\(\int \frac {20 e^{2 x} x+4 x^2+(8+5 e^{2 x}+x^2) \log (\frac {1}{5} (8+5 e^{2 x}+x^2))}{(8 x+5 e^{2 x} x+x^3) \log (\frac {1}{5} (8+5 e^{2 x}+x^2))} \, dx\) [2278]

Optimal result

Integrand size = 81, antiderivative size = 21 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x \log ^2\left (e^{2 x}+\frac {1}{5} \left (8+x^2\right )\right )\right ) \]

[Out]

ln(x*ln(exp(x)^2+1/5*x^2+8/5)^2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6820, 6816} \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=2 \log \left (\log \left (\frac {1}{5} \left (x^2+5 e^{2 x}+8\right )\right )\right )+\log (x) \]

[In]

Int[(20*E^(2*x)*x + 4*x^2 + (8 + 5*E^(2*x) + x^2)*Log[(8 + 5*E^(2*x) + x^2)/5])/((8*x + 5*E^(2*x)*x + x^3)*Log
[(8 + 5*E^(2*x) + x^2)/5]),x]

[Out]

Log[x] + 2*Log[Log[(8 + 5*E^(2*x) + x^2)/5]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log (x)+2 \log \left (\log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )\right ) \]

[In]

Integrate[(20*E^(2*x)*x + 4*x^2 + (8 + 5*E^(2*x) + x^2)*Log[(8 + 5*E^(2*x) + x^2)/5])/((8*x + 5*E^(2*x)*x + x^
3)*Log[(8 + 5*E^(2*x) + x^2)/5]),x]

[Out]

Log[x] + 2*Log[Log[(8 + 5*E^(2*x) + x^2)/5]]

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\(\ln \left (x \right )+2 \ln \left (\ln \left ({\mathrm e}^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )\right )\)	\(19\)
parallelrisch	\(\ln \left (x \right )+2 \ln \left (\ln \left ({\mathrm e}^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )\right )\)	\(19\)

[In]

int(((5*exp(x)^2+x^2+8)*ln(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^2)/(5*x*exp(x)^2+x^3+8*x)/ln(exp(x)^2+1/5*x
^2+8/5),x,method=_RETURNVERBOSE)

[Out]

ln(x)+2*ln(ln(exp(2*x)+1/5*x^2+8/5))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x\right ) + 2 \, \log \left (\log \left (\frac {1}{5} \, x^{2} + e^{\left (2 \, x\right )} + \frac {8}{5}\right )\right ) \]

[In]

integrate(((5*exp(x)^2+x^2+8)*log(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^2)/(5*x*exp(x)^2+x^3+8*x)/log(exp(x)
^2+1/5*x^2+8/5),x, algorithm="fricas")

[Out]

log(x) + 2*log(log(1/5*x^2 + e^(2*x) + 8/5))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log {\left (x \right )} + 2 \log {\left (\log {\left (\frac {x^{2}}{5} + e^{2 x} + \frac {8}{5} \right )} \right )} \]

[In]

integrate(((5*exp(x)**2+x**2+8)*ln(exp(x)**2+1/5*x**2+8/5)+20*x*exp(x)**2+4*x**2)/(5*x*exp(x)**2+x**3+8*x)/ln(
exp(x)**2+1/5*x**2+8/5),x)

[Out]

log(x) + 2*log(log(x**2/5 + exp(2*x) + 8/5))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x\right ) + 2 \, \log \left (-\log \left (5\right ) + \log \left (x^{2} + 5 \, e^{\left (2 \, x\right )} + 8\right )\right ) \]

[In]

integrate(((5*exp(x)^2+x^2+8)*log(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^2)/(5*x*exp(x)^2+x^3+8*x)/log(exp(x)
^2+1/5*x^2+8/5),x, algorithm="maxima")

[Out]

log(x) + 2*log(-log(5) + log(x^2 + 5*e^(2*x) + 8))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x\right ) + 2 \, \log \left (\log \left (\frac {1}{5} \, x^{2} + e^{\left (2 \, x\right )} + \frac {8}{5}\right )\right ) \]

[In]

integrate(((5*exp(x)^2+x^2+8)*log(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^2)/(5*x*exp(x)^2+x^3+8*x)/log(exp(x)
^2+1/5*x^2+8/5),x, algorithm="giac")

[Out]

log(x) + 2*log(log(1/5*x^2 + e^(2*x) + 8/5))

Mupad [B] (verification not implemented)

Time = 8.42 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=2\,\ln \left (\ln \left ({\mathrm {e}}^{2\,x}+\frac {x^2}{5}+\frac {8}{5}\right )\right )+\ln \left (x\right ) \]

[In]

int((20*x*exp(2*x) + log(exp(2*x) + x^2/5 + 8/5)*(5*exp(2*x) + x^2 + 8) + 4*x^2)/(log(exp(2*x) + x^2/5 + 8/5)*
(8*x + 5*x*exp(2*x) + x^3)),x)

[Out]

2*log(log(exp(2*x) + x^2/5 + 8/5)) + log(x)