3.47.77 \(\int \frac {-192+64 x+192 \log (\frac {x}{4})}{(225-150 x+25 x^2+e^2 (-90+60 x-10 x^2)+e^4 (9-6 x+x^2)) \log ^2(\frac {x}{4})} \, dx\)

Optimal. Leaf size=27 \[ \frac {64 x}{\left (5-e^2\right )^2 (3-x) \log \left (\frac {x}{4}\right )} \]

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Rubi [F]  time = 0.34, 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 {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-192 + 64*x + 192*Log[x/4])/((225 - 150*x + 25*x^2 + E^2*(-90 + 60*x - 10*x^2) + E^4*(9 - 6*x + x^2))*Log
[x/4]^2),x]

[Out]

(256*Defer[Subst][Defer[Int][1/((-3 + 4*x)*Log[x]^2), x], x, x/4])/(5 - E^2)^2 + (768*Defer[Subst][Defer[Int][
1/((-3 + 4*x)^2*Log[x]), x], x, x/4])/(5 - E^2)^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {64 \left (-3+x+3 \log \left (\frac {x}{4}\right )\right )}{\left (5-e^2\right )^2 (3-x)^2 \log ^2\left (\frac {x}{4}\right )} \, dx\\ &=\frac {64 \int \frac {-3+x+3 \log \left (\frac {x}{4}\right )}{(3-x)^2 \log ^2\left (\frac {x}{4}\right )} \, dx}{\left (5-e^2\right )^2}\\ &=\frac {256 \operatorname {Subst}\left (\int \frac {-3+4 x+3 \log (x)}{(3-4 x)^2 \log ^2(x)} \, dx,x,\frac {x}{4}\right )}{\left (5-e^2\right )^2}\\ &=\frac {256 \operatorname {Subst}\left (\int \left (\frac {1}{(-3+4 x) \log ^2(x)}+\frac {3}{(-3+4 x)^2 \log (x)}\right ) \, dx,x,\frac {x}{4}\right )}{\left (5-e^2\right )^2}\\ &=\frac {256 \operatorname {Subst}\left (\int \frac {1}{(-3+4 x) \log ^2(x)} \, dx,x,\frac {x}{4}\right )}{\left (5-e^2\right )^2}+\frac {768 \operatorname {Subst}\left (\int \frac {1}{(-3+4 x)^2 \log (x)} \, dx,x,\frac {x}{4}\right )}{\left (5-e^2\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 0.85 \begin {gather*} -\frac {64 x}{\left (-5+e^2\right )^2 (-3+x) \log \left (\frac {x}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-192 + 64*x + 192*Log[x/4])/((225 - 150*x + 25*x^2 + E^2*(-90 + 60*x - 10*x^2) + E^4*(9 - 6*x + x^2
))*Log[x/4]^2),x]

[Out]

(-64*x)/((-5 + E^2)^2*(-3 + x)*Log[x/4])

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fricas [A]  time = 0.72, size = 29, normalized size = 1.07 \begin {gather*} -\frac {64 \, x}{{\left ({\left (x - 3\right )} e^{4} - 10 \, {\left (x - 3\right )} e^{2} + 25 \, x - 75\right )} \log \left (\frac {1}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*log(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90)*exp(2)+25*x^2-150*x+225)/log(1/4*x
)^2,x, algorithm="fricas")

[Out]

-64*x/(((x - 3)*e^4 - 10*(x - 3)*e^2 + 25*x - 75)*log(1/4*x))

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giac [B]  time = 0.15, size = 52, normalized size = 1.93 \begin {gather*} -\frac {64 \, x}{x e^{4} \log \left (\frac {1}{4} \, x\right ) - 10 \, x e^{2} \log \left (\frac {1}{4} \, x\right ) + 25 \, x \log \left (\frac {1}{4} \, x\right ) - 3 \, e^{4} \log \left (\frac {1}{4} \, x\right ) + 30 \, e^{2} \log \left (\frac {1}{4} \, x\right ) - 75 \, \log \left (\frac {1}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*log(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90)*exp(2)+25*x^2-150*x+225)/log(1/4*x
)^2,x, algorithm="giac")

[Out]

-64*x/(x*e^4*log(1/4*x) - 10*x*e^2*log(1/4*x) + 25*x*log(1/4*x) - 3*e^4*log(1/4*x) + 30*e^2*log(1/4*x) - 75*lo
g(1/4*x))

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maple [A]  time = 0.31, size = 21, normalized size = 0.78

method	result	size
norman	\(-\frac {64 x}{\left ({\mathrm e}^{2}-5\right )^{2} \left (x -3\right ) \ln \left (\frac {x}{4}\right )}\)	\(21\)
risch	\(-\frac {64 x}{\left (x \,{\mathrm e}^{4}-3 \,{\mathrm e}^{4}-10 \,{\mathrm e}^{2} x +30 \,{\mathrm e}^{2}+25 x -75\right ) \ln \left (\frac {x}{4}\right )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((192*ln(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90)*exp(2)+25*x^2-150*x+225)/ln(1/4*x)^2,x,me
thod=_RETURNVERBOSE)

[Out]

-64/(exp(2)-5)^2*x/(x-3)/ln(1/4*x)

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maxima [B]  time = 0.49, size = 65, normalized size = 2.41 \begin {gather*} \frac {64 \, x}{2 \, {\left (e^{4} \log \relax (2) - 10 \, e^{2} \log \relax (2) + 25 \, \log \relax (2)\right )} x - 6 \, e^{4} \log \relax (2) + 60 \, e^{2} \log \relax (2) - {\left (x {\left (e^{4} - 10 \, e^{2} + 25\right )} - 3 \, e^{4} + 30 \, e^{2} - 75\right )} \log \relax (x) - 150 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*log(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90)*exp(2)+25*x^2-150*x+225)/log(1/4*x
)^2,x, algorithm="maxima")

[Out]

64*x/(2*(e^4*log(2) - 10*e^2*log(2) + 25*log(2))*x - 6*e^4*log(2) + 60*e^2*log(2) - (x*(e^4 - 10*e^2 + 25) - 3
*e^4 + 30*e^2 - 75)*log(x) - 150*log(2))

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mupad [B]  time = 3.84, size = 20, normalized size = 0.74 \begin {gather*} -\frac {64\,x}{\ln \left (\frac {x}{4}\right )\,{\left ({\mathrm {e}}^2-5\right )}^2\,\left (x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*x + 192*log(x/4) - 192)/(log(x/4)^2*(exp(4)*(x^2 - 6*x + 9) - exp(2)*(10*x^2 - 60*x + 90) - 150*x + 25
*x^2 + 225)),x)

[Out]

-(64*x)/(log(x/4)*(exp(2) - 5)^2*(x - 3))

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sympy [A]  time = 0.21, size = 36, normalized size = 1.33 \begin {gather*} - \frac {64 x}{\left (- 10 x e^{2} + 25 x + x e^{4} - 3 e^{4} - 75 + 30 e^{2}\right ) \log {\left (\frac {x}{4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((192*ln(1/4*x)+64*x-192)/((x**2-6*x+9)*exp(2)**2+(-10*x**2+60*x-90)*exp(2)+25*x**2-150*x+225)/ln(1/4
*x)**2,x)

[Out]

-64*x/((-10*x*exp(2) + 25*x + x*exp(4) - 3*exp(4) - 75 + 30*exp(2))*log(x/4))

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