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

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

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Rubi [A]  time = 0.13, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.06, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^3 \tan ^{-1}(x)^2}{\left (1+x^2\right )^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 0.94, size = 51, normalized size = 0.65 \[ \frac {{\left (5 \, x^{4} - 6 \, x^{2} - 3\right )} \arctan \relax (x)^{2} + 5 \, x^{2} + 2 \, {\left (5 \, x^{3} + 3 \, x\right )} \arctan \relax (x) + 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arctan \relax (x)^{2}}{{\left (x^{2} + 1\right )}^{3}}\,{d x} \]

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)

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maple [A]  time = 0.42, size = 78, normalized size = 0.99

method	result	size
default	\(\frac {\arctan \relax (x )^{2}}{4 \left (x^{2}+1\right )^{2}}-\frac {\arctan \relax (x )^{2}}{2 \left (x^{2}+1\right )}+\frac {5 x^{3} \arctan \relax (x )}{16 \left (x^{2}+1\right )^{2}}+\frac {3 x \arctan \relax (x )}{16 \left (x^{2}+1\right )^{2}}+\frac {5 \arctan \relax (x )^{2}}{32}-\frac {1}{32 \left (x^{2}+1\right )^{2}}+\frac {5}{32 \left (x^{2}+1\right )}\)	\(78\)
risch	\(-\frac {\left (5 x^{4}-6 x^{2}-3\right ) \ln \left (i x +1\right )^{2}}{128 \left (x^{2}+1\right )^{2}}+\frac {\left (-6 x^{2} \ln \left (-i x +1\right )-3 \ln \left (-i x +1\right )+5 x^{4} \ln \left (-i x +1\right )-10 i x^{3}-6 i x \right ) \ln \left (i x +1\right )}{64 \left (x +i\right )^{2} \left (x -i\right )^{2}}-\frac {5 x^{4} \ln \left (-i x +1\right )^{2}-6 x^{2} \ln \left (-i x +1\right )^{2}-3 \ln \left (-i x +1\right )^{2}-20 i x^{3} \ln \left (-i x +1\right )-12 i \ln \left (-i x +1\right ) x -20 x^{2}-16}{128 \left (x +i\right )^{2} \left (x -i\right )^{2}}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 1.04, size = 94, normalized size = 1.19 \[ \frac {1}{16} \, {\left (\frac {5 \, x^{3} + 3 \, x}{x^{4} + 2 \, x^{2} + 1} + 5 \, \arctan \relax (x)\right )} \arctan \relax (x) - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \relax (x)^{2}}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {5 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \relax (x)^{2} - 5 \, x^{2} - 4}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

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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mupad [B]  time = 0.36, size = 56, normalized size = 0.71 \[ -\frac {-5\,x^4\,{\mathrm {atan}\relax (x)}^2+4\,x^4-10\,x^3\,\mathrm {atan}\relax (x)+6\,x^2\,{\mathrm {atan}\relax (x)}^2+3\,x^2-6\,x\,\mathrm {atan}\relax (x)+3\,{\mathrm {atan}\relax (x)}^2}{32\,{\left (x^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atan(x)^2)/(x^2 + 1)^3,x)

[Out]

-(3*atan(x)^2 - 10*x^3*atan(x) + 6*x^2*atan(x)^2 - 5*x^4*atan(x)^2 - 6*x*atan(x) + 3*x^2 + 4*x^4)/(32*(x^2 + 1
)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \operatorname {atan}^{2}{\relax (x )}}{\left (x^{2} + 1\right )^{3}}\, dx \]

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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