3.354 \(\int \frac{4+x+3 x^2}{x+x^3} \, dx\)

Optimal. Leaf size=17 \[ -\frac{1}{2} \log \left (x^2+1\right )+4 \log (x)+\tan ^{-1}(x) \]

[Out]

ArcTan[x] + 4*Log[x] - Log[1 + x^2]/2

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Rubi [A]  time = 0.0370974, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {1593, 1802, 635, 203, 260} \[ -\frac{1}{2} \log \left (x^2+1\right )+4 \log (x)+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(4 + x + 3*x^2)/(x + x^3),x]

[Out]

ArcTan[x] + 4*Log[x] - Log[1 + x^2]/2

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0046738, size = 17, normalized size = 1. \[ -\frac{1}{2} \log \left (x^2+1\right )+4 \log (x)+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + x + 3*x^2)/(x + x^3),x]

[Out]

ArcTan[x] + 4*Log[x] - Log[1 + x^2]/2

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Maple [A]  time = 0.004, size = 16, normalized size = 0.9 \begin{align*} \arctan \left ( x \right ) +4\,\ln \left ( x \right ) -{\frac{\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+x+4)/(x^3+x),x)

[Out]

arctan(x)+4*ln(x)-1/2*ln(x^2+1)

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Maxima [A]  time = 1.4965, size = 20, normalized size = 1.18 \begin{align*} \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x^{2} + 1\right ) + 4 \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+x+4)/(x^3+x),x, algorithm="maxima")

[Out]

arctan(x) - 1/2*log(x^2 + 1) + 4*log(x)

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Fricas [A]  time = 1.78526, size = 55, normalized size = 3.24 \begin{align*} \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x^{2} + 1\right ) + 4 \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+x+4)/(x^3+x),x, algorithm="fricas")

[Out]

arctan(x) - 1/2*log(x^2 + 1) + 4*log(x)

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Sympy [A]  time = 0.115043, size = 15, normalized size = 0.88 \begin{align*} 4 \log{\left (x \right )} - \frac{\log{\left (x^{2} + 1 \right )}}{2} + \operatorname{atan}{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+x+4)/(x**3+x),x)

[Out]

4*log(x) - log(x**2 + 1)/2 + atan(x)

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Giac [A]  time = 1.12893, size = 22, normalized size = 1.29 \begin{align*} \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x^{2} + 1\right ) + 4 \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+x+4)/(x^3+x),x, algorithm="giac")

[Out]

arctan(x) - 1/2*log(x^2 + 1) + 4*log(abs(x))