\(\int e^{-40 x^4+20 x^5+(-8 x^3+4 x^4) \log (2)-4 \log ^2(2)} (-160 x^3+100 x^4+(-24 x^2+16 x^3) \log (2)) \, dx\) [7764]

Optimal result

Integrand size = 59, antiderivative size = 26 \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=-3+e^{4 \left (-\log ^2(2)+(-2+x) x^3 (5 x+\log (2))\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6838} \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=2^{4 x^4-8 x^3} e^{20 x^5-40 x^4-4 \log ^2(2)} \]

[In]

Int[E^(-40*x^4 + 20*x^5 + (-8*x^3 + 4*x^4)*Log[2] - 4*Log[2]^2)*(-160*x^3 + 100*x^4 + (-24*x^2 + 16*x^3)*Log[2
]),x]

[Out]

2^(-8*x^3 + 4*x^4)*E^(-40*x^4 + 20*x^5 - 4*Log[2]^2)

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}& = 2^{-8 x^3+4 x^4} e^{-40 x^4+20 x^5-4 \log ^2(2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=2^{4 (-2+x) x^3} e^{-40 x^4+20 x^5-4 \log ^2(2)} \]

[In]

Integrate[E^(-40*x^4 + 20*x^5 + (-8*x^3 + 4*x^4)*Log[2] - 4*Log[2]^2)*(-160*x^3 + 100*x^4 + (-24*x^2 + 16*x^3)
*Log[2]),x]

[Out]

2^(4*(-2 + x)*x^3)*E^(-40*x^4 + 20*x^5 - 4*Log[2]^2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12

method	result	size
risch	\(16^{\left (-2+x \right ) x^{3}} {\mathrm e}^{-4 \ln \left (2\right )^{2}+20 x^{5}-40 x^{4}}\)	\(29\)
gosper	\({\mathrm e}^{4 x^{4} \ln \left (2\right )+20 x^{5}-8 x^{3} \ln \left (2\right )-40 x^{4}-4 \ln \left (2\right )^{2}}\)	\(33\)
derivativedivides	\({\mathrm e}^{-4 \ln \left (2\right )^{2}+\left (4 x^{4}-8 x^{3}\right ) \ln \left (2\right )+20 x^{5}-40 x^{4}}\)	\(33\)
default	\({\mathrm e}^{-4 \ln \left (2\right )^{2}+\left (4 x^{4}-8 x^{3}\right ) \ln \left (2\right )+20 x^{5}-40 x^{4}}\)	\(33\)
norman	\({\mathrm e}^{-4 \ln \left (2\right )^{2}+\left (4 x^{4}-8 x^{3}\right ) \ln \left (2\right )+20 x^{5}-40 x^{4}}\)	\(33\)
parallelrisch	\({\mathrm e}^{4 x^{4} \ln \left (2\right )+20 x^{5}-8 x^{3} \ln \left (2\right )-40 x^{4}-4 \ln \left (2\right )^{2}}\)	\(33\)

[In]

int(((16*x^3-24*x^2)*ln(2)+100*x^4-160*x^3)*exp(-4*ln(2)^2+(4*x^4-8*x^3)*ln(2)+20*x^5-40*x^4),x,method=_RETURN
VERBOSE)

[Out]

16^((-2+x)*x^3)*exp(-4*ln(2)^2+20*x^5-40*x^4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=e^{\left (20 \, x^{5} - 40 \, x^{4} + 4 \, {\left (x^{4} - 2 \, x^{3}\right )} \log \left (2\right ) - 4 \, \log \left (2\right )^{2}\right )} \]

[In]

integrate(((16*x^3-24*x^2)*log(2)+100*x^4-160*x^3)*exp(-4*log(2)^2+(4*x^4-8*x^3)*log(2)+20*x^5-40*x^4),x, algo
rithm="fricas")

[Out]

e^(20*x^5 - 40*x^4 + 4*(x^4 - 2*x^3)*log(2) - 4*log(2)^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=e^{20 x^{5} - 40 x^{4} + \left (4 x^{4} - 8 x^{3}\right ) \log {\left (2 \right )} - 4 \log {\left (2 \right )}^{2}} \]

[In]

integrate(((16*x**3-24*x**2)*ln(2)+100*x**4-160*x**3)*exp(-4*ln(2)**2+(4*x**4-8*x**3)*ln(2)+20*x**5-40*x**4),x
)

[Out]

exp(20*x**5 - 40*x**4 + (4*x**4 - 8*x**3)*log(2) - 4*log(2)**2)

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=e^{\left (20 \, x^{5} + 4 \, x^{4} \log \left (2\right ) - 40 \, x^{4} - 8 \, x^{3} \log \left (2\right ) - 4 \, \log \left (2\right )^{2}\right )} \]

[In]

integrate(((16*x^3-24*x^2)*log(2)+100*x^4-160*x^3)*exp(-4*log(2)^2+(4*x^4-8*x^3)*log(2)+20*x^5-40*x^4),x, algo
rithm="maxima")

[Out]

e^(20*x^5 + 4*x^4*log(2) - 40*x^4 - 8*x^3*log(2) - 4*log(2)^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=e^{\left (20 \, x^{5} + 4 \, x^{4} \log \left (2\right ) - 40 \, x^{4} - 8 \, x^{3} \log \left (2\right ) - 4 \, \log \left (2\right )^{2}\right )} \]

[In]

integrate(((16*x^3-24*x^2)*log(2)+100*x^4-160*x^3)*exp(-4*log(2)^2+(4*x^4-8*x^3)*log(2)+20*x^5-40*x^4),x, algo
rithm="giac")

[Out]

e^(20*x^5 + 4*x^4*log(2) - 40*x^4 - 8*x^3*log(2) - 4*log(2)^2)

Mupad [B] (verification not implemented)

Time = 13.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int e^{-40 x^4+20 x^5+\left (-8 x^3+4 x^4\right ) \log (2)-4 \log ^2(2)} \left (-160 x^3+100 x^4+\left (-24 x^2+16 x^3\right ) \log (2)\right ) \, dx=\frac {2^{4\,x^4}\,{\mathrm {e}}^{-4\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{20\,x^5}\,{\mathrm {e}}^{-40\,x^4}}{2^{8\,x^3}} \]

[In]

int(-exp(20*x^5 - 4*log(2)^2 - 40*x^4 - log(2)*(8*x^3 - 4*x^4))*(log(2)*(24*x^2 - 16*x^3) + 160*x^3 - 100*x^4)
,x)

[Out]

(2^(4*x^4)*exp(-4*log(2)^2)*exp(20*x^5)*exp(-40*x^4))/2^(8*x^3)