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

Optimal. Leaf size=129 \[ -\frac{3}{8 a c^2 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)^3}{2 c^2 \left (a^2 x^2+1\right )}+\frac{3 \tan ^{-1}(a x)^2}{4 a c^2 \left (a^2 x^2+1\right )}-\frac{3 x \tan ^{-1}(a x)}{4 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^4}{8 a c^2}-\frac{3 \tan ^{-1}(a x)^2}{8 a c^2} \]

[Out]

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

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Rubi [A]  time = 0.104387, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4892, 4930, 261} \[ -\frac{3}{8 a c^2 \left (a^2 x^2+1\right )}+\frac{x \tan ^{-1}(a x)^3}{2 c^2 \left (a^2 x^2+1\right )}+\frac{3 \tan ^{-1}(a x)^2}{4 a c^2 \left (a^2 x^2+1\right )}-\frac{3 x \tan ^{-1}(a x)}{4 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^4}{8 a c^2}-\frac{3 \tan ^{-1}(a x)^2}{8 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0307542, size = 71, normalized size = 0.55 \[ \frac{\left (a^2 x^2+1\right ) \tan ^{-1}(a x)^4+\left (3-3 a^2 x^2\right ) \tan ^{-1}(a x)^2+4 a x \tan ^{-1}(a x)^3-6 a x \tan ^{-1}(a x)-3}{8 c^2 \left (a^3 x^2+a\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.144, size = 118, normalized size = 0.9 \begin{align*} -{\frac{3}{8\,a{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}-{\frac{3\,x\arctan \left ( ax \right ) }{4\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}-{\frac{3\, \left ( \arctan \left ( ax \right ) \right ) ^{2}}{8\,a{c}^{2}}}+{\frac{3\, \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,a{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{x \left ( \arctan \left ( ax \right ) \right ) ^{3}}{2\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{4}}{8\,a{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.76755, size = 288, normalized size = 2.23 \begin{align*} \frac{1}{2} \,{\left (\frac{x}{a^{2} c^{2} x^{2} + c^{2}} + \frac{\arctan \left (a x\right )}{a c^{2}}\right )} \arctan \left (a x\right )^{3} - \frac{3 \,{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 1\right )} a \arctan \left (a x\right )^{2}}{4 \,{\left (a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}} - \frac{1}{8} \,{\left (\frac{{\left ({\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{4} - 3 \,{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 3\right )} a^{2}}{a^{6} c^{2} x^{2} + a^{4} c^{2}} - \frac{2 \,{\left (2 \,{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 3 \, a x - 3 \,{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a \arctan \left (a x\right )}{a^{5} c^{2} x^{2} + a^{3} c^{2}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.85608, size = 182, normalized size = 1.41 \begin{align*} \frac{4 \, a x \arctan \left (a x\right )^{3} +{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{4} - 6 \, a x \arctan \left (a x\right ) - 3 \,{\left (a^{2} x^{2} - 1\right )} \arctan \left (a x\right )^{2} - 3}{8 \,{\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atan}^{3}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**3/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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