### 3.189 $$\int \frac{x^2}{(-3+x) (2+x)^2} \, dx$$

Optimal. Leaf size=28 $\frac{4}{5 (x+2)}+\frac{9}{25} \log (3-x)+\frac{16}{25} \log (x+2)$

[Out]

4/(5*(2 + x)) + (9*Log[3 - x])/25 + (16*Log[2 + x])/25

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Rubi [A]  time = 0.0100071, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.071, Rules used = {88} $\frac{4}{5 (x+2)}+\frac{9}{25} \log (3-x)+\frac{16}{25} \log (x+2)$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2/((-3 + x)*(2 + x)^2),x]

[Out]

4/(5*(2 + x)) + (9*Log[3 - x])/25 + (16*Log[2 + x])/25

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{x^2}{(-3+x) (2+x)^2} \, dx &=\int \left (\frac{9}{25 (-3+x)}-\frac{4}{5 (2+x)^2}+\frac{16}{25 (2+x)}\right ) \, dx\\ &=\frac{4}{5 (2+x)}+\frac{9}{25} \log (3-x)+\frac{16}{25} \log (2+x)\\ \end{align*}

Mathematica [A]  time = 0.0158948, size = 26, normalized size = 0.93 $\frac{4}{5 (x+2)}+\frac{9}{25} \log (x-3)+\frac{16}{25} \log (x+2)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2/((-3 + x)*(2 + x)^2),x]

[Out]

4/(5*(2 + x)) + (9*Log[-3 + x])/25 + (16*Log[2 + x])/25

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Maple [A]  time = 0.006, size = 21, normalized size = 0.8 \begin{align*}{\frac{4}{10+5\,x}}+{\frac{16\,\ln \left ( 2+x \right ) }{25}}+{\frac{9\,\ln \left ( -3+x \right ) }{25}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5/(2+x)+16/25*ln(2+x)+9/25*ln(-3+x)

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Maxima [A]  time = 0.926318, size = 27, normalized size = 0.96 \begin{align*} \frac{4}{5 \,{\left (x + 2\right )}} + \frac{16}{25} \, \log \left (x + 2\right ) + \frac{9}{25} \, \log \left (x - 3\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5/(x + 2) + 16/25*log(x + 2) + 9/25*log(x - 3)

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Fricas [A]  time = 1.83164, size = 89, normalized size = 3.18 \begin{align*} \frac{16 \,{\left (x + 2\right )} \log \left (x + 2\right ) + 9 \,{\left (x + 2\right )} \log \left (x - 3\right ) + 20}{25 \,{\left (x + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*(16*(x + 2)*log(x + 2) + 9*(x + 2)*log(x - 3) + 20)/(x + 2)

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Sympy [A]  time = 0.113319, size = 22, normalized size = 0.79 \begin{align*} \frac{9 \log{\left (x - 3 \right )}}{25} + \frac{16 \log{\left (x + 2 \right )}}{25} + \frac{4}{5 x + 10} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9*log(x - 3)/25 + 16*log(x + 2)/25 + 4/(5*x + 10)

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Giac [A]  time = 1.04837, size = 35, normalized size = 1.25 \begin{align*} \frac{4}{5 \,{\left (x + 2\right )}} + \log \left ({\left | x + 2 \right |}\right ) + \frac{9}{25} \, \log \left ({\left | -\frac{5}{x + 2} + 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5/(x + 2) + log(abs(x + 2)) + 9/25*log(abs(-5/(x + 2) + 1))