3.2.89 \(\int \frac {x^2}{(-3+x) (2+x)^2} \, dx\) [189]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 = {90} \begin {gather*} \frac {4}{5 (x+2)}+\frac {9}{25} \log (3-x)+\frac {16}{25} \log (x+2) \end {gather*}

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 90

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*}

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Mathematica [A]
time = 0.01, size = 26, normalized size = 0.93 \begin {gather*} \frac {4}{5 (2+x)}+\frac {9}{25} \log (-3+x)+\frac {16}{25} \log (2+x) \end {gather*}

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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Mathics [A]
time = 1.75, size = 26, normalized size = 0.93 \begin {gather*} \frac {20+\left (2+x\right ) \left (9 \text {Log}\left [-3+x\right ]+16 \text {Log}\left [2+x\right ]\right )}{50+25 x} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^2/((-3 + x)*(2 + x)^2),x]')

[Out]

(20 + (2 + x) (9 Log[-3 + x] + 16 Log[2 + x])) / (25 (2 + x))

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Maple [A]
time = 0.05, size = 21, normalized size = 0.75

method	result	size
default	\(\frac {9 \ln \left (-3+x \right )}{25}+\frac {4}{5 \left (2+x \right )}+\frac {16 \ln \left (2+x \right )}{25}\)	\(21\)
norman	\(\frac {9 \ln \left (-3+x \right )}{25}+\frac {4}{5 \left (2+x \right )}+\frac {16 \ln \left (2+x \right )}{25}\)	\(21\)
risch	\(\frac {9 \ln \left (-3+x \right )}{25}+\frac {4}{5 \left (2+x \right )}+\frac {16 \ln \left (2+x \right )}{25}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-3+x)/(2+x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 20, normalized size = 0.71 \begin {gather*} \frac {4}{5 \, {\left (x + 2\right )}} + \frac {16}{25} \, \log \left (x + 2\right ) + \frac {9}{25} \, \log \left (x - 3\right ) \end {gather*}

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 = 0.33, size = 27, normalized size = 0.96 \begin {gather*} \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 {gather*}

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 = 0.06, size = 22, normalized size = 0.79 \begin {gather*} \frac {9 \log {\left (x - 3 \right )}}{25} + \frac {16 \log {\left (x + 2 \right )}}{25} + \frac {4}{5 x + 10} \end {gather*}

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 = 0.00, size = 28, normalized size = 1.00 \begin {gather*} \frac {9}{25} \ln \left |x-3\right |+\frac {16}{25} \ln \left |x+2\right |+\frac {\frac {1}{25}\cdot 20}{x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.21, size = 22, normalized size = 0.79 \begin {gather*} \frac {16\,\ln \left (x+2\right )}{25}+\frac {9\,\ln \left (x-3\right )}{25}+\frac {4}{5\,\left (x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((x + 2)^2*(x - 3)),x)

[Out]

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

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