3.70.36 \(\int \frac {12+5 x-24 x^2-x^3+x \log (x)}{5 x-x^3+x \log (x)} \, dx\)

Optimal. Leaf size=25 \[ 3+e^5+x+6 \left (4+e^5+\log \left (\left (5-x^2+\log (x)\right )^2\right )\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 14, normalized size of antiderivative = 0.56, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6742, 6684} \begin {gather*} 12 \log \left (-x^2+\log (x)+5\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 5*x - 24*x^2 - x^3 + x*Log[x])/(5*x - x^3 + x*Log[x]),x]

[Out]

x + 12*Log[5 - x^2 + Log[x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {12 \left (-1+2 x^2\right )}{x \left (-5+x^2-\log (x)\right )}\right ) \, dx\\ &=x+12 \int \frac {-1+2 x^2}{x \left (-5+x^2-\log (x)\right )} \, dx\\ &=x+12 \log \left (5-x^2+\log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 14, normalized size = 0.56 \begin {gather*} x+12 \log \left (5-x^2+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + 5*x - 24*x^2 - x^3 + x*Log[x])/(5*x - x^3 + x*Log[x]),x]

[Out]

x + 12*Log[5 - x^2 + Log[x]]

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fricas [A]  time = 0.88, size = 14, normalized size = 0.56 \begin {gather*} x + 12 \, \log \left (-x^{2} + \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)-x^3-24*x^2+5*x+12)/(x*log(x)-x^3+5*x),x, algorithm="fricas")

[Out]

x + 12*log(-x^2 + log(x) + 5)

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giac [A]  time = 0.23, size = 14, normalized size = 0.56 \begin {gather*} x + 12 \, \log \left (-x^{2} + \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)-x^3-24*x^2+5*x+12)/(x*log(x)-x^3+5*x),x, algorithm="giac")

[Out]

x + 12*log(-x^2 + log(x) + 5)

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maple [A]  time = 0.02, size = 15, normalized size = 0.60

method	result	size
norman	\(x +12 \ln \left (-\ln \relax (x )+x^{2}-5\right )\)	\(15\)
risch	\(x +12 \ln \left (\ln \relax (x )-x^{2}+5\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x)-x^3-24*x^2+5*x+12)/(x*ln(x)-x^3+5*x),x,method=_RETURNVERBOSE)

[Out]

x+12*ln(-ln(x)+x^2-5)

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maxima [A]  time = 0.40, size = 14, normalized size = 0.56 \begin {gather*} x + 12 \, \log \left (-x^{2} + \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)-x^3-24*x^2+5*x+12)/(x*log(x)-x^3+5*x),x, algorithm="maxima")

[Out]

x + 12*log(-x^2 + log(x) + 5)

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mupad [B]  time = 4.25, size = 14, normalized size = 0.56 \begin {gather*} x+12\,\ln \left (x^2-\ln \relax (x)-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + x*log(x) - 24*x^2 - x^3 + 12)/(5*x + x*log(x) - x^3),x)

[Out]

x + 12*log(x^2 - log(x) - 5)

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sympy [A]  time = 0.12, size = 12, normalized size = 0.48 \begin {gather*} x + 12 \log {\left (- x^{2} + \log {\relax (x )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x)-x**3-24*x**2+5*x+12)/(x*ln(x)-x**3+5*x),x)

[Out]

x + 12*log(-x**2 + log(x) + 5)

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