3.92.12 \(\int \frac {-2 x+14 e^3 x-24 e^6 x+(-2 e^3+6 e^6) \log (5 e^x)}{x^2-8 e^3 x^2+e^6 (-3+16 x^2)+(2 e^3 x-8 e^6 x) \log (5 e^x)+e^6 \log ^2(5 e^x)} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\frac {1}{4 \left (-3+\left (-4 x+\frac {x}{e^3}+\log \left (5 e^x\right )\right )^2\right )}\right ) \]

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Rubi [F]  time = 0.67, 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 {-2 x+14 e^3 x-24 e^6 x+\left (-2 e^3+6 e^6\right ) \log \left (5 e^x\right )}{x^2-8 e^3 x^2+e^6 \left (-3+16 x^2\right )+\left (2 e^3 x-8 e^6 x\right ) \log \left (5 e^x\right )+e^6 \log ^2\left (5 e^x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x + 14*E^3*x - 24*E^6*x + (-2*E^3 + 6*E^6)*Log[5*E^x])/(x^2 - 8*E^3*x^2 + E^6*(-3 + 16*x^2) + (2*E^3*x
- 8*E^6*x)*Log[5*E^x] + E^6*Log[5*E^x]^2),x]

[Out]

2*(1 - 4*E^3)*(1 - 3*E^3)*Defer[Int][x/(3*E^6 - (1 + 8*E^3*(-1 + 2*E^3))*x^2 - 2*E^3*(1 - 4*E^3)*x*Log[5*E^x]
- E^6*Log[5*E^x]^2), x] + 2*E^3*(1 - 3*E^3)*Defer[Int][Log[5*E^x]/(3*E^6 - (1 + 8*E^3*(-1 + 2*E^3))*x^2 - 2*E^
3*(1 - 4*E^3)*x*Log[5*E^x] - E^6*Log[5*E^x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-24 e^6 x+\left (-2+14 e^3\right ) x+\left (-2 e^3+6 e^6\right ) \log \left (5 e^x\right )}{x^2-8 e^3 x^2+e^6 \left (-3+16 x^2\right )+\left (2 e^3 x-8 e^6 x\right ) \log \left (5 e^x\right )+e^6 \log ^2\left (5 e^x\right )} \, dx\\ &=\int \frac {\left (-2+14 e^3-24 e^6\right ) x+\left (-2 e^3+6 e^6\right ) \log \left (5 e^x\right )}{x^2-8 e^3 x^2+e^6 \left (-3+16 x^2\right )+\left (2 e^3 x-8 e^6 x\right ) \log \left (5 e^x\right )+e^6 \log ^2\left (5 e^x\right )} \, dx\\ &=\int \frac {\left (-2+14 e^3-24 e^6\right ) x+\left (-2 e^3+6 e^6\right ) \log \left (5 e^x\right )}{\left (1-8 e^3\right ) x^2+e^6 \left (-3+16 x^2\right )+\left (2 e^3 x-8 e^6 x\right ) \log \left (5 e^x\right )+e^6 \log ^2\left (5 e^x\right )} \, dx\\ &=\int \frac {2 \left (1-3 e^3\right ) \left (-\left (\left (1-4 e^3\right ) x\right )-e^3 \log \left (5 e^x\right )\right )}{\left (1-8 e^3\right ) x^2+e^6 \left (-3+16 x^2\right )+\left (2 e^3 x-8 e^6 x\right ) \log \left (5 e^x\right )+e^6 \log ^2\left (5 e^x\right )} \, dx\\ &=\left (2 \left (1-3 e^3\right )\right ) \int \frac {-\left (\left (1-4 e^3\right ) x\right )-e^3 \log \left (5 e^x\right )}{\left (1-8 e^3\right ) x^2+e^6 \left (-3+16 x^2\right )+\left (2 e^3 x-8 e^6 x\right ) \log \left (5 e^x\right )+e^6 \log ^2\left (5 e^x\right )} \, dx\\ &=\left (2 \left (1-3 e^3\right )\right ) \int \left (\frac {\left (1-4 e^3\right ) x}{3 e^6-\left (1+8 e^3 \left (-1+2 e^3\right )\right ) x^2-2 e^3 \left (1-4 e^3\right ) x \log \left (5 e^x\right )-e^6 \log ^2\left (5 e^x\right )}+\frac {e^3 \log \left (5 e^x\right )}{3 e^6-\left (1+8 e^3 \left (-1+2 e^3\right )\right ) x^2-2 e^3 \left (1-4 e^3\right ) x \log \left (5 e^x\right )-e^6 \log ^2\left (5 e^x\right )}\right ) \, dx\\ &=\left (2 e^3 \left (1-3 e^3\right )\right ) \int \frac {\log \left (5 e^x\right )}{3 e^6-\left (1+8 e^3 \left (-1+2 e^3\right )\right ) x^2-2 e^3 \left (1-4 e^3\right ) x \log \left (5 e^x\right )-e^6 \log ^2\left (5 e^x\right )} \, dx+\left (2 \left (1-4 e^3\right ) \left (1-3 e^3\right )\right ) \int \frac {x}{3 e^6-\left (1+8 e^3 \left (-1+2 e^3\right )\right ) x^2-2 e^3 \left (1-4 e^3\right ) x \log \left (5 e^x\right )-e^6 \log ^2\left (5 e^x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.22, size = 74, normalized size = 2.85 \begin {gather*} -\frac {2 \left (-1+3 e^3\right ) \log \left (x^2-8 e^3 x^2+e^6 \left (-3+16 x^2\right )+\left (2 e^3 x-8 e^6 x\right ) \log \left (5 e^x\right )+e^6 \log ^2\left (5 e^x\right )\right )}{-2+6 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 14*E^3*x - 24*E^6*x + (-2*E^3 + 6*E^6)*Log[5*E^x])/(x^2 - 8*E^3*x^2 + E^6*(-3 + 16*x^2) + (2
*E^3*x - 8*E^6*x)*Log[5*E^x] + E^6*Log[5*E^x]^2),x]

