3.56.99 \(\int \frac {x^3+3 e^2 \log (5)+(6 e^2 x^2-2 x^3-x \log (5)) \log (x)+(9 e^4 x-6 e^2 x^2+x^3) \log ^2(x)}{x^3+(6 e^2 x^2-2 x^3) \log (x)+(9 e^4 x-6 e^2 x^2+x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ x-\log (5)-\frac {\log (5)}{x+\left (3 e^2-x\right ) \log (x)} \]

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Rubi [F]  time = 1.08, 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 {x^3+3 e^2 \log (5)+\left (6 e^2 x^2-2 x^3-x \log (5)\right ) \log (x)+\left (9 e^4 x-6 e^2 x^2+x^3\right ) \log ^2(x)}{x^3+\left (6 e^2 x^2-2 x^3\right ) \log (x)+\left (9 e^4 x-6 e^2 x^2+x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^3 + 3*E^2*Log[5] + (6*E^2*x^2 - 2*x^3 - x*Log[5])*Log[x] + (9*E^4*x - 6*E^2*x^2 + x^3)*Log[x]^2)/(x^3 +
(6*E^2*x^2 - 2*x^3)*Log[x] + (9*E^4*x - 6*E^2*x^2 + x^3)*Log[x]^2),x]

[Out]

x + 3*E^2*Log[5]*Defer[Int][1/((3*E^2 - x)*(x + 3*E^2*Log[x] - x*Log[x])^2), x] - Log[5]*Defer[Int][1/((3*E^2
- x)*(x + 3*E^2*Log[x] - x*Log[x])), x] - Log[5]*Defer[Int][(-x - 3*E^2*Log[x] + x*Log[x])^(-2), x] + 3*E^2*Lo
g[5]*Defer[Int][1/(x*(-x - 3*E^2*Log[x] + x*Log[x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^3+3 e^2 \log (5)-x \left (-6 e^2 x+2 x^2+\log (5)\right ) \log (x)+x \left (-3 e^2+x\right )^2 \log ^2(x)}{x \left (x+\left (3 e^2-x\right ) \log (x)\right )^2} \, dx\\ &=\int \left (1+\frac {\left (9 e^4-3 e^2 x+x^2\right ) \log (5)}{\left (3 e^2-x\right ) x \left (x+3 e^2 \log (x)-x \log (x)\right )^2}-\frac {\log (5)}{\left (3 e^2-x\right ) \left (x+3 e^2 \log (x)-x \log (x)\right )}\right ) \, dx\\ &=x+\log (5) \int \frac {9 e^4-3 e^2 x+x^2}{\left (3 e^2-x\right ) x \left (x+3 e^2 \log (x)-x \log (x)\right )^2} \, dx-\log (5) \int \frac {1}{\left (3 e^2-x\right ) \left (x+3 e^2 \log (x)-x \log (x)\right )} \, dx\\ &=x-\log (5) \int \frac {1}{\left (3 e^2-x\right ) \left (x+3 e^2 \log (x)-x \log (x)\right )} \, dx+\log (5) \int \left (\frac {3 e^2}{\left (3 e^2-x\right ) \left (x+3 e^2 \log (x)-x \log (x)\right )^2}-\frac {1}{\left (-x-3 e^2 \log (x)+x \log (x)\right )^2}+\frac {3 e^2}{x \left (-x-3 e^2 \log (x)+x \log (x)\right )^2}\right ) \, dx\\ &=x-\log (5) \int \frac {1}{\left (3 e^2-x\right ) \left (x+3 e^2 \log (x)-x \log (x)\right )} \, dx-\log (5) \int \frac {1}{\left (-x-3 e^2 \log (x)+x \log (x)\right )^2} \, dx+\left (3 e^2 \log (5)\right ) \int \frac {1}{\left (3 e^2-x\right ) \left (x+3 e^2 \log (x)-x \log (x)\right )^2} \, dx+\left (3 e^2 \log (5)\right ) \int \frac {1}{x \left (-x-3 e^2 \log (x)+x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 22, normalized size = 0.85 \begin {gather*} x-\frac {\log (5)}{x+3 e^2 \log (x)-x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3 + 3*E^2*Log[5] + (6*E^2*x^2 - 2*x^3 - x*Log[5])*Log[x] + (9*E^4*x - 6*E^2*x^2 + x^3)*Log[x]^2)/
(x^3 + (6*E^2*x^2 - 2*x^3)*Log[x] + (9*E^4*x - 6*E^2*x^2 + x^3)*Log[x]^2),x]

[Out]

x - Log[5]/(x + 3*E^2*Log[x] - x*Log[x])

