\(\int \frac {-4 \log ^2(4 x)+e^{2 e^x} (1+(1-2 e^x x) \log (4 x))}{3 x^2 \log ^2(4 x)} \, dx\) [7873]

Optimal result

Integrand size = 45, antiderivative size = 24 \[ \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{3 x^2 \log ^2(4 x)} \, dx=\frac {4-\frac {e^{2 e^x}}{\log (4 x)}}{3 x} \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

4/(3*x) + Defer[Int][E^(2*E^x)/(x^2*Log[4*x]^2), x]/3 + Defer[Int][E^(2*E^x)/(x^2*Log[4*x]), x]/3 - (2*Defer[I
nt][E^(2*E^x + x)/(x*Log[4*x]), x])/3

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{x^2 \log ^2(4 x)} \, dx \\ & = \frac {1}{3} \int \left (-\frac {2 e^{2 e^x+x}}{x \log (4 x)}-\frac {-e^{2 e^x}-e^{2 e^x} \log (4 x)+4 \log ^2(4 x)}{x^2 \log ^2(4 x)}\right ) \, dx \\ & = -\left (\frac {1}{3} \int \frac {-e^{2 e^x}-e^{2 e^x} \log (4 x)+4 \log ^2(4 x)}{x^2 \log ^2(4 x)} \, dx\right )-\frac {2}{3} \int \frac {e^{2 e^x+x}}{x \log (4 x)} \, dx \\ & = -\left (\frac {1}{3} \int \left (\frac {4}{x^2}-\frac {e^{2 e^x} (1+\log (4 x))}{x^2 \log ^2(4 x)}\right ) \, dx\right )-\frac {2}{3} \int \frac {e^{2 e^x+x}}{x \log (4 x)} \, dx \\ & = \frac {4}{3 x}+\frac {1}{3} \int \frac {e^{2 e^x} (1+\log (4 x))}{x^2 \log ^2(4 x)} \, dx-\frac {2}{3} \int \frac {e^{2 e^x+x}}{x \log (4 x)} \, dx \\ & = \frac {4}{3 x}+\frac {1}{3} \int \left (\frac {e^{2 e^x}}{x^2 \log ^2(4 x)}+\frac {e^{2 e^x}}{x^2 \log (4 x)}\right ) \, dx-\frac {2}{3} \int \frac {e^{2 e^x+x}}{x \log (4 x)} \, dx \\ & = \frac {4}{3 x}+\frac {1}{3} \int \frac {e^{2 e^x}}{x^2 \log ^2(4 x)} \, dx+\frac {1}{3} \int \frac {e^{2 e^x}}{x^2 \log (4 x)} \, dx-\frac {2}{3} \int \frac {e^{2 e^x+x}}{x \log (4 x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{3 x^2 \log ^2(4 x)} \, dx=\frac {4-\frac {e^{2 e^x}}{\log (4 x)}}{3 x} \]

[In]

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

[Out]

(4 - E^(2*E^x)/Log[4*x])/(3*x)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {4}{3 x}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{3 \ln \left (4 x \right ) x}\)	\(23\)
parallelrisch	\(-\frac {-4 \ln \left (4 x \right )+{\mathrm e}^{2 \,{\mathrm e}^{x}}}{3 x \ln \left (4 x \right )}\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{3 x^2 \log ^2(4 x)} \, dx=-\frac {e^{\left (2 \, e^{x}\right )} - 4 \, \log \left (4 \, x\right )}{3 \, x \log \left (4 \, x\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{3 x^2 \log ^2(4 x)} \, dx=- \frac {e^{2 e^{x}}}{3 x \log {\left (4 x \right )}} + \frac {4}{3 x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{3 x^2 \log ^2(4 x)} \, dx=-\frac {e^{\left (2 \, e^{x}\right )}}{3 \, {\left (2 \, x \log \left (2\right ) + x \log \left (x\right )\right )}} + \frac {4}{3 \, x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{3 x^2 \log ^2(4 x)} \, dx=-\frac {e^{\left (2 \, e^{x}\right )} - 4 \, \log \left (4 \, x\right )}{3 \, x \log \left (4 \, x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-4 \log ^2(4 x)+e^{2 e^x} \left (1+\left (1-2 e^x x\right ) \log (4 x)\right )}{3 x^2 \log ^2(4 x)} \, dx=\frac {4}{3\,x}-\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^x}}{3\,x\,\ln \left (4\,x\right )} \]

[In]

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

[Out]

4/(3*x) - exp(2*exp(x))/(3*x*log(4*x))