3.86.14 \(\int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} (-5 x^2+2 x^3-2 x \log (5))}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} (x^2+2 x^2 \log (5))} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\frac {8 \left (\frac {x}{e^{(5-x) x}+x}+\log (5)\right )}{x}\right ) \]

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Rubi [F]  time = 7.37, 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^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-x^2 - E^(10*x - 2*x^2)*Log[5] - x^2*Log[5] + E^(5*x - x^2)*(-5*x^2 + 2*x^3 - 2*x*Log[5]))/(x^3 + E^(10*x
- 2*x^2)*x*Log[5] + x^3*Log[5] + E^(5*x - x^2)*(x^2 + 2*x^2*Log[5])),x]

[Out]

(E^(-5*x + x^2)*(5*x - 2*x^2))/((5 - 2*x)*Log[5]) - Log[x] + Defer[Int][(E^(2*x^2)*x)/(E^(10*x) + E^(x*(5 + x)
)*x), x] - 5*Defer[Int][(E^(2*x^2)*x^2)/(E^(10*x) + E^(x*(5 + x))*x), x] + 2*Defer[Int][(E^(2*x^2)*x^3)/(E^(10
*x) + E^(x*(5 + x))*x), x] + ((1 + Log[5])^2*Defer[Int][(E^(-5*x + 2*x^2)*x)/(-(E^(5*x)*Log[5]) - E^x^2*x*(1 +
Log[5])), x])/Log[5] + (2*(1 + Log[5])^2*Defer[Int][(E^(-5*x + 2*x^2)*x^3)/(-(E^(5*x)*Log[5]) - E^x^2*x*(1 +
Log[5])), x])/Log[5] + (5*(1 + Log[5])^2*Defer[Int][(E^(-5*x + 2*x^2)*x^2)/(E^(5*x)*Log[5] + E^x^2*x*(1 + Log[
5])), x])/Log[5]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx\\ &=\int \frac {x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{e^{10 x-2 x^2} x \log (5)+x^3 (1+\log (5))+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx\\ &=\int \frac {e^{2 x^2} \left (x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )\right )}{x \left (e^{5 x}+e^{x^2} x\right ) \left (e^{5 x} \log (5)+e^{x^2} x (1+\log (5))\right )} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {e^{2 x^2} x \left (1-5 x+2 x^2\right )}{e^{10 x}+e^{x (5+x)} x}+\frac {e^{-5 x+x^2} \left (1-5 x+2 x^2\right )}{\log (5)}+\frac {e^{-5 x+2 x^2} x \left (-1+5 x-2 x^2\right ) (1+\log (5))^2}{\log (5) \left (e^{5 x} \log (5)+e^{x^2} x (1+\log (5))\right )}\right ) \, dx\\ &=-\log (x)+\frac {\int e^{-5 x+x^2} \left (1-5 x+2 x^2\right ) \, dx}{\log (5)}+\frac {(1+\log (5))^2 \int \frac {e^{-5 x+2 x^2} x \left (-1+5 x-2 x^2\right )}{e^{5 x} \log (5)+e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\int \frac {e^{2 x^2} x \left (1-5 x+2 x^2\right )}{e^{10 x}+e^{x (5+x)} x} \, dx\\ &=\frac {e^{-5 x+x^2} \left (5 x-2 x^2\right )}{(5-2 x) \log (5)}-\log (x)+\frac {(1+\log (5))^2 \int \left (\frac {e^{-5 x+2 x^2} x}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))}+\frac {2 e^{-5 x+2 x^2} x^3}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))}+\frac {5 e^{-5 x+2 x^2} x^2}{e^{5 x} \log (5)+e^{x^2} x (1+\log (5))}\right ) \, dx}{\log (5)}+\int \left (\frac {e^{2 x^2} x}{e^{10 x}+e^{x (5+x)} x}-\frac {5 e^{2 x^2} x^2}{e^{10 x}+e^{x (5+x)} x}+\frac {2 e^{2 x^2} x^3}{e^{10 x}+e^{x (5+x)} x}\right ) \, dx\\ &=\frac {e^{-5 x+x^2} \left (5 x-2 x^2\right )}{(5-2 x) \log (5)}-\log (x)+2 \int \frac {e^{2 x^2} x^3}{e^{10 x}+e^{x (5+x)} x} \, dx-5 \int \frac {e^{2 x^2} x^2}{e^{10 x}+e^{x (5+x)} x} \, dx+\frac {(1+\log (5))^2 \int \frac {e^{-5 x+2 x^2} x}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\frac {\left (2 (1+\log (5))^2\right ) \int \frac {e^{-5 x+2 x^2} x^3}{-e^{5 x} \log (5)-e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\frac {\left (5 (1+\log (5))^2\right ) \int \frac {e^{-5 x+2 x^2} x^2}{e^{5 x} \log (5)+e^{x^2} x (1+\log (5))} \, dx}{\log (5)}+\int \frac {e^{2 x^2} x}{e^{10 x}+e^{x (5+x)} x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 47, normalized size = 1.96 \begin {gather*} -\log (x)-\log \left (e^{5 x}+e^{x^2} x\right )+\log \left (e^{x^2} x+e^{5 x} \log (5)+e^{x^2} x \log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 - E^(10*x - 2*x^2)*Log[5] - x^2*Log[5] + E^(5*x - x^2)*(-5*x^2 + 2*x^3 - 2*x*Log[5]))/(x^3 + E
^(10*x - 2*x^2)*x*Log[5] + x^3*Log[5] + E^(5*x - x^2)*(x^2 + 2*x^2*Log[5])),x]