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fricas [A]  time = 0.81, size = 38, normalized size = 1.46 \begin {gather*} -\frac {x^{2} - {\left (x^{2} - 3 \, x e^{2}\right )} \log \relax (x) - \log \relax (5)}{{\left (x - 3 \, e^{2}\right )} \log \relax (x) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x*exp(2)^2-6*x^2*exp(2)+x^3)*log(x)^2+(-x*log(5)+6*x^2*exp(2)-2*x^3)*log(x)+3*exp(2)*log(5)+x^3)
/((9*x*exp(2)^2-6*x^2*exp(2)+x^3)*log(x)^2+(6*x^2*exp(2)-2*x^3)*log(x)+x^3),x, algorithm="fricas")

[Out]

-(x^2 - (x^2 - 3*x*e^2)*log(x) - log(5))/((x - 3*e^2)*log(x) - x)

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giac [A]  time = 0.16, size = 38, normalized size = 1.46 \begin {gather*} \frac {x^{2} \log \relax (x) - 3 \, x e^{2} \log \relax (x) - x^{2} + \log \relax (5)}{x \log \relax (x) - 3 \, e^{2} \log \relax (x) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x*exp(2)^2-6*x^2*exp(2)+x^3)*log(x)^2+(-x*log(5)+6*x^2*exp(2)-2*x^3)*log(x)+3*exp(2)*log(5)+x^3)
/((9*x*exp(2)^2-6*x^2*exp(2)+x^3)*log(x)^2+(6*x^2*exp(2)-2*x^3)*log(x)+x^3),x, algorithm="giac")

[Out]

(x^2*log(x) - 3*x*e^2*log(x) - x^2 + log(5))/(x*log(x) - 3*e^2*log(x) - x)

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maple [A]  time = 0.23, size = 22, normalized size = 0.85

method	result	size
default	\(x -\frac {\ln \relax (5)}{3 \,{\mathrm e}^{2} \ln \relax (x )-x \ln \relax (x )+x}\)	\(22\)
risch	\(x -\frac {\ln \relax (5)}{3 \,{\mathrm e}^{2} \ln \relax (x )-x \ln \relax (x )+x}\)	\(22\)
norman	\(\frac {x^{2}+3 \,{\mathrm e}^{2} x -x^{2} \ln \relax (x )+9 \,{\mathrm e}^{4} \ln \relax (x )-\ln \relax (5)}{3 \,{\mathrm e}^{2} \ln \relax (x )-x \ln \relax (x )+x}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9*x*exp(2)^2-6*x^2*exp(2)+x^3)*ln(x)^2+(-x*ln(5)+6*x^2*exp(2)-2*x^3)*ln(x)+3*exp(2)*ln(5)+x^3)/((9*x*exp
(2)^2-6*x^2*exp(2)+x^3)*ln(x)^2+(6*x^2*exp(2)-2*x^3)*ln(x)+x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 38, normalized size = 1.46 \begin {gather*} -\frac {x^{2} - {\left (x^{2} - 3 \, x e^{2}\right )} \log \relax (x) - \log \relax (5)}{{\left (x - 3 \, e^{2}\right )} \log \relax (x) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x*exp(2)^2-6*x^2*exp(2)+x^3)*log(x)^2+(-x*log(5)+6*x^2*exp(2)-2*x^3)*log(x)+3*exp(2)*log(5)+x^3)
/((9*x*exp(2)^2-6*x^2*exp(2)+x^3)*log(x)^2+(6*x^2*exp(2)-2*x^3)*log(x)+x^3),x, algorithm="maxima")

[Out]

-(x^2 - (x^2 - 3*x*e^2)*log(x) - log(5))/((x - 3*e^2)*log(x) - x)

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mupad [B]  time = 4.97, size = 38, normalized size = 1.46 \begin {gather*} -\frac {\ln \relax (5)+x^2\,\ln \relax (x)-x^2-3\,x\,{\mathrm {e}}^2\,\ln \relax (x)}{x+3\,{\mathrm {e}}^2\,\ln \relax (x)-x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(2)*log(5) + log(x)^2*(9*x*exp(4) - 6*x^2*exp(2) + x^3) - log(x)*(x*log(5) - 6*x^2*exp(2) + 2*x^3) +
x^3)/(log(x)*(6*x^2*exp(2) - 2*x^3) + log(x)^2*(9*x*exp(4) - 6*x^2*exp(2) + x^3) + x^3),x)

[Out]

-(log(5) + x^2*log(x) - x^2 - 3*x*exp(2)*log(x))/(x + 3*exp(2)*log(x) - x*log(x))

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sympy [A]  time = 0.19, size = 15, normalized size = 0.58 \begin {gather*} x + \frac {\log {\relax (5 )}}{- x + \left (x - 3 e^{2}\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*x*exp(2)**2-6*x**2*exp(2)+x**3)*ln(x)**2+(-x*ln(5)+6*x**2*exp(2)-2*x**3)*ln(x)+3*exp(2)*ln(5)+x*
*3)/((9*x*exp(2)**2-6*x**2*exp(2)+x**3)*ln(x)**2+(6*x**2*exp(2)-2*x**3)*ln(x)+x**3),x)

[Out]

x + log(5)/(-x + (x - 3*exp(2))*log(x))

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