[Out]

(-2*(-1 + 3*E^3)*Log[x^2 - 8*E^3*x^2 + E^6*(-3 + 16*x^2) + (2*E^3*x - 8*E^6*x)*Log[5*E^x] + E^6*Log[5*E^x]^2])
/(-2 + 6*E^3)

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fricas [B]  time = 0.60, size = 47, normalized size = 1.81 \begin {gather*} -\log \left (-6 \, x^{2} e^{3} + e^{6} \log \relax (5)^{2} + x^{2} + 3 \, {\left (3 \, x^{2} - 1\right )} e^{6} - 2 \, {\left (3 \, x e^{6} - x e^{3}\right )} \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(3)^2-2*exp(3))*log(5*exp(x))-24*x*exp(3)^2+14*x*exp(3)-2*x)/(exp(3)^2*log(5*exp(x))^2+(-8*x*
exp(3)^2+2*x*exp(3))*log(5*exp(x))+(16*x^2-3)*exp(3)^2-8*x^2*exp(3)+x^2),x, algorithm="fricas")

[Out]

-log(-6*x^2*e^3 + e^6*log(5)^2 + x^2 + 3*(3*x^2 - 1)*e^6 - 2*(3*x*e^6 - x*e^3)*log(5))

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giac [B]  time = 0.15, size = 50, normalized size = 1.92 \begin {gather*} -\log \left ({\left | -9 \, x^{2} e^{6} + 6 \, x^{2} e^{3} + 6 \, x e^{6} \log \relax (5) - 2 \, x e^{3} \log \relax (5) - e^{6} \log \relax (5)^{2} - x^{2} + 3 \, e^{6} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(3)^2-2*exp(3))*log(5*exp(x))-24*x*exp(3)^2+14*x*exp(3)-2*x)/(exp(3)^2*log(5*exp(x))^2+(-8*x*
exp(3)^2+2*x*exp(3))*log(5*exp(x))+(16*x^2-3)*exp(3)^2-8*x^2*exp(3)+x^2),x, algorithm="giac")

[Out]

-log(abs(-9*x^2*e^6 + 6*x^2*e^3 + 6*x*e^6*log(5) - 2*x*e^3*log(5) - e^6*log(5)^2 - x^2 + 3*e^6))

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maple [B]  time = 0.23, size = 64, normalized size = 2.46

