3.1.92 \(\int \frac {1}{2-7 x+3 x^2} \, dx\) [92]

Optimal. Leaf size=21 \[ -\frac {1}{5} \log (1-3 x)+\frac {1}{5} \log (2-x) \]

[Out]

-1/5*ln(1-3*x)+1/5*ln(2-x)

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Rubi [A]
time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {630, 31} \begin {gather*} \frac {1}{5} \log (2-x)-\frac {1}{5} \log (1-3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 7*x + 3*x^2)^(-1),x]

[Out]

-1/5*Log[1 - 3*x] + Log[2 - x]/5

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {1}{2-7 x+3 x^2} \, dx &=\frac {3}{5} \int \frac {1}{-6+3 x} \, dx-\frac {3}{5} \int \frac {1}{-1+3 x} \, dx\\ &=-\frac {1}{5} \log (1-3 x)+\frac {1}{5} \log (2-x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 21, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \log (1-3 x)+\frac {1}{5} \log (2-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 7*x + 3*x^2)^(-1),x]

[Out]

-1/5*Log[1 - 3*x] + Log[2 - x]/5

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Maple [A]
time = 0.08, size = 16, normalized size = 0.76

method	result	size
default	\(\frac {\ln \left (-2+x \right )}{5}-\frac {\ln \left (3 x -1\right )}{5}\)	\(16\)
norman	\(\frac {\ln \left (-2+x \right )}{5}-\frac {\ln \left (3 x -1\right )}{5}\)	\(16\)
risch	\(\frac {\ln \left (-2+x \right )}{5}-\frac {\ln \left (3 x -1\right )}{5}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2-7*x+2),x,method=_RETURNVERBOSE)

[Out]

1/5*ln(-2+x)-1/5*ln(3*x-1)

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Maxima [A]
time = 2.57, size = 15, normalized size = 0.71 \begin {gather*} -\frac {1}{5} \, \log \left (3 \, x - 1\right ) + \frac {1}{5} \, \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2-7*x+2),x, algorithm="maxima")

[Out]

-1/5*log(3*x - 1) + 1/5*log(x - 2)

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Fricas [A]
time = 0.40, size = 15, normalized size = 0.71 \begin {gather*} -\frac {1}{5} \, \log \left (3 \, x - 1\right ) + \frac {1}{5} \, \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2-7*x+2),x, algorithm="fricas")

[Out]

-1/5*log(3*x - 1) + 1/5*log(x - 2)

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Sympy [A]
time = 0.03, size = 14, normalized size = 0.67 \begin {gather*} \frac {\log {\left (x - 2 \right )}}{5} - \frac {\log {\left (x - \frac {1}{3} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x**2-7*x+2),x)

[Out]

log(x - 2)/5 - log(x - 1/3)/5

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Giac [A]
time = 1.09, size = 17, normalized size = 0.81 \begin {gather*} -\frac {1}{5} \, \log \left ({\left | 3 \, x - 1 \right |}\right ) + \frac {1}{5} \, \log \left ({\left | x - 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2-7*x+2),x, algorithm="giac")

[Out]

-1/5*log(abs(3*x - 1)) + 1/5*log(abs(x - 2))

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Mupad [B]
time = 0.09, size = 8, normalized size = 0.38 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\frac {6\,x}{5}-\frac {7}{5}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2 - 7*x + 2),x)

[Out]

-(2*atanh((6*x)/5 - 7/5))/5

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