3.260 \(\int \frac{1+2 x^2+x^5}{-x+x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac{x^3}{3}+x+2 \log (1-x)-\log (x)+\log (x+1) \]

[Out]

x + x^3/3 + 2*Log[1 - x] - Log[x] + Log[1 + x]

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Rubi [A]  time = 0.0372468, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1593, 1802} \[ \frac{x^3}{3}+x+2 \log (1-x)-\log (x)+\log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^2 + x^5)/(-x + x^3),x]

[Out]

x + x^3/3 + 2*Log[1 - x] - Log[x] + Log[1 + x]

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{1+2 x^2+x^5}{-x+x^3} \, dx &=\int \frac{1+2 x^2+x^5}{x \left (-1+x^2\right )} \, dx\\ &=\int \left (1+\frac{2}{-1+x}-\frac{1}{x}+x^2+\frac{1}{1+x}\right ) \, dx\\ &=x+\frac{x^3}{3}+2 \log (1-x)-\log (x)+\log (1+x)\\ \end{align*}

Mathematica [A]  time = 0.0053016, size = 25, normalized size = 1. \[ \frac{x^3}{3}+x+2 \log (1-x)-\log (x)+\log (x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^2 + x^5)/(-x + x^3),x]

[Out]

x + x^3/3 + 2*Log[1 - x] - Log[x] + Log[1 + x]

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Maple [A]  time = 0.007, size = 22, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3}}+x+2\,\ln \left ( x-1 \right ) -\ln \left ( x \right ) +\ln \left ( 1+x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5+2*x^2+1)/(x^3-x),x)

[Out]

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

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Maxima [A]  time = 0.988588, size = 28, normalized size = 1.12 \begin{align*} \frac{1}{3} \, x^{3} + x + \log \left (x + 1\right ) + 2 \, \log \left (x - 1\right ) - \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+2*x^2+1)/(x^3-x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.39819, size = 68, normalized size = 2.72 \begin{align*} \frac{1}{3} \, x^{3} + x + \log \left (x + 1\right ) + 2 \, \log \left (x - 1\right ) - \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+2*x^2+1)/(x^3-x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.115942, size = 20, normalized size = 0.8 \begin{align*} \frac{x^{3}}{3} + x - \log{\left (x \right )} + 2 \log{\left (x - 1 \right )} + \log{\left (x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5+2*x**2+1)/(x**3-x),x)

[Out]

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

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Giac [A]  time = 1.14471, size = 32, normalized size = 1.28 \begin{align*} \frac{1}{3} \, x^{3} + x + \log \left ({\left | x + 1 \right |}\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) - \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+2*x^2+1)/(x^3-x),x, algorithm="giac")

[Out]

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