method	result	size
norman	\(-\ln \left ({\mathrm e}^{6} \ln \left (5 \,{\mathrm e}^{x}\right )^{2}-8 \ln \left (5 \,{\mathrm e}^{x}\right ) x \,{\mathrm e}^{6}+16 x^{2} {\mathrm e}^{6}+2 \ln \left (5 \,{\mathrm e}^{x}\right ) x \,{\mathrm e}^{3}-8 x^{2} {\mathrm e}^{3}-3 \,{\mathrm e}^{6}+x^{2}\right )\)	\(64\)
risch	\(-\ln \left (\ln \left ({\mathrm e}^{x}\right )^{2}+{\mathrm e}^{-3} \left (2 \,{\mathrm e}^{3} \ln \relax (5)-8 x \,{\mathrm e}^{3}+2 x \right ) \ln \left ({\mathrm e}^{x}\right )-\frac {\left (-8 x \,{\mathrm e}^{3} \ln \relax (5)+12 \,{\mathrm e}^{6}-4 x^{2}-64 x^{2} {\mathrm e}^{6}+32 x^{2} {\mathrm e}^{3}-4 \,{\mathrm e}^{6} \ln \relax (5)^{2}+32 \ln \relax (5) {\mathrm e}^{6} x \right ) {\mathrm e}^{-6}}{4}\right )\)	\(81\)
default	\(\frac {\ln \left (9 x^{2} {\mathrm e}^{6}-6 \,{\mathrm e}^{6} x \left (\ln \left (5 \,{\mathrm e}^{x}\right )-x \right )+{\mathrm e}^{6} \left (\ln \left (5 \,{\mathrm e}^{x}\right )-x \right )^{2}-6 x^{2} {\mathrm e}^{3}+2 \,{\mathrm e}^{3} x \left (\ln \left (5 \,{\mathrm e}^{x}\right )-x \right )-3 \,{\mathrm e}^{6}+x^{2}\right )}{-1+3 \,{\mathrm e}^{3}}-\frac {3 \ln \left (9 x^{2} {\mathrm e}^{6}-6 \,{\mathrm e}^{6} x \left (\ln \left (5 \,{\mathrm e}^{x}\right )-x \right )+{\mathrm e}^{6} \left (\ln \left (5 \,{\mathrm e}^{x}\right )-x \right )^{2}-6 x^{2} {\mathrm e}^{3}+2 \,{\mathrm e}^{3} x \left (\ln \left (5 \,{\mathrm e}^{x}\right )-x \right )-3 \,{\mathrm e}^{6}+x^{2}\right ) {\mathrm e}^{3}}{-1+3 \,{\mathrm e}^{3}}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*exp(3)^2-2*exp(3))*ln(5*exp(x))-24*x*exp(3)^2+14*x*exp(3)-2*x)/(exp(3)^2*ln(5*exp(x))^2+(-8*x*exp(3)^2
+2*x*exp(3))*ln(5*exp(x))+(16*x^2-3)*exp(3)^2-8*x^2*exp(3)+x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(exp(3)^2*ln(5*exp(x))^2-8*ln(5*exp(x))*x*exp(3)^2+16*x^2*exp(3)^2+2*ln(5*exp(x))*x*exp(3)-8*x^2*exp(3)-3*e
xp(3)^2+x^2)

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maxima [B]  time = 0.51, size = 1021, normalized size = 39.27 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(3)^2-2*exp(3))*log(5*exp(x))-24*x*exp(3)^2+14*x*exp(3)-2*x)/(exp(3)^2*log(5*exp(x))^2+(-8*x*
exp(3)^2+2*x*exp(3))*log(5*exp(x))+(16*x^2-3)*exp(3)^2-8*x^2*exp(3)+x^2),x, algorithm="maxima")

[Out]

