\(\int \frac {e^{\frac {5+20 x^2-5 \log (\frac {4+e^{2 x}}{5 x^2})}{x}} (20+80 x^2+e^{2 x} (5-10 x+20 x^2)+(20+5 e^{2 x}) \log (\frac {4+e^{2 x}}{5 x^2}))}{4 x^2+e^{2 x} x^2} \, dx\) [768]

Optimal result

Integrand size = 96, antiderivative size = 31 \[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx=e^{\frac {5 \left (1+4 x^2-\log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6838} \[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx=5^{5/x} e^{\frac {5 \left (4 x^2+1\right )}{x}} \left (\frac {e^{2 x}+4}{x^2}\right )^{-5/x} \]

[In]

Int[(E^((5 + 20*x^2 - 5*Log[(4 + E^(2*x))/(5*x^2)])/x)*(20 + 80*x^2 + E^(2*x)*(5 - 10*x + 20*x^2) + (20 + 5*E^
(2*x))*Log[(4 + E^(2*x))/(5*x^2)]))/(4*x^2 + E^(2*x)*x^2),x]

[Out]

(5^(5/x)*E^((5*(1 + 4*x^2))/x))/((4 + E^(2*x))/x^2)^(5/x)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 5^{5/x} e^{\frac {5 \left (1+4 x^2\right )}{x}} \left (\frac {4+e^{2 x}}{x^2}\right )^{-5/x} \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx=\int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx \]

[In]

Integrate[(E^((5 + 20*x^2 - 5*Log[(4 + E^(2*x))/(5*x^2)])/x)*(20 + 80*x^2 + E^(2*x)*(5 - 10*x + 20*x^2) + (20
+ 5*E^(2*x))*Log[(4 + E^(2*x))/(5*x^2)]))/(4*x^2 + E^(2*x)*x^2),x]

[Out]

Integrate[(E^((5 + 20*x^2 - 5*Log[(4 + E^(2*x))/(5*x^2)])/x)*(20 + 80*x^2 + E^(2*x)*(5 - 10*x + 20*x^2) + (20
+ 5*E^(2*x))*Log[(4 + E^(2*x))/(5*x^2)]))/(4*x^2 + E^(2*x)*x^2), x]

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\({\mathrm e}^{-\frac {5 \left (-4 x^{2}+\ln \left (\frac {{\mathrm e}^{2 x}+4}{5 x^{2}}\right )-1\right )}{x}}\)	\(26\)
risch	\(x^{\frac {10}{x}} 3125^{\frac {1}{x}} \left ({\mathrm e}^{2 x}+4\right )^{-\frac {5}{x}} {\mathrm e}^{\frac {-\frac {5 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}+4\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}+4\right )}{x^{2}}\right )}^{2}}{2}+\frac {5 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x}+4\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}+4\right )}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{2}-\frac {5 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+5 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-\frac {5 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{2}+\frac {5 i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}+4\right )}{x^{2}}\right )}^{3}}{2}-\frac {5 i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{2 x}+4\right )}{x^{2}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{2}+20 x^{2}+5}{x}}\)	\(196\)

[In]

int(((5*exp(x)^2+20)*ln(1/5*(exp(x)^2+4)/x^2)+(20*x^2-10*x+5)*exp(x)^2+80*x^2+20)*exp((-5*ln(1/5*(exp(x)^2+4)/
x^2)+20*x^2+5)/x)/(exp(x)^2*x^2+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx=e^{\left (\frac {5 \, {\left (4 \, x^{2} - \log \left (\frac {e^{\left (2 \, x\right )} + 4}{5 \, x^{2}}\right ) + 1\right )}}{x}\right )} \]

[In]

integrate(((5*exp(x)^2+20)*log(1/5*(exp(x)^2+4)/x^2)+(20*x^2-10*x+5)*exp(x)^2+80*x^2+20)*exp((-5*log(1/5*(exp(
x)^2+4)/x^2)+20*x^2+5)/x)/(exp(x)^2*x^2+4*x^2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx=\text {Timed out} \]

[In]

integrate(((5*exp(x)**2+20)*ln(1/5*(exp(x)**2+4)/x**2)+(20*x**2-10*x+5)*exp(x)**2+80*x**2+20)*exp((-5*ln(1/5*(
exp(x)**2+4)/x**2)+20*x**2+5)/x)/(exp(x)**2*x**2+4*x**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx=e^{\left (20 \, x + \frac {5 \, \log \left (5\right )}{x} + \frac {10 \, \log \left (x\right )}{x} - \frac {5 \, \log \left (e^{\left (2 \, x\right )} + 4\right )}{x} + \frac {5}{x}\right )} \]

[In]

integrate(((5*exp(x)^2+20)*log(1/5*(exp(x)^2+4)/x^2)+(20*x^2-10*x+5)*exp(x)^2+80*x^2+20)*exp((-5*log(1/5*(exp(
x)^2+4)/x^2)+20*x^2+5)/x)/(exp(x)^2*x^2+4*x^2),x, algorithm="maxima")

[Out]

e^(20*x + 5*log(5)/x + 10*log(x)/x - 5*log(e^(2*x) + 4)/x + 5/x)

Giac [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx=e^{\left (20 \, x - \frac {5 \, \log \left (\frac {e^{\left (2 \, x\right )}}{5 \, x^{2}} + \frac {4}{5 \, x^{2}}\right )}{x} + \frac {5}{x}\right )} \]

[In]

integrate(((5*exp(x)^2+20)*log(1/5*(exp(x)^2+4)/x^2)+(20*x^2-10*x+5)*exp(x)^2+80*x^2+20)*exp((-5*log(1/5*(exp(
x)^2+4)/x^2)+20*x^2+5)/x)/(exp(x)^2*x^2+4*x^2),x, algorithm="giac")

[Out]

e^(20*x - 5*log(1/5*e^(2*x)/x^2 + 4/5/x^2)/x + 5/x)

Mupad [B] (verification not implemented)

Time = 7.84 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {5+20 x^2-5 \log \left (\frac {4+e^{2 x}}{5 x^2}\right )}{x}} \left (20+80 x^2+e^{2 x} \left (5-10 x+20 x^2\right )+\left (20+5 e^{2 x}\right ) \log \left (\frac {4+e^{2 x}}{5 x^2}\right )\right )}{4 x^2+e^{2 x} x^2} \, dx={\mathrm {e}}^{20\,x+\frac {5}{x}}\,{\left (\frac {3125\,x^{10}}{{\left ({\mathrm {e}}^{2\,x}+4\right )}^5}\right )}^{1/x} \]

[In]

int((exp((20*x^2 - 5*log((exp(2*x)/5 + 4/5)/x^2) + 5)/x)*(exp(2*x)*(20*x^2 - 10*x + 5) + 80*x^2 + log((exp(2*x
)/5 + 4/5)/x^2)*(5*exp(2*x) + 20) + 20))/(x^2*exp(2*x) + 4*x^2),x)

[Out]

exp(20*x + 5/x)*((3125*x^10)/(exp(2*x) + 4)^5)^(1/x)