3.60.77 \(\int \frac {-40+32 x}{-5 x+2 x^2} \, dx\)

Optimal. Leaf size=16 \[ 4 \log \left ((-x+x (-4+2 x))^2\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 0.75, number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {628} \begin {gather*} 8 \log \left (5 x-2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40 + 32*x)/(-5*x + 2*x^2),x]

[Out]

8*Log[5*x - 2*x^2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=8 \log \left (5 x-2 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.69 \begin {gather*} 8 (\log (5-2 x)+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40 + 32*x)/(-5*x + 2*x^2),x]

[Out]

8*(Log[5 - 2*x] + Log[x])

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fricas [A]  time = 0.81, size = 12, normalized size = 0.75 \begin {gather*} 8 \, \log \left (2 \, x^{2} - 5 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x-40)/(2*x^2-5*x),x, algorithm="fricas")

[Out]

8*log(2*x^2 - 5*x)

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giac [A]  time = 0.15, size = 13, normalized size = 0.81 \begin {gather*} 8 \, \log \left ({\left | 2 \, x^{2} - 5 \, x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x-40)/(2*x^2-5*x),x, algorithm="giac")

[Out]

8*log(abs(2*x^2 - 5*x))

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maple [A]  time = 0.29, size = 11, normalized size = 0.69

method	result	size
default	\(8 \ln \left (x \left (2 x -5\right )\right )\)	\(11\)
risch	\(8 \ln \left (2 x^{2}-5 x \right )\)	\(13\)
norman	\(8 \ln \relax (x )+8 \ln \left (2 x -5\right )\)	\(14\)
meijerg	\(8 \ln \left (-\frac {2 x}{5}+1\right )+8 \ln \relax (x )+8 \ln \relax (2)-8 \ln \relax (5)+8 i \pi \)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x-40)/(2*x^2-5*x),x,method=_RETURNVERBOSE)

[Out]

8*ln(x*(2*x-5))

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maxima [A]  time = 0.35, size = 12, normalized size = 0.75 \begin {gather*} 8 \, \log \left (2 \, x^{2} - 5 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x-40)/(2*x^2-5*x),x, algorithm="maxima")

[Out]

8*log(2*x^2 - 5*x)

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mupad [B]  time = 0.07, size = 10, normalized size = 0.62 \begin {gather*} 8\,\ln \left (x\,\left (2\,x-5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x - 40)/(5*x - 2*x^2),x)

[Out]

8*log(x*(2*x - 5))

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sympy [A]  time = 0.09, size = 10, normalized size = 0.62 \begin {gather*} 8 \log {\left (2 x^{2} - 5 x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x-40)/(2*x**2-5*x),x)

[Out]

8*log(2*x**2 - 5*x)

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