\(\int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} (40 x^2+40 x^3)-\log (4)+x \log (\frac {5}{x})}{5 x} \, dx\) [6752]

Optimal result

Integrand size = 53, antiderivative size = 30 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=e^{3+4 e^{2 x} x^2}+\frac {1}{5} (1+x+\log (4)) \log \left (\frac {5}{x}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx \]

[In]

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]

[Out]

(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
+ 2*x + 4*E^(2*x)*x^2)*x^2, x]

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\frac {1}{5} \left (5 e^{3+4 e^{2 x} x^2}+x \log \left (\frac {5}{x}\right )-(1+\log (4)) \log (x)\right ) \]

[In]

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]

[Out]

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

Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
default	\(-\frac {\left (1+2 \ln \left (2\right )\right ) \ln \left (x \right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}+\frac {x \ln \left (\frac {5}{x}\right )}{5}\)	\(33\)
parts	\(-\frac {\left (1+2 \ln \left (2\right )\right ) \ln \left (x \right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}+\frac {x \ln \left (\frac {5}{x}\right )}{5}\)	\(33\)
risch	\(-\frac {x \ln \left (x \right )}{5}+\frac {x \ln \left (5\right )}{5}-\frac {2 \ln \left (2\right ) \ln \left (x \right )}{5}-\frac {\ln \left (x \right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}\)	\(34\)
parallelrisch	\(\frac {2 \ln \left (\frac {5}{x}\right ) \ln \left (2\right )}{5}+\frac {x \ln \left (\frac {5}{x}\right )}{5}+\frac {\ln \left (\frac {5}{x}\right )}{5}+{\mathrm e}^{4 \,{\mathrm e}^{2 x} x^{2}+3}\)	\(41\)

[In]

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)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\frac {1}{5} \, {\left ({\left (x + 2 \, \log \left (2\right ) + 1\right )} e^{\left (2 \, x\right )} \log \left (\frac {5}{x}\right ) + 5 \, e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 2 \, x + 3\right )}\right )} e^{\left (-2 \, x\right )} \]

[In]

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"
)

[Out]

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)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\frac {x \log {\left (\frac {5}{x} \right )}}{5} + e^{4 x^{2} e^{2 x} + 3} + \left (- \frac {2 \log {\left (2 \right )}}{5} - \frac {1}{5}\right ) \log {\left (x \right )} \]

[In]

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)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=-\frac {2}{5} \, \log \left (2\right ) \log \left (x\right ) + \frac {1}{5} \, x \log \left (\frac {5}{x}\right ) + e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 3\right )} - \frac {1}{5} \, \log \left (x\right ) \]

[In]

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"
)

[Out]

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

Giac [F]

\[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx=\int { \frac {40 \, {\left (x^{3} + x^{2}\right )} e^{\left (4 \, x^{2} e^{\left (2 \, x\right )} + 2 \, x + 3\right )} + x \log \left (\frac {5}{x}\right ) - x - 2 \, \log \left (2\right ) - 1}{5 \, x} \,d x } \]

[In]

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")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 12.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-1-x+e^{3+2 x+4 e^{2 x} x^2} \left (40 x^2+40 x^3\right )-\log (4)+x \log \left (\frac {5}{x}\right )}{5 x} \, dx={\mathrm {e}}^{4\,x^2\,{\mathrm {e}}^{2\,x}+3}-\ln \left (x\right )\,\left (\frac {\ln \left (4\right )}{5}+\frac {1}{5}\right )+\frac {x\,\ln \left (\frac {5}{x}\right )}{5} \]

[In]

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)

[Out]

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