\(\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} (20+15 x^4 \log (2)))}{5 x^3 \log (2)} \, dx\) [6631]

Optimal result

Integrand size = 72, antiderivative size = 29 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {1}{5 x}+\frac {e^{e^{x^3-\frac {4}{x \log (2)}}}}{x} \]

[Out]

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

Rubi [F]

\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx \]

[In]

Int[(-(x*Log[2]) + E^E^((-4 + x^4*Log[2])/(x*Log[2]))*(-5*x*Log[2] + E^((-4 + x^4*Log[2])/(x*Log[2]))*(20 + 15
*x^4*Log[2])))/(5*x^3*Log[2]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{x^3} \, dx}{5 \log (2)} \\ & = \frac {\int \left (-\frac {\left (1+5 e^{e^{x^3-\frac {4}{x \log (2)}}}\right ) \log (2)}{x^2}+\frac {5 e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} \left (4+x^4 \log (8)\right )}{x^3}\right ) \, dx}{5 \log (2)} \\ & = -\left (\frac {1}{5} \int \frac {1+5 e^{e^{x^3-\frac {4}{x \log (2)}}}}{x^2} \, dx\right )+\frac {\int \frac {e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} \left (4+x^4 \log (8)\right )}{x^3} \, dx}{\log (2)} \\ & = -\left (\frac {1}{5} \int \left (\frac {1}{x^2}+\frac {5 e^{e^{\frac {x^4-\frac {4}{\log (2)}}{x}}}}{x^2}\right ) \, dx\right )+\frac {\int \left (\frac {4 e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}}}{x^3}+e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} x \log (8)\right ) \, dx}{\log (2)} \\ & = \frac {1}{5 x}+\frac {4 \int \frac {e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}}}{x^3} \, dx}{\log (2)}+\frac {\log (8) \int e^{e^{x^3-\frac {4}{x \log (2)}}+x^3-\frac {4}{x \log (2)}} x \, dx}{\log (2)}-\int \frac {e^{e^{\frac {x^4-\frac {4}{\log (2)}}{x}}}}{x^2} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx \]

[In]

Integrate[(-(x*Log[2]) + E^E^((-4 + x^4*Log[2])/(x*Log[2]))*(-5*x*Log[2] + E^((-4 + x^4*Log[2])/(x*Log[2]))*(2
0 + 15*x^4*Log[2])))/(5*x^3*Log[2]),x]

[Out]

Integrate[(-(x*Log[2]) + E^E^((-4 + x^4*Log[2])/(x*Log[2]))*(-5*x*Log[2] + E^((-4 + x^4*Log[2])/(x*Log[2]))*(2
0 + 15*x^4*Log[2])))/x^3, x]/(5*Log[2])

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {x \,{\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \left (2\right )-4}{x \ln \left (2\right )}}}+\frac {x}{5}}{x^{2}}\)	\(29\)
risch	\(\frac {1}{5 x}+\frac {{\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \left (2\right )-4}{x \ln \left (2\right )}}}}{x}\)	\(29\)
parallelrisch	\(\frac {5 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{4} \ln \left (2\right )-4}{x \ln \left (2\right )}}}+\ln \left (2\right )}{5 \ln \left (2\right ) x}\)	\(35\)

[In]

int(1/5*(((15*x^4*ln(2)+20)*exp((x^4*ln(2)-4)/x/ln(2))-5*x*ln(2))*exp(exp((x^4*ln(2)-4)/x/ln(2)))-x*ln(2))/x^3
/ln(2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {5 \, e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} + 1}{5 \, x} \]

[In]

integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2))*exp(exp((x^4*log(2)-4)/x/log(2)))-
x*log(2))/x^3/log(2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {e^{e^{\frac {x^{4} \log {\left (2 \right )} - 4}{x \log {\left (2 \right )}}}}}{x} + \frac {1}{5 x} \]

[In]

integrate(1/5*(((15*x**4*ln(2)+20)*exp((x**4*ln(2)-4)/x/ln(2))-5*x*ln(2))*exp(exp((x**4*ln(2)-4)/x/ln(2)))-x*l
n(2))/x**3/ln(2),x)

[Out]

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

Maxima [F]

\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int { \frac {5 \, {\left ({\left (3 \, x^{4} \log \left (2\right ) + 4\right )} e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )} - x \log \left (2\right )\right )} e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} - x \log \left (2\right )}{5 \, x^{3} \log \left (2\right )} \,d x } \]

[In]

integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2))*exp(exp((x^4*log(2)-4)/x/log(2)))-
x*log(2))/x^3/log(2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\int { \frac {5 \, {\left ({\left (3 \, x^{4} \log \left (2\right ) + 4\right )} e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )} - x \log \left (2\right )\right )} e^{\left (e^{\left (\frac {x^{4} \log \left (2\right ) - 4}{x \log \left (2\right )}\right )}\right )} - x \log \left (2\right )}{5 \, x^{3} \log \left (2\right )} \,d x } \]

[In]

integrate(1/5*(((15*x^4*log(2)+20)*exp((x^4*log(2)-4)/x/log(2))-5*x*log(2))*exp(exp((x^4*log(2)-4)/x/log(2)))-
x*log(2))/x^3/log(2),x, algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 11.62 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-x \log (2)+e^{e^{\frac {-4+x^4 \log (2)}{x \log (2)}}} \left (-5 x \log (2)+e^{\frac {-4+x^4 \log (2)}{x \log (2)}} \left (20+15 x^4 \log (2)\right )\right )}{5 x^3 \log (2)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-\frac {4}{x\,\ln \left (2\right )}}}+\frac {1}{5}}{x} \]

[In]

int(-((x*log(2))/5 + (exp(exp((x^4*log(2) - 4)/(x*log(2))))*(5*x*log(2) - exp((x^4*log(2) - 4)/(x*log(2)))*(15
*x^4*log(2) + 20)))/5)/(x^3*log(2)),x)

[Out]

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