3.48.72 \(\int \frac {-1-2 x+2 x^2}{-2+2 x} \, dx\)

Optimal. Leaf size=16 \[ \frac {1}{2} \left (x^2-\log (1-x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {698} \begin {gather*} \frac {x^2}{2}-\frac {1}{2} \log (1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 2*x + 2*x^2)/(-2 + 2*x),x]

[Out]

x^2/2 - Log[1 - x]/2

Rule 698

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{2 (-1+x)}+x\right ) \, dx\\ &=\frac {x^2}{2}-\frac {1}{2} \log (1-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.06 \begin {gather*} \frac {1}{2} \left (1+x^2-\log (2-2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + x^2 - Log[2 - 2*x])/2

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fricas [A]  time = 0.71, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{2} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - 1/2*log(x - 1)

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giac [A]  time = 0.12, size = 13, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - 1/2*log(abs(x - 1))

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maple [A]  time = 0.09, size = 13, normalized size = 0.81

method	result	size
default	\(\frac {x^{2}}{2}-\frac {\ln \left (x -1\right )}{2}\)	\(13\)
risch	\(\frac {x^{2}}{2}-\frac {\ln \left (x -1\right )}{2}\)	\(13\)
norman	\(\frac {x^{2}}{2}-\frac {\ln \left (2 x -2\right )}{2}\)	\(15\)
meijerg	\(-\frac {\ln \left (1-x \right )}{2}+\frac {x \left (6+3 x \right )}{6}-x\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2-2*x-1)/(2*x-2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{2} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - 1/2*log(x - 1)

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mupad [B]  time = 0.04, size = 12, normalized size = 0.75 \begin {gather*} \frac {x^2}{2}-\frac {\ln \left (x-1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - 2*x^2 + 1)/(2*x - 2),x)

[Out]

x^2/2 - log(x - 1)/2

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sympy [A]  time = 0.07, size = 10, normalized size = 0.62 \begin {gather*} \frac {x^{2}}{2} - \frac {\log {\left (x - 1 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2-2*x-1)/(2*x-2),x)

[Out]

x**2/2 - log(x - 1)/2

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