3.2.38 \(\int \frac {-1+4 x+(1+x-2 x^2) \log (5 x)}{-2+x \log (5 x)} \, dx\)

Optimal. Leaf size=18 \[ 4+x-x^2+\log (2-x \log (5 x)) \]

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Rubi [F]  time = 0.26, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1+4 x+\left (1+x-2 x^2\right ) \log (5 x)}{-2+x \log (5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + 4*x + (1 + x - 2*x^2)*Log[5*x])/(-2 + x*Log[5*x]),x]

[Out]

x - x^2 + Log[x] + Defer[Int][(-2 + x*Log[5*x])^(-1), x] + 2*Defer[Int][1/(x*(-2 + x*Log[5*x])), x]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 17, normalized size = 0.94 \begin {gather*} x-x^2+\log (2-x \log (5 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 4*x + (1 + x - 2*x^2)*Log[5*x])/(-2 + x*Log[5*x]),x]

[Out]

x - x^2 + Log[2 - x*Log[5*x]]

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fricas [A]  time = 0.64, size = 24, normalized size = 1.33 \begin {gather*} -x^{2} + x + \log \left (5 \, x\right ) + \log \left (\frac {x \log \left (5 \, x\right ) - 2}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+x+1)*log(5*x)+4*x-1)/(x*log(5*x)-2),x, algorithm="fricas")

[Out]

-x^2 + x + log(5*x) + log((x*log(5*x) - 2)/x)

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giac [A]  time = 0.26, size = 17, normalized size = 0.94 \begin {gather*} -x^{2} + x + \log \left (-x \log \left (5 \, x\right ) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+x+1)*log(5*x)+4*x-1)/(x*log(5*x)-2),x, algorithm="giac")

[Out]

-x^2 + x + log(-x*log(5*x) + 2)

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maple [A]  time = 0.03, size = 17, normalized size = 0.94

method	result	size
norman	\(-x^{2}+x +\ln \left (x \ln \left (5 x \right )-2\right )\)	\(17\)
risch	\(\ln \relax (x )+x -x^{2}+\ln \left (\ln \left (5 x \right )-\frac {2}{x}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2+x+1)*ln(5*x)+4*x-1)/(x*ln(5*x)-2),x,method=_RETURNVERBOSE)

[Out]

-x^2+x+ln(x*ln(5*x)-2)

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maxima [A]  time = 0.83, size = 24, normalized size = 1.33 \begin {gather*} -x^{2} + x + \log \relax (x) + \log \left (\frac {x \log \relax (5) + x \log \relax (x) - 2}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+x+1)*log(5*x)+4*x-1)/(x*log(5*x)-2),x, algorithm="maxima")

[Out]

-x^2 + x + log(x) + log((x*log(5) + x*log(x) - 2)/x)

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mupad [B]  time = 0.34, size = 23, normalized size = 1.28 \begin {gather*} \ln \left (x\,\ln \left (5\,x\right )-2\right )+\frac {x^2-x^3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + log(5*x)*(x - 2*x^2 + 1) - 1)/(x*log(5*x) - 2),x)

[Out]

log(x*log(5*x) - 2) + (x^2 - x^3)/x

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sympy [A]  time = 0.16, size = 17, normalized size = 0.94 \begin {gather*} - x^{2} + x + \log {\relax (x )} + \log {\left (\log {\left (5 x \right )} - \frac {2}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2+x+1)*ln(5*x)+4*x-1)/(x*ln(5*x)-2),x)

[Out]

-x**2 + x + log(x) + log(log(5*x) - 2/x)

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