\(\int e^{-\frac {4 (-3+\log (3))}{\log (2)}} (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3) \, dx\) [5505]

Optimal result

Integrand size = 33, antiderivative size = 23 \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=-x-e^{\frac {4 (3-\log (3))}{\log (2)}} x^4 \]

[Out]

-x-x^4*exp((3-ln(3))/ln(2))^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {12} \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=x^4 \left (-e^{\frac {4 (3-\log (3))}{\log (2)}}\right )-x \]

[In]

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

[Out]

-x - E^((4*(3 - Log[3]))/Log[2])*x^4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=-x-3^{-\frac {4}{\log (2)}} e^{\frac {12}{\log (2)}} x^4 \]

[In]

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

[Out]

-x - (E^(12/Log[2])*x^4)/3^(4/Log[2])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
risch	\(-3^{-\frac {4}{\ln \left (2\right )}} x^{4} {\mathrm e}^{\frac {12}{\ln \left (2\right )}}-x\)	\(25\)
gosper	\(-x \left ({\mathrm e}^{\frac {4 \ln \left (3\right )-12}{\ln \left (2\right )}}+x^{3}\right ) {\mathrm e}^{-\frac {4 \left (\ln \left (3\right )-3\right )}{\ln \left (2\right )}}\)	\(32\)
default	\({\mathrm e}^{-\frac {4 \left (\ln \left (3\right )-3\right )}{\ln \left (2\right )}} \left (-{\mathrm e}^{\frac {4 \ln \left (3\right )-12}{\ln \left (2\right )}} x -x^{4}\right )\)	\(35\)
parallelrisch	\({\mathrm e}^{-\frac {4 \left (\ln \left (3\right )-3\right )}{\ln \left (2\right )}} \left (-{\mathrm e}^{\frac {4 \ln \left (3\right )-12}{\ln \left (2\right )}} x -x^{4}\right )\)	\(35\)
norman	\(\left (-3^{-\frac {1}{\ln \left (2\right )}} {\mathrm e}^{\frac {3}{\ln \left (2\right )}} x^{4}-3^{\frac {3}{\ln \left (2\right )}} {\mathrm e}^{-\frac {9}{\ln \left (2\right )}} x \right ) {\mathrm e}^{-\frac {3 \left (\ln \left (3\right )-3\right )}{\ln \left (2\right )}}\)	\(53\)

[In]

int((-exp((ln(3)-3)/ln(2))^4-4*x^3)/exp((ln(3)-3)/ln(2))^4,x,method=_RETURNVERBOSE)

[Out]

-1/(3^(1/ln(2)))^4*x^4*exp(12/ln(2))-x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=-{\left (x^{4} + x e^{\left (\frac {4 \, {\left (\log \left (3\right ) - 3\right )}}{\log \left (2\right )}\right )}\right )} e^{\left (-\frac {4 \, {\left (\log \left (3\right ) - 3\right )}}{\log \left (2\right )}\right )} \]

[In]

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

[Out]

-(x^4 + x*e^(4*(log(3) - 3)/log(2)))*e^(-4*(log(3) - 3)/log(2))

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=- \frac {x^{4} e^{\frac {12}{\log {\left (2 \right )}}}}{3^{\frac {4}{\log {\left (2 \right )}}}} - x \]

[In]

integrate((-exp((ln(3)-3)/ln(2))**4-4*x**3)/exp((ln(3)-3)/ln(2))**4,x)

[Out]

-x**4*exp(12/log(2))/3**(4/log(2)) - x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=-{\left (x^{4} + x e^{\left (\frac {4 \, {\left (\log \left (3\right ) - 3\right )}}{\log \left (2\right )}\right )}\right )} e^{\left (-\frac {4 \, {\left (\log \left (3\right ) - 3\right )}}{\log \left (2\right )}\right )} \]

[In]

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

[Out]

-(x^4 + x*e^(4*(log(3) - 3)/log(2)))*e^(-4*(log(3) - 3)/log(2))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=-{\left (x^{4} + x e^{\left (\frac {4 \, {\left (\log \left (3\right ) - 3\right )}}{\log \left (2\right )}\right )}\right )} e^{\left (-\frac {4 \, {\left (\log \left (3\right ) - 3\right )}}{\log \left (2\right )}\right )} \]

[In]

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

[Out]

-(x^4 + x*e^(4*(log(3) - 3)/log(2)))*e^(-4*(log(3) - 3)/log(2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int e^{-\frac {4 (-3+\log (3))}{\log (2)}} \left (-e^{\frac {4 (-3+\log (3))}{\log (2)}}-4 x^3\right ) \, dx=-{\mathrm {e}}^{-\frac {\ln \left (81\right )-12}{\ln \left (2\right )}}\,x^4-x \]

[In]

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

[Out]

- x - x^4*exp(-(log(81) - 12)/log(2))