3.323 \(\int \frac {2 x+x^4}{1+x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {x^3}{3}+\log \left (x^2+1\right )-x+\tan ^{-1}(x) \]

[Out]

-x+1/3*x^3+arctan(x)+ln(x^2+1)

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Rubi [A]  time = 0.03, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1593, 1802, 635, 203, 260} \[ \frac {x^3}{3}+\log \left (x^2+1\right )-x+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(2*x + x^4)/(1 + x^2),x]

[Out]

-x + x^3/3 + ArcTan[x] + Log[1 + x^2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.00 \[ \frac {x^3}{3}+\log \left (x^2+1\right )-x+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(2*x + x^4)/(1 + x^2),x]

[Out]

-x + x^3/3 + ArcTan[x] + Log[1 + x^2]

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fricas [A]  time = 0.91, size = 17, normalized size = 0.89 \[ \frac {1}{3} \, x^{3} - x + \arctan \relax (x) + \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+2*x)/(x^2+1),x, algorithm="fricas")

[Out]

1/3*x^3 - x + arctan(x) + log(x^2 + 1)

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giac [A]  time = 0.27, size = 17, normalized size = 0.89 \[ \frac {1}{3} \, x^{3} - x + \arctan \relax (x) + \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+2*x)/(x^2+1),x, algorithm="giac")

[Out]

1/3*x^3 - x + arctan(x) + log(x^2 + 1)

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maple [A]  time = 0.00, size = 18, normalized size = 0.95 \[ \frac {x^{3}}{3}-x +\arctan \relax (x )+\ln \left (x^{2}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+2*x)/(x^2+1),x)

[Out]

-x+1/3*x^3+arctan(x)+ln(x^2+1)

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maxima [A]  time = 2.03, size = 17, normalized size = 0.89 \[ \frac {1}{3} \, x^{3} - x + \arctan \relax (x) + \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+2*x)/(x^2+1),x, algorithm="maxima")

[Out]

1/3*x^3 - x + arctan(x) + log(x^2 + 1)

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mupad [B]  time = 2.10, size = 17, normalized size = 0.89 \[ \ln \left (x^2+1\right )-x+\mathrm {atan}\relax (x)+\frac {x^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + x^4)/(x^2 + 1),x)

[Out]

log(x^2 + 1) - x + atan(x) + x^3/3

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sympy [A]  time = 0.10, size = 15, normalized size = 0.79 \[ \frac {x^{3}}{3} - x + \log {\left (x^{2} + 1 \right )} + \operatorname {atan}{\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+2*x)/(x**2+1),x)

[Out]

x**3/3 - x + log(x**2 + 1) + atan(x)

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