∫ −1−x+e3+2x+4e2xx2 (40x2+40x3)−log(4)+x log( 5 x) ___________________________________________________________________

Optimal. Leaf size=30 \[ e^{3+4 e^{2 x} x^2}+\frac {1}{5} (1+x+\log (4)) \log \left (\frac {5}{x}\right ) \]

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Rubi [F] time = 0.29, 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 {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx \end {gather*}

Int[(-1 - x + E^(3 + 2*x + 4*E^(2*x)*x^2)*(40*x^2 + 40*x^3) - Log[4] + x*Log[5/x])/(5*x),x]

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

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (40 e^{3+2 x+4 e^{2 x} x^2} x (1+x)+\frac {-1-x-\log (4)+x \log \left (\frac {5}{x}\right )}{x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-1-x-\log (4)+x \log \left (\frac {5}{x}\right )}{x} \, dx+8 \int e^{3+2 x+4 e^{2 x} x^2} x (1+x) \, dx\\ &=\frac {1}{5} \int \left (\frac {-1-x-\log (4)}{x}+\log \left (\frac {5}{x}\right )\right ) \, dx+8 \int \left (e^{3+2 x+4 e^{2 x} x^2} x+e^{3+2 x+4 e^{2 x} x^2} x^2\right ) \, dx\\ &=\frac {1}{5} \int \frac {-1-x-\log (4)}{x} \, dx+\frac {1}{5} \int \log \left (\frac {5}{x}\right ) \, dx+8 \int e^{3+2 x+4 e^{2 x} x^2} x \, dx+8 \int e^{3+2 x+4 e^{2 x} x^2} x^2 \, dx\\ &=\frac {x}{5}+\frac {1}{5} x \log \left (\frac {5}{x}\right )+\frac {1}{5} \int \left (-1+\frac {-1-\log (4)}{x}\right ) \, dx+8 \int e^{3+2 x+4 e^{2 x} x^2} x \, dx+8 \int e^{3+2 x+4 e^{2 x} x^2} x^2 \, dx\\ &=\frac {1}{5} x \log \left (\frac {5}{x}\right )-\frac {1}{5} (1+\log (4)) \log (x)+8 \int e^{3+2 x+4 e^{2 x} x^2} x \, dx+8 \int e^{3+2 x+4 e^{2 x} x^2} x^2 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.15, size = 37, normalized size = 1.23 \begin {gather*} \frac {1}{5} \left (5 e^{3+4 e^{2 x} x^2}+x \log \left (\frac {5}{x}\right )-(1+\log (4)) \log (x)\right ) \end {gather*}

Integrate[(-1 - x + E^(3 + 2*x + 4*E^(2*x)*x^2)*(40*x^2 + 40*x^3) - Log[4] + x*Log[5/x])/(5*x),x]

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fricas [A] time = 0.50, size = 42, normalized size = 1.40 \begin {gather*} \frac {1}{5} \, {\left ({\left (x + 2 \, \log \relax (2) + 1\right )} e^{\left (2 \, x\right )} \log \left (\frac {5}{x}\right ) + 5 \, e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 2 \, x + 3\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

integrate(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*log(5/x)-2*log(2)-x-1)/x,x, algorithm="fricas"

1/5*((x + 2*log(2) + 1)*e^(2*x)*log(5/x) + 5*e^(4*x^2*e^(2*x) + 2*x + 3))*e^(-2*x)

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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {40 \, {\left (x^{3} + x^{2}\right )} e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 2 \, x + 3\right )} + x \log \left (\frac {5}{x}\right ) - x - 2 \, \log \relax (2) - 1}{5 \, x}\,{d x} \end {gather*}

integrate(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*log(5/x)-2*log(2)-x-1)/x,x, algorithm="giac")

integrate(1/5*(40*(x^3 + x^2)*e^(4*x^2*e^(2*x) + 2*x + 3) + x*log(5/x) - x - 2*log(2) - 1)/x, x)

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method	result	size

default	\(-\frac {2 \ln \relax (2) \ln \relax (x )}{5}-\frac {\ln \relax (x )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}+\frac {x \ln \left (\frac {5}{x}\right )}{5}\)	\(33\)
risch	\(-\frac {x \ln \relax (x )}{5}+\frac {x \ln \relax (5)}{5}-\frac {2 \ln \relax (2) \ln \relax (x )}{5}-\frac {\ln \relax (x )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}\)	\(34\)

int(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*ln(5/x)-2*ln(2)-x-1)/x,x,method=_RETURNVERBOSE)

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maxima [A] time = 0.46, size = 32, normalized size = 1.07 \begin {gather*} -\frac {2}{5} \, \log \relax (2) \log \relax (x) + \frac {1}{5} \, x \log \left (\frac {5}{x}\right ) + e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 3\right )} - \frac {1}{5} \, \log \relax (x) \end {gather*}

integrate(1/5*((40*x^3+40*x^2)*exp(x)^2*exp(4*exp(x)^2*x^2+3)+x*log(5/x)-2*log(2)-x-1)/x,x, algorithm="maxima"

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mupad [B] time = 4.14, size = 32, normalized size = 1.07 \begin {gather*} {\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^{2\,x}+3}-\ln \relax (x)\,\left (\frac {\ln \relax (4)}{5}+\frac {1}{5}\right )+\frac {x\,\ln \left (\frac {5}{x}\right )}{5} \end {gather*}

int(-(x/5 + (2*log(2))/5 - (x*log(5/x))/5 - (exp(4*x^2*exp(2*x) + 3)*exp(2*x)*(40*x^2 + 40*x^3))/5 + 1/5)/x,x)

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sympy [A] time = 0.42, size = 36, normalized size = 1.20 \begin {gather*} \frac {x \log {\left (\frac {5}{x} \right )}}{5} + e^{4 x^{2} e^{2 x} + 3} + \left (- \frac {2 \log {\relax (2 )}}{5} - \frac {1}{5}\right ) \log {\relax (x )} \end {gather*}

integrate(1/5*((40*x**3+40*x**2)*exp(x)**2*exp(4*exp(x)**2*x**2+3)+x*ln(5/x)-2*ln(2)-x-1)/x,x)

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3.68.52 \(\int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} (40 x^2+40 x^3)-\log (4)+x \log (\frac {5}{x})}{5 x} \, dx\)