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

Optimal. Leaf size=79 \[ \frac{5}{32 \left (x^2+1\right )}-\frac{1}{32 \left (x^2+1\right )^2}+\frac{x^4 \tan ^{-1}(x)^2}{4 \left (x^2+1\right )^2}+\frac{x^3 \tan ^{-1}(x)}{8 \left (x^2+1\right )^2}+\frac{3 x \tan ^{-1}(x)}{16 \left (x^2+1\right )}-\frac{3}{32} \tan ^{-1}(x)^2 \]

[Out]

-1/(32*(1 + x^2)^2) + 5/(32*(1 + x^2)) + (x^3*ArcTan[x])/(8*(1 + x^2)^2) + (3*x*ArcTan[x])/(16*(1 + x^2)) - (3
*ArcTan[x]^2)/32 + (x^4*ArcTan[x]^2)/(4*(1 + x^2)^2)

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Rubi [A]  time = 0.133031, antiderivative size = 82, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4944, 4938, 4934, 4884} \[ -\frac{x^4}{32 \left (x^2+1\right )^2}+\frac{3}{32 \left (x^2+1\right )}+\frac{x^4 \tan ^{-1}(x)^2}{4 \left (x^2+1\right )^2}+\frac{x^3 \tan ^{-1}(x)}{8 \left (x^2+1\right )^2}+\frac{3 x \tan ^{-1}(x)}{16 \left (x^2+1\right )}-\frac{3}{32} \tan ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[x]^2)/(1 + x^2)^3,x]

[Out]

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

Mathematica [A]  time = 0.0561438, size = 47, normalized size = 0.59 \[ \frac{5 x^2+2 \left (5 x^2+3\right ) x \tan ^{-1}(x)+\left (5 x^4-6 x^2-3\right ) \tan ^{-1}(x)^2+4}{32 \left (x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcTan[x]^2)/(1 + x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.025, size = 78, normalized size = 1. \begin{align*}{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{4\, \left ({x}^{2}+1 \right ) ^{2}}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{2\,{x}^{2}+2}}+{\frac{5\,{x}^{3}\arctan \left ( x \right ) }{16\, \left ({x}^{2}+1 \right ) ^{2}}}+{\frac{3\,x\arctan \left ( x \right ) }{16\, \left ({x}^{2}+1 \right ) ^{2}}}+{\frac{5\, \left ( \arctan \left ( x \right ) \right ) ^{2}}{32}}-{\frac{1}{32\, \left ({x}^{2}+1 \right ) ^{2}}}+{\frac{5}{32\,{x}^{2}+32}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(x)^2/(x^2+1)^3,x)

[Out]

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

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Maxima [A]  time = 1.48649, size = 127, normalized size = 1.61 \begin{align*} \frac{1}{16} \,{\left (\frac{5 \, x^{3} + 3 \, x}{x^{4} + 2 \, x^{2} + 1} + 5 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac{{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2}}{4 \,{\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac{5 \,{\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2} - 5 \, x^{2} - 4}{32 \,{\left (x^{4} + 2 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)^2/(x^2+1)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.53098, size = 132, normalized size = 1.67 \begin{align*} \frac{{\left (5 \, x^{4} - 6 \, x^{2} - 3\right )} \arctan \left (x\right )^{2} + 5 \, x^{2} + 2 \,{\left (5 \, x^{3} + 3 \, x\right )} \arctan \left (x\right ) + 4}{32 \,{\left (x^{4} + 2 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)^2/(x^2+1)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(x)**2/(x**2+1)**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \arctan \left (x\right )^{2}}{{\left (x^{2} + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)^2/(x^2+1)^3,x, algorithm="giac")

[Out]

integrate(x^3*arctan(x)^2/(x^2 + 1)^3, x)