)^2}{32 a^4 c^3} \]

[Out]

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

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Rubi [A]  time = 0.185193, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4944, 4938, 4934, 4884} \[ -\frac{x^4}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac{3}{32 a^4 c^3 \left (a^2 x^2+1\right )}+\frac{x^4 \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{x^3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac{3 x \tan ^{-1}(a x)}{16 a^3 c^3 \left (a^2 x^2+1\right )}-\frac{3 \tan ^{-1}(a x)^2}{32 a^4 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 4938

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b*(f*x
)^m*(d + e*x^2)^(q + 1))/(c*d*m^2), x] + (Dist[(f^2*(m - 1))/(c^2*d*m), Int[(f*x)^(m - 2)*(d + e*x^2)^(q + 1)*
(a + b*ArcTan[c*x]), x], x] - Simp[(f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(c^2*d*m), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -1]

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 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0856415, size = 74, normalized size = 0.53 \[ \frac{5 a^2 x^2+2 a x \left (5 a^2 x^2+3\right ) \tan ^{-1}(a x)+\left (5 a^4 x^4-6 a^2 x^2-3\right ) \tan ^{-1}(a x)^2+4}{32 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 154, normalized size = 1.1 \begin{align*}{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{c}^{3}{a}^{4} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,{c}^{3}{a}^{4} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{5\,{x}^{3}\arctan \left ( ax \right ) }{16\,a{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{3\,\arctan \left ( ax \right ) x}{16\,{c}^{3}{a}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{5\, \left ( \arctan \left ( ax \right ) \right ) ^{2}}{32\,{c}^{3}{a}^{4}}}-{\frac{1}{32\,{c}^{3}{a}^{4} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{5}{32\,{c}^{3}{a}^{4} \left ({a}^{2}{x}^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.58772, size = 250, normalized size = 1.79 \begin{align*} \frac{1}{16} \, a{\left (\frac{5 \, a^{2} x^{3} + 3 \, x}{a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}} + \frac{5 \, \arctan \left (a x\right )}{a^{5} c^{3}}\right )} \arctan \left (a x\right ) + \frac{{\left (5 \, a^{2} x^{2} - 5 \,{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a^{2}}{32 \,{\left (a^{10} c^{3} x^{4} + 2 \, a^{8} c^{3} x^{2} + a^{6} c^{3}\right )}} - \frac{{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2}}{4 \,{\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^3*arctan(a*x)^2/(a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

1/16*a*((5*a^2*x^3 + 3*x)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3) + 5*arctan(a*x)/(a^5*c^3))*arctan(a*x) + 1/3
2*(5*a^2*x^2 - 5*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 4)*a^2/(a^10*c^3*x^4 + 2*a^8*c^3*x^2 + a^6*c^3) - 1
/4*(2*a^2*x^2 + 1)*arctan(a*x)^2/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

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Fricas [A]  time = 2.18483, size = 192, normalized size = 1.37 \begin{align*} \frac{5 \, a^{2} x^{2} +{\left (5 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \arctan \left (a x\right )^{2} + 2 \,{\left (5 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right ) + 4}{32 \,{\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^3*arctan(a*x)^2/(a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

1/32*(5*a^2*x^2 + (5*a^4*x^4 - 6*a^2*x^2 - 3)*arctan(a*x)^2 + 2*(5*a^3*x^3 + 3*a*x)*arctan(a*x) + 4)/(a^8*c^3*
x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

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

[Out]

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

[Out]

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