3.24.80 \(\int \frac {-3 e^4 x^2+2 x^4-x^4 \log (\frac {64}{25 x})}{e^8+9 x^4+6 x^5+x^6+e^4 (-6 x^2-2 x^3)+(2 e^4 x^2-6 x^4-2 x^5) \log (\frac {64}{25 x})+x^4 \log ^2(\frac {64}{25 x})} \, dx\) [2380]

Optimal. Leaf size=25 \[ \frac {x}{3-\frac {e^4}{x^2}+x-\log \left (\frac {64}{25 x}\right )} \]

[Out]

x/(3+x-exp(2)^2/x^2-ln(64/25/x))

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Rubi [F]
time = 0.65, 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 {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*E^4*x^2 + 2*x^4 - x^4*Log[64/(25*x)])/(E^8 + 9*x^4 + 6*x^5 + x^6 + E^4*(-6*x^2 - 2*x^3) + (2*E^4*x^2 -
6*x^4 - 2*x^5)*Log[64/(25*x)] + x^4*Log[64/(25*x)]^2),x]

[Out]

Defer[Int][x^2/(-E^4 + 3*x^2 + x^3 - x^2*Log[64/(25*x)]), x] - 2*E^4*Defer[Int][x^2/(E^4 - 3*x^2 - x^3 + x^2*L
og[64/(25*x)])^2, x] - Defer[Int][x^4/(E^4 - 3*x^2 - x^3 + x^2*Log[64/(25*x)])^2, x] - Defer[Int][x^5/(E^4 - 3
*x^2 - x^3 + x^2*Log[64/(25*x)])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{\left (e^4-x^2 (3+x)+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx\\ &=\int \left (\frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )}-\frac {x^2 \left (2 e^4+x^2+x^3\right )}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}\right ) \, dx\\ &=\int \frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )} \, dx-\int \frac {x^2 \left (2 e^4+x^2+x^3\right )}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx\\ &=\int \frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )} \, dx-\int \left (\frac {2 e^4 x^2}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}+\frac {x^4}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}+\frac {x^5}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}\right ) \, dx\\ &=-\left (\left (2 e^4\right ) \int \frac {x^2}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx\right )+\int \frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )} \, dx-\int \frac {x^4}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx-\int \frac {x^5}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.10, size = 33, normalized size = 1.32 \begin {gather*} -\frac {x^3}{e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^4*x^2 + 2*x^4 - x^4*Log[64/(25*x)])/(E^8 + 9*x^4 + 6*x^5 + x^6 + E^4*(-6*x^2 - 2*x^3) + (2*E^4
*x^2 - 6*x^4 - 2*x^5)*Log[64/(25*x)] + x^4*Log[64/(25*x)]^2),x]

[Out]

-(x^3/(E^4 - 3*x^2 - x^3 + x^2*Log[64/(25*x)]))

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Maple [A]
time = 1.84, size = 32, normalized size = 1.28

method	result	size
risch	\(-\frac {x^{3}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\)	\(31\)
derivativedivides	\(-\frac {262144}{\frac {262144 \,{\mathrm e}^{4}}{x^{3}}+\frac {262144 \ln \left (\frac {64}{25 x}\right )}{x}-\frac {786432}{x}-262144}\)	\(32\)
default	\(-\frac {262144}{\frac {262144 \,{\mathrm e}^{4}}{x^{3}}+\frac {262144 \ln \left (\frac {64}{25 x}\right )}{x}-\frac {786432}{x}-262144}\)	\(32\)
norman	\(\frac {3 x^{2}-x^{2} \ln \left (\frac {64}{25 x}\right )-{\mathrm e}^{4}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4*ln(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*ln(64/25/x)^2+(2*x^2*exp(2)^2-2*x^5-6*x^4)*ln(64/25/x)+exp(2)
^4+(-2*x^3-6*x^2)*exp(2)^2+x^6+6*x^5+9*x^4),x,method=_RETURNVERBOSE)

[Out]

-262144/(262144*exp(2)^2/x^3+262144*ln(64/25/x)/x-786432/x-262144)

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Maxima [A]
time = 0.50, size = 34, normalized size = 1.36 \begin {gather*} \frac {x^{3}}{x^{3} + x^{2} {\left (2 \, \log \left (5\right ) - 6 \, \log \left (2\right ) + 3\right )} + x^{2} \log \left (x\right ) - e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4*log(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*log(64/25/x)^2+(2*x^2*exp(2)^2-2*x^5-6*x^4)*log(64/25/
x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^6+6*x^5+9*x^4),x, algorithm="maxima")

[Out]

x^3/(x^3 + x^2*(2*log(5) - 6*log(2) + 3) + x^2*log(x) - e^4)

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Fricas [A]
time = 0.37, size = 30, normalized size = 1.20 \begin {gather*} \frac {x^{3}}{x^{3} - x^{2} \log \left (\frac {64}{25 \, x}\right ) + 3 \, x^{2} - e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4*log(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*log(64/25/x)^2+(2*x^2*exp(2)^2-2*x^5-6*x^4)*log(64/25/
x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^6+6*x^5+9*x^4),x, algorithm="fricas")

[Out]

x^3/(x^3 - x^2*log(64/25/x) + 3*x^2 - e^4)

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Sympy [A]
time = 0.09, size = 26, normalized size = 1.04 \begin {gather*} - \frac {x^{3}}{- x^{3} + x^{2} \log {\left (\frac {64}{25 x} \right )} - 3 x^{2} + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4*ln(64/25/x)-3*x**2*exp(2)**2+2*x**4)/(x**4*ln(64/25/x)**2+(2*x**2*exp(2)**2-2*x**5-6*x**4)*ln
(64/25/x)+exp(2)**4+(-2*x**3-6*x**2)*exp(2)**2+x**6+6*x**5+9*x**4),x)

[Out]

-x**3/(-x**3 + x**2*log(64/(25*x)) - 3*x**2 + exp(4))

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Giac [A]
time = 0.43, size = 27, normalized size = 1.08 \begin {gather*} -\frac {1}{\frac {\log \left (\frac {64}{25 \, x}\right )}{x} - \frac {3}{x} + \frac {e^{4}}{x^{3}} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4*log(64/25/x)-3*x^2*exp(2)^2+2*x^4)/(x^4*log(64/25/x)^2+(2*x^2*exp(2)^2-2*x^5-6*x^4)*log(64/25/
x)+exp(2)^4+(-2*x^3-6*x^2)*exp(2)^2+x^6+6*x^5+9*x^4),x, algorithm="giac")

[Out]

-1/(log(64/25/x)/x - 3/x + e^4/x^3 - 1)

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Mupad [B]
time = 1.79, size = 30, normalized size = 1.20 \begin {gather*} -\frac {x^3}{{\mathrm {e}}^4-3\,x^2-x^3+x^2\,\ln \left (\frac {64}{25\,x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^2*exp(4) - 2*x^4 + x^4*log(64/(25*x)))/(exp(8) + x^4*log(64/(25*x))^2 - log(64/(25*x))*(6*x^4 - 2*x^
2*exp(4) + 2*x^5) - exp(4)*(6*x^2 + 2*x^3) + 9*x^4 + 6*x^5 + x^6),x)

[Out]

-x^3/(exp(4) - 3*x^2 - x^3 + x^2*log(64/(25*x)))

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