∫ −12−300e5−x2 −24x2 ____________________________________________

Optimal. Leaf size=20 \[ \log \left (-4 \left (3+\frac {3}{25} e^{-5+x^2}\right ) x+\log (5)\right ) \]

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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 {-12-300 e^{5-x^2}-24 x^2}{-12 x+25 e^{5-x^2} (-12 x+\log (5))} \, dx \end {gather*}

Int[(-12 - 300*E^(5 - x^2) - 24*x^2)/(-12*x + 25*E^(5 - x^2)*(-12*x + Log[5])),x]

x^2 + Log[x] + 25*E^5*Log[5]*Defer[Int][1/(x*(300*E^5*x + 12*E^x^2*x - 25*E^5*Log[5])), x] + 50*E^5*Log[5]*Def

er[Int][x/(300*E^5*x + 12*E^x^2*x - 25*E^5*Log[5]), x] - 600*E^5*Defer[Int][x^2/(300*E^5*x + 12*E^x^2*x - 25*E

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1+2 x^2}{x}-\frac {25 e^5 \left (24 x^3-\log (5)-2 x^2 \log (5)\right )}{x \left (300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)\right )}\right ) \, dx\\ &=-\left (\left (25 e^5\right ) \int \frac {24 x^3-\log (5)-2 x^2 \log (5)}{x \left (300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)\right )} \, dx\right )+\int \frac {1+2 x^2}{x} \, dx\\ &=-\left (\left (25 e^5\right ) \int \left (\frac {24 x^2}{300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)}-\frac {\log (5)}{x \left (300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)\right )}-\frac {2 x \log (5)}{300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)}\right ) \, dx\right )+\int \left (\frac {1}{x}+2 x\right ) \, dx\\ &=x^2+\log (x)-\left (600 e^5\right ) \int \frac {x^2}{300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)} \, dx+\left (25 e^5 \log (5)\right ) \int \frac {1}{x \left (300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)\right )} \, dx+\left (50 e^5 \log (5)\right ) \int \frac {x}{300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.07, size = 23, normalized size = 1.15 \begin {gather*} \log \left (300 e^5 x+12 e^{x^2} x-25 e^5 \log (5)\right ) \end {gather*}

Integrate[(-12 - 300*E^(5 - x^2) - 24*x^2)/(-12*x + 25*E^(5 - x^2)*(-12*x + Log[5])),x]

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method	result	size

norman	\(x^{2}+\ln \left ({\mathrm e}^{2 \ln \left (5\right )-x^{2}+5} \ln \left (5\right )-12 x \,{\mathrm e}^{2 \ln \left (5\right )-x^{2}+5}-12 x \right )\)	\(40\)
risch	\(\ln \left (-\ln \left (5\right )+12 x \right )+x^{2}-2 \ln \left (5\right )-5+\ln \left (25 \,{\mathrm e}^{-x^{2}+5}-\frac {12 x}{\ln \left (5\right )-12 x}\right )\)	\(42\)

int((-12*exp(2*ln(5)-x^2+5)-24*x^2-12)/((ln(5)-12*x)*exp(2*ln(5)-x^2+5)-12*x),x,method=_RETURNVERBOSE)

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Maxima [A]
time = 0.53, size = 28, normalized size = 1.40 \begin {gather*} \log \left (x\right ) + \log \left (\frac {300 \, x e^{5} + 12 \, x e^{\left (x^{2}\right )} - 25 \, e^{5} \log \left (5\right )}{12 \, x}\right ) \end {gather*}

integrate((-12*exp(2*log(5)-x^2+5)-24*x^2-12)/((log(5)-12*x)*exp(2*log(5)-x^2+5)-12*x),x, algorithm="maxima")

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (19) = 38\).
time = 0.39, size = 51, normalized size = 2.55 \begin {gather*} x^{2} + \log \left (12 \, x - \log \left (5\right )\right ) + \log \left (-\frac {{\left (12 \, x - \log \left (5\right )\right )} e^{\left (-x^{2} + 2 \, \log \left (5\right ) + 5\right )} + 12 \, x}{12 \, x - \log \left (5\right )}\right ) \end {gather*}

integrate((-12*exp(2*log(5)-x^2+5)-24*x^2-12)/((log(5)-12*x)*exp(2*log(5)-x^2+5)-12*x),x, algorithm="fricas")

x^2 + log(12*x - log(5)) + log(-((12*x - log(5))*e^(-x^2 + 2*log(5) + 5) + 12*x)/(12*x - log(5)))

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Sympy [A]
time = 0.19, size = 31, normalized size = 1.55 \begin {gather*} x^{2} + \log {\left (12 x - \log {\left (5 \right )} \right )} + \log {\left (\frac {12 x}{300 x - 25 \log {\left (5 \right )}} + e^{5 - x^{2}} \right )} \end {gather*}

integrate((-12*exp(2*ln(5)-x**2+5)-24*x**2-12)/((ln(5)-12*x)*exp(2*ln(5)-x**2+5)-12*x),x)

x**2 + log(12*x - log(5)) + log(12*x/(300*x - 25*log(5)) + exp(5 - x**2))

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Giac [A]
time = 0.40, size = 32, normalized size = 1.60 \begin {gather*} x^{2} + \log \left (300 \, x e^{\left (-x^{2} + 5\right )} - 25 \, e^{\left (-x^{2} + 5\right )} \log \left (5\right ) + 12 \, x\right ) \end {gather*}

integrate((-12*exp(2*log(5)-x^2+5)-24*x^2-12)/((log(5)-12*x)*exp(2*log(5)-x^2+5)-12*x),x, algorithm="giac")

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Mupad [B]
time = 0.20, size = 32, normalized size = 1.60 \begin {gather*} \ln \left (12\,x-25\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x^2}\,\ln \left (5\right )+300\,x\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x^2}\right )+x^2 \end {gather*}

int((12*exp(2*log(5) - x^2 + 5) + 24*x^2 + 12)/(12*x + exp(2*log(5) - x^2 + 5)*(12*x - log(5))),x)

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3.19.21 \(\int \frac {-12-300 e^{5-x^2}-24 x^2}{-12 x+25 e^{5-x^2} (-12 x+\log (5))} \, dx\) [1821]