\(\int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{(-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+(e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)) \log (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)})) \log (-15+\log (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}))} \, dx\) [65]

Optimal result

Integrand size = 113, antiderivative size = 21 \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=4 \log \left (\log \left (-15+\log \left (e^{\frac {1}{x^2}}+\frac {e^x}{\log ^2(2)}\right )\right )\right ) \]

[Out]

4*ln(ln(ln(exp(x)/ln(2)^2+exp(1/x^2))-15))

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6816} \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=4 \log \left (\log \left (\log \left (\frac {e^{\frac {1}{x^2}} \log ^2(2)+e^x}{\log ^2(2)}\right )-15\right )\right ) \]

[In]

Int[(4*E^x*x^3 - 8*E^x^(-2)*Log[2]^2)/((-15*E^x*x^3 - 15*E^x^(-2)*x^3*Log[2]^2 + (E^x*x^3 + E^x^(-2)*x^3*Log[2
]^2)*Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2])*Log[-15 + Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2]]),x]

[Out]

4*Log[Log[-15 + Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2]]]

Rule 6816

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

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=4 \log \left (\log \left (-15+\log \left (e^{\frac {1}{x^2}}+\frac {e^x}{\log ^2(2)}\right )\right )\right ) \]

[In]

Integrate[(4*E^x*x^3 - 8*E^x^(-2)*Log[2]^2)/((-15*E^x*x^3 - 15*E^x^(-2)*x^3*Log[2]^2 + (E^x*x^3 + E^x^(-2)*x^3
*Log[2]^2)*Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2])*Log[-15 + Log[(E^x + E^x^(-2)*Log[2]^2)/Log[2]^2]]),x]

[Out]

4*Log[Log[-15 + Log[E^x^(-2) + E^x/Log[2]^2]]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19

\[4 \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{x}+\ln \left (2\right )^{2} {\mathrm e}^{\frac {1}{x^{2}}}}{\ln \left (2\right )^{2}}\right )-15\right )\right )\]

[In]

int((4*exp(x)*x^3-8*ln(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*ln(2)^2*exp(1/x^2))*ln((exp(x)+ln(2)^2*exp(1/x^2))/ln
(2)^2)-15*exp(x)*x^3-15*x^3*ln(2)^2*exp(1/x^2))/ln(ln((exp(x)+ln(2)^2*exp(1/x^2))/ln(2)^2)-15),x)

[Out]

4*ln(ln(ln((exp(x)+ln(2)^2*exp(1/x^2))/ln(2)^2)-15))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=4 \, \log \left (\log \left (\log \left (\frac {e^{\left (\frac {1}{x^{2}}\right )} \log \left (2\right )^{2} + e^{x}}{\log \left (2\right )^{2}}\right ) - 15\right )\right ) \]

[In]

integrate((4*exp(x)*x^3-8*log(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*log(2)^2*exp(1/x^2))*log((exp(x)+log(2)^2*exp(
1/x^2))/log(2)^2)-15*exp(x)*x^3-15*x^3*log(2)^2*exp(1/x^2))/log(log((exp(x)+log(2)^2*exp(1/x^2))/log(2)^2)-15)
,x, algorithm="fricas")

[Out]

4*log(log(log((e^(x^(-2))*log(2)^2 + e^x)/log(2)^2) - 15))

Sympy [F(-1)]

Timed out. \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=\text {Timed out} \]

[In]

integrate((4*exp(x)*x**3-8*ln(2)**2*exp(1/x**2))/((exp(x)*x**3+x**3*ln(2)**2*exp(1/x**2))*ln((exp(x)+ln(2)**2*
exp(1/x**2))/ln(2)**2)-15*exp(x)*x**3-15*x**3*ln(2)**2*exp(1/x**2))/ln(ln((exp(x)+ln(2)**2*exp(1/x**2))/ln(2)*
*2)-15),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=4 \, \log \left (\log \left (\log \left (e^{\left (\frac {1}{x^{2}}\right )} \log \left (2\right )^{2} + e^{x}\right ) - 2 \, \log \left (\log \left (2\right )\right ) - 15\right )\right ) \]

[In]

integrate((4*exp(x)*x^3-8*log(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*log(2)^2*exp(1/x^2))*log((exp(x)+log(2)^2*exp(
1/x^2))/log(2)^2)-15*exp(x)*x^3-15*x^3*log(2)^2*exp(1/x^2))/log(log((exp(x)+log(2)^2*exp(1/x^2))/log(2)^2)-15)
,x, algorithm="maxima")

[Out]

4*log(log(log(e^(x^(-2))*log(2)^2 + e^x) - 2*log(log(2)) - 15))

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=4 \, \log \left (\log \left (\log \left (e^{\left (\frac {1}{x^{2}}\right )} \log \left (2\right )^{2} + e^{x}\right ) - 2 \, \log \left (\log \left (2\right )\right ) - 15\right )\right ) \]

[In]

integrate((4*exp(x)*x^3-8*log(2)^2*exp(1/x^2))/((exp(x)*x^3+x^3*log(2)^2*exp(1/x^2))*log((exp(x)+log(2)^2*exp(
1/x^2))/log(2)^2)-15*exp(x)*x^3-15*x^3*log(2)^2*exp(1/x^2))/log(log((exp(x)+log(2)^2*exp(1/x^2))/log(2)^2)-15)
,x, algorithm="giac")

[Out]

4*log(log(log(e^(x^(-2))*log(2)^2 + e^x) - 2*log(log(2)) - 15))

Mupad [B] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {4 e^x x^3-8 e^{\frac {1}{x^2}} \log ^2(2)}{\left (-15 e^x x^3-15 e^{\frac {1}{x^2}} x^3 \log ^2(2)+\left (e^x x^3+e^{\frac {1}{x^2}} x^3 \log ^2(2)\right ) \log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right ) \log \left (-15+\log \left (\frac {e^x+e^{\frac {1}{x^2}} \log ^2(2)}{\log ^2(2)}\right )\right )} \, dx=4\,\ln \left (\ln \left (\ln \left (\frac {{\mathrm {e}}^x+{\mathrm {e}}^{\frac {1}{x^2}}\,{\ln \left (2\right )}^2}{{\ln \left (2\right )}^2}\right )-15\right )\right ) \]

[In]

int(-(4*x^3*exp(x) - 8*exp(1/x^2)*log(2)^2)/(log(log((exp(x) + exp(1/x^2)*log(2)^2)/log(2)^2) - 15)*(15*x^3*ex
p(x) - log((exp(x) + exp(1/x^2)*log(2)^2)/log(2)^2)*(x^3*exp(x) + x^3*exp(1/x^2)*log(2)^2) + 15*x^3*exp(1/x^2)
*log(2)^2)),x)

[Out]

4*log(log(log((exp(x) + exp(1/x^2)*log(2)^2)/log(2)^2) - 15))