3.299 \(\int \frac{2+x^2}{(-5+x) (-3+x) (4+x)} \, dx\)

Optimal. Leaf size=29 \[ -\frac{11}{14} \log (3-x)+\frac{3}{2} \log (5-x)+\frac{2}{7} \log (x+4) \]

[Out]

(-11*Log[3 - x])/14 + (3*Log[5 - x])/2 + (2*Log[4 + x])/7

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Rubi [A]  time = 0.0528289, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {1612} \[ -\frac{11}{14} \log (3-x)+\frac{3}{2} \log (5-x)+\frac{2}{7} \log (x+4) \]

Antiderivative was successfully verified.

[In]

Int[(2 + x^2)/((-5 + x)*(-3 + x)*(4 + x)),x]

[Out]

(-11*Log[3 - x])/14 + (3*Log[5 - x])/2 + (2*Log[4 + x])/7

Rule 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

Rubi steps

\begin{align*} \int \frac{2+x^2}{(-5+x) (-3+x) (4+x)} \, dx &=\int \left (\frac{3}{2 (-5+x)}-\frac{11}{14 (-3+x)}+\frac{2}{7 (4+x)}\right ) \, dx\\ &=-\frac{11}{14} \log (3-x)+\frac{3}{2} \log (5-x)+\frac{2}{7} \log (4+x)\\ \end{align*}

Mathematica [A]  time = 0.0074461, size = 29, normalized size = 1. \[ -\frac{11}{14} \log (3-x)+\frac{3}{2} \log (5-x)+\frac{2}{7} \log (x+4) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x^2)/((-5 + x)*(-3 + x)*(4 + x)),x]

[Out]

(-11*Log[3 - x])/14 + (3*Log[5 - x])/2 + (2*Log[4 + x])/7

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Maple [A]  time = 0.009, size = 20, normalized size = 0.7 \begin{align*} -{\frac{11\,\ln \left ( -3+x \right ) }{14}}+{\frac{2\,\ln \left ( 4+x \right ) }{7}}+{\frac{3\,\ln \left ( -5+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2)/(-5+x)/(-3+x)/(4+x),x)

[Out]

-11/14*ln(-3+x)+2/7*ln(4+x)+3/2*ln(-5+x)

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Maxima [A]  time = 0.971552, size = 26, normalized size = 0.9 \begin{align*} \frac{2}{7} \, \log \left (x + 4\right ) - \frac{11}{14} \, \log \left (x - 3\right ) + \frac{3}{2} \, \log \left (x - 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)/(-5+x)/(-3+x)/(4+x),x, algorithm="maxima")

[Out]

2/7*log(x + 4) - 11/14*log(x - 3) + 3/2*log(x - 5)

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Fricas [A]  time = 1.5611, size = 70, normalized size = 2.41 \begin{align*} \frac{2}{7} \, \log \left (x + 4\right ) - \frac{11}{14} \, \log \left (x - 3\right ) + \frac{3}{2} \, \log \left (x - 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)/(-5+x)/(-3+x)/(4+x),x, algorithm="fricas")

[Out]

2/7*log(x + 4) - 11/14*log(x - 3) + 3/2*log(x - 5)

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Sympy [A]  time = 0.130081, size = 24, normalized size = 0.83 \begin{align*} \frac{3 \log{\left (x - 5 \right )}}{2} - \frac{11 \log{\left (x - 3 \right )}}{14} + \frac{2 \log{\left (x + 4 \right )}}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+2)/(-5+x)/(-3+x)/(4+x),x)

[Out]

3*log(x - 5)/2 - 11*log(x - 3)/14 + 2*log(x + 4)/7

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Giac [A]  time = 1.15394, size = 30, normalized size = 1.03 \begin{align*} \frac{2}{7} \, \log \left ({\left | x + 4 \right |}\right ) - \frac{11}{14} \, \log \left ({\left | x - 3 \right |}\right ) + \frac{3}{2} \, \log \left ({\left | x - 5 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)/(-5+x)/(-3+x)/(4+x),x, algorithm="giac")

[Out]

2/7*log(abs(x + 4)) - 11/14*log(abs(x - 3)) + 3/2*log(abs(x - 5))