sqrt(3)*e^3*log(5*e^x)*log(-(sqrt(3)*(3*e^3 - 1)*e^3 - x*(9*e^6 - 6*e^3 + 1) + 3*e^6*log(5) - e^3*log(5))/(sqr
t(3)*(3*e^3 - 1)*e^3 + x*(9*e^6 - 6*e^3 + 1) - 3*e^6*log(5) + e^3*log(5)))/(3*e^3 - 1) - 4*(sqrt(3)*log(5)*log
(-(sqrt(3)*(3*e^3 - 1)*e^3 - x*(9*e^6 - 6*e^3 + 1) + 3*e^6*log(5) - e^3*log(5))/(sqrt(3)*(3*e^3 - 1)*e^3 + x*(
9*e^6 - 6*e^3 + 1) - 3*e^6*log(5) + e^3*log(5)))/(3*e^3 - 1)^2 + 3*log(x^2*(9*e^6 - 6*e^3 + 1) - 2*(3*e^6*log(
5) - e^3*log(5))*x + (log(5)^2 - 3)*e^6)/(9*e^6 - 6*e^3 + 1))*e^6 + 7/3*(sqrt(3)*log(5)*log(-(sqrt(3)*(3*e^3 -
1)*e^3 - x*(9*e^6 - 6*e^3 + 1) + 3*e^6*log(5) - e^3*log(5))/(sqrt(3)*(3*e^3 - 1)*e^3 + x*(9*e^6 - 6*e^3 + 1)
- 3*e^6*log(5) + e^3*log(5)))/(3*e^3 - 1)^2 + 3*log(x^2*(9*e^6 - 6*e^3 + 1) - 2*(3*e^6*log(5) - e^3*log(5))*x
+ (log(5)^2 - 3)*e^6)/(9*e^6 - 6*e^3 + 1))*e^3 - sqrt(3)*(x*log(-(sqrt(3)*(3*e^3 - 1)*e^3 - x*(9*e^6 - 6*e^3 +
1) + 3*e^6*log(5) - e^3*log(5))/(sqrt(3)*(3*e^3 - 1)*e^3 + x*(9*e^6 - 6*e^3 + 1) - 3*e^6*log(5) + e^3*log(5))
) + (e^3*log(5) - sqrt(3)*e^3)*log(x*(3*e^3 - 1) - e^3*log(5) + sqrt(3)*e^3)/(3*e^3 - 1) - (e^3*log(5) + sqrt(
3)*e^3)*log(x*(3*e^3 - 1) - e^3*log(5) - sqrt(3)*e^3)/(3*e^3 - 1))*e^3/(3*e^3 - 1) - 1/3*sqrt(3)*log(5*e^x)*lo
g(-(sqrt(3)*(3*e^3 - 1)*e^3 - x*(9*e^6 - 6*e^3 + 1) + 3*e^6*log(5) - e^3*log(5))/(sqrt(3)*(3*e^3 - 1)*e^3 + x*
(9*e^6 - 6*e^3 + 1) - 3*e^6*log(5) + e^3*log(5)))/(3*e^3 - 1) + 1/3*sqrt(3)*(x*log(-(sqrt(3)*(3*e^3 - 1)*e^3 -
x*(9*e^6 - 6*e^3 + 1) + 3*e^6*log(5) - e^3*log(5))/(sqrt(3)*(3*e^3 - 1)*e^3 + x*(9*e^6 - 6*e^3 + 1) - 3*e^6*l
og(5) + e^3*log(5))) + (e^3*log(5) - sqrt(3)*e^3)*log(x*(3*e^3 - 1) - e^3*log(5) + sqrt(3)*e^3)/(3*e^3 - 1) -
(e^3*log(5) + sqrt(3)*e^3)*log(x*(3*e^3 - 1) - e^3*log(5) - sqrt(3)*e^3)/(3*e^3 - 1))/(3*e^3 - 1) - 1/3*sqrt(3
)*log(5)*log(-(sqrt(3)*(3*e^3 - 1)*e^3 - x*(9*e^6 - 6*e^3 + 1) + 3*e^6*log(5) - e^3*log(5))/(sqrt(3)*(3*e^3 -
1)*e^3 + x*(9*e^6 - 6*e^3 + 1) - 3*e^6*log(5) + e^3*log(5)))/(3*e^3 - 1)^2 - log(x^2*(9*e^6 - 6*e^3 + 1) - 2*(
3*e^6*log(5) - e^3*log(5))*x + (log(5)^2 - 3)*e^6)/(9*e^6 - 6*e^3 + 1)

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mupad [B]  time = 10.04, size = 48, normalized size = 1.85 \begin {gather*} -\ln \left (3\,{\mathrm {e}}^6-{\mathrm {e}}^6\,{\ln \relax (5)}^2+6\,x^2\,{\mathrm {e}}^3-9\,x^2\,{\mathrm {e}}^6-x^2+2\,x\,{\mathrm {e}}^3\,\ln \relax (5)\,\left (3\,{\mathrm {e}}^3-1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - 14*x*exp(3) + 24*x*exp(6) + log(5*exp(x))*(2*exp(3) - 6*exp(6)))/(exp(6)*(16*x^2 - 3) - 8*x^2*exp(
3) + exp(6)*log(5*exp(x))^2 + x^2 + log(5*exp(x))*(2*x*exp(3) - 8*x*exp(6))),x)

[Out]

-log(3*exp(6) - exp(6)*log(5)^2 + 6*x^2*exp(3) - 9*x^2*exp(6) - x^2 + 2*x*exp(3)*log(5)*(3*exp(3) - 1))

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sympy [B]  time = 0.55, size = 49, normalized size = 1.88 \begin {gather*} - \log {\left (x^{2} \left (- 6 e^{3} + 1 + 9 e^{6}\right ) + x \left (- 6 e^{6} \log {\relax (5 )} + 2 e^{3} \log {\relax (5 )}\right ) - 3 e^{6} + e^{6} \log {\relax (5 )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*exp(3)**2-2*exp(3))*ln(5*exp(x))-24*x*exp(3)**2+14*x*exp(3)-2*x)/(exp(3)**2*ln(5*exp(x))**2+(-8*
x*exp(3)**2+2*x*exp(3))*ln(5*exp(x))+(16*x**2-3)*exp(3)**2-8*x**2*exp(3)+x**2),x)

[Out]

-log(x**2*(-6*exp(3) + 1 + 9*exp(6)) + x*(-6*exp(6)*log(5) + 2*exp(3)*log(5)) - 3*exp(6) + exp(6)*log(5)**2)

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