[Out]

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

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fricas [A]  time = 0.74, size = 40, normalized size = 1.67 \begin {gather*} \log \left (x \log \relax (5) + e^{\left (-x^{2} + 5 \, x\right )} \log \relax (5) + x\right ) - \log \left (x + e^{\left (-x^{2} + 5 \, x\right )}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 44, normalized size = 1.83 \begin {gather*} \log \left (-x \log \relax (5) - e^{\left (-x^{2} + 5 \, x\right )} \log \relax (5) - x\right ) - \log \left (x + e^{\left (-x^{2} + 5 \, x\right )}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 37, normalized size = 1.54

method	result	size
risch	\(-\ln \relax (x )+\ln \left ({\mathrm e}^{-\left (x -5\right ) x}+\frac {\left (\ln \relax (5)+1\right ) x}{\ln \relax (5)}\right )-\ln \left (x +{\mathrm e}^{-\left (x -5\right ) x}\right )\)	\(37\)
norman	\(-\ln \relax (x )-\ln \left (x +{\mathrm e}^{-x^{2}+5 x}\right )+\ln \left (x \ln \relax (5)+{\mathrm e}^{-x^{2}+5 x} \ln \relax (5)+x \right )\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(x)+ln(exp(-(x-5)*x)+(ln(5)+1)*x/ln(5))-ln(x+exp(-(x-5)*x))

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maxima [B]  time = 0.50, size = 52, normalized size = 2.17 \begin {gather*} -\log \relax (x) - \log \left (\frac {x e^{\left (x^{2}\right )} + e^{\left (5 \, x\right )}}{x}\right ) + \log \left (\frac {x {\left (\log \relax (5) + 1\right )} e^{\left (x^{2}\right )} + e^{\left (5 \, x\right )} \log \relax (5)}{x {\left (\log \relax (5) + 1\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.13, size = 34, normalized size = 1.42 \begin {gather*} \ln \left (x+x\,\ln \relax (5)+{\mathrm {e}}^{-x\,\left (x-5\right )}\,\ln \relax (5)\right )-\ln \left (x+{\mathrm {e}}^{-x\,\left (x-5\right )}\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5*x - x^2)*(2*x*log(5) + 5*x^2 - 2*x^3) + exp(10*x - 2*x^2)*log(5) + x^2*log(5) + x^2)/(exp(5*x - x^
2)*(2*x^2*log(5) + x^2) + x^3*log(5) + x^3 + x*exp(10*x - 2*x^2)*log(5)),x)

[Out]

log(x + x*log(5) + exp(-x*(x - 5))*log(5)) - log(x + exp(-x*(x - 5))) - log(x)

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sympy [B]  time = 0.45, size = 39, normalized size = 1.62 \begin {gather*} - \log {\relax (x )} - \log {\left (x + e^{- x^{2} + 5 x} \right )} + \log {\left (\frac {2 x + 2 x \log {\relax (5 )}}{2 \log {\relax (5 )}} + e^{- x^{2} + 5 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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