3.3.67 \(\int \frac {-2-3 x+5 x^2}{(-2+x) x^2} \, dx\) [267]

Optimal. Leaf size=18 \[ -\frac {1}{x}+3 \log (2-x)+2 \log (x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {907} \begin {gather*} -\frac {1}{x}+3 \log (2-x)+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + 3*Log[2 - x] + 2*Log[x]

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} -\frac {1}{x}+3 \log (2-x)+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-1) + 3*Log[2 - x] + 2*Log[x]

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Maple [A]
time = 0.04, size = 17, normalized size = 0.94

method	result	size
default	\(-\frac {1}{x}+2 \ln \left (x \right )+3 \ln \left (-2+x \right )\)	\(17\)
norman	\(-\frac {1}{x}+2 \ln \left (x \right )+3 \ln \left (-2+x \right )\)	\(17\)
risch	\(-\frac {1}{x}+2 \ln \left (x \right )+3 \ln \left (-2+x \right )\)	\(17\)
meijerg	\(-\frac {1}{x}+2 \ln \left (x \right )-2 \ln \left (2\right )+2 i \pi +3 \ln \left (1-\frac {x}{2}\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2-3*x-2)/(-2+x)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 2.54, size = 16, normalized size = 0.89 \begin {gather*} -\frac {1}{x} + 3 \, \log \left (x - 2\right ) + 2 \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x + 3*log(x - 2) + 2*log(x)

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Fricas [A]
time = 1.46, size = 18, normalized size = 1.00 \begin {gather*} \frac {3 \, x \log \left (x - 2\right ) + 2 \, x \log \left (x\right ) - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x*log(x - 2) + 2*x*log(x) - 1)/x

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Sympy [A]
time = 0.04, size = 14, normalized size = 0.78 \begin {gather*} 2 \log {\left (x \right )} + 3 \log {\left (x - 2 \right )} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x) + 3*log(x - 2) - 1/x

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Giac [A]
time = 0.45, size = 18, normalized size = 1.00 \begin {gather*} -\frac {1}{x} + 3 \, \log \left ({\left | x - 2 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.15, size = 16, normalized size = 0.89 \begin {gather*} 3\,\ln \left (x-2\right )+2\,\ln \left (x\right )-\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*log(x - 2) + 2*log(x) - 1/x

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