\(\int e^{(4 x+3 x^2+4 x^3) \log (2)-3 x \log (2) \log (4)} ((4+6 x+12 x^2) \log (2)-3 \log (2) \log (4)) \, dx\) [6851]

Optimal result

Integrand size = 48, antiderivative size = 20 \[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=2^{x \left (4 \left (1+x^2\right )-3 (-x+\log (4))\right )} \]

[Out]

exp(x*(4*x^2+4-6*ln(2)+3*x)*ln(2))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6838} \[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=2^{4 x^3+3 x^2+4 x} e^{-3 x \log (2) \log (4)} \]

[In]

Int[E^((4*x + 3*x^2 + 4*x^3)*Log[2] - 3*x*Log[2]*Log[4])*((4 + 6*x + 12*x^2)*Log[2] - 3*Log[2]*Log[4]),x]

[Out]

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

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^{4 x+3 x^2+4 x^3} e^{-3 x \log (2) \log (4)} \\ \end{align*}

Mathematica [F]

\[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=\int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx \]

[In]

Integrate[E^((4*x + 3*x^2 + 4*x^3)*Log[2] - 3*x*Log[2]*Log[4])*((4 + 6*x + 12*x^2)*Log[2] - 3*Log[2]*Log[4]),x
]

[Out]

Integrate[E^((4*x + 3*x^2 + 4*x^3)*Log[2] - 3*x*Log[2]*Log[4])*((4 + 6*x + 12*x^2)*Log[2] - 3*Log[2]*Log[4]),
x]

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
parallelrisch	\({\mathrm e}^{-x \ln \left (2\right ) \left (-4 x^{2}+6 \ln \left (2\right )-3 x -4\right )}\)	\(21\)
risch	\(2^{x \left (4 x^{2}+3 x +4\right )} {\mathrm e}^{-6 x \ln \left (2\right )^{2}}\)	\(24\)
derivativedivides	\({\mathrm e}^{-6 x \ln \left (2\right )^{2}+\left (4 x^{3}+3 x^{2}+4 x \right ) \ln \left (2\right )}\)	\(27\)
default	\({\mathrm e}^{-6 x \ln \left (2\right )^{2}+\left (4 x^{3}+3 x^{2}+4 x \right ) \ln \left (2\right )}\)	\(27\)
norman	\({\mathrm e}^{-6 x \ln \left (2\right )^{2}+\left (4 x^{3}+3 x^{2}+4 x \right ) \ln \left (2\right )}\)	\(27\)
gosper	\({\mathrm e}^{4 x^{3} \ln \left (2\right )-6 x \ln \left (2\right )^{2}+3 x^{2} \ln \left (2\right )+4 x \ln \left (2\right )}\)	\(29\)

[In]

int((-6*ln(2)^2+(12*x^2+6*x+4)*ln(2))*exp(-6*x*ln(2)^2+(4*x^3+3*x^2+4*x)*ln(2)),x,method=_RETURNVERBOSE)

[Out]

exp(-x*ln(2)*(-4*x^2+6*ln(2)-3*x-4))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=e^{\left (-6 \, x \log \left (2\right )^{2} + {\left (4 \, x^{3} + 3 \, x^{2} + 4 \, x\right )} \log \left (2\right )\right )} \]

[In]

integrate((-6*log(2)^2+(12*x^2+6*x+4)*log(2))*exp(-6*x*log(2)^2+(4*x^3+3*x^2+4*x)*log(2)),x, algorithm="fricas
")

[Out]

e^(-6*x*log(2)^2 + (4*x^3 + 3*x^2 + 4*x)*log(2))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=e^{- 6 x \log {\left (2 \right )}^{2} + \left (4 x^{3} + 3 x^{2} + 4 x\right ) \log {\left (2 \right )}} \]

[In]

integrate((-6*ln(2)**2+(12*x**2+6*x+4)*ln(2))*exp(-6*x*ln(2)**2+(4*x**3+3*x**2+4*x)*ln(2)),x)

[Out]

exp(-6*x*log(2)**2 + (4*x**3 + 3*x**2 + 4*x)*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=e^{\left (4 \, x^{3} \log \left (2\right ) + 3 \, x^{2} \log \left (2\right ) - 6 \, x \log \left (2\right )^{2} + 4 \, x \log \left (2\right )\right )} \]

[In]

integrate((-6*log(2)^2+(12*x^2+6*x+4)*log(2))*exp(-6*x*log(2)^2+(4*x^3+3*x^2+4*x)*log(2)),x, algorithm="maxima
")

[Out]

e^(4*x^3*log(2) + 3*x^2*log(2) - 6*x*log(2)^2 + 4*x*log(2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=e^{\left (4 \, x^{3} \log \left (2\right ) + 3 \, x^{2} \log \left (2\right ) - 6 \, x \log \left (2\right )^{2} + 4 \, x \log \left (2\right )\right )} \]

[In]

integrate((-6*log(2)^2+(12*x^2+6*x+4)*log(2))*exp(-6*x*log(2)^2+(4*x^3+3*x^2+4*x)*log(2)),x, algorithm="giac")

[Out]

e^(4*x^3*log(2) + 3*x^2*log(2) - 6*x*log(2)^2 + 4*x*log(2))

Mupad [B] (verification not implemented)

Time = 11.51 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int e^{\left (4 x+3 x^2+4 x^3\right ) \log (2)-3 x \log (2) \log (4)} \left (\left (4+6 x+12 x^2\right ) \log (2)-3 \log (2) \log (4)\right ) \, dx=2^{4\,x}\,2^{3\,x^2}\,2^{4\,x^3}\,{\mathrm {e}}^{-6\,x\,{\ln \left (2\right )}^2} \]

[In]

int(exp(log(2)*(4*x + 3*x^2 + 4*x^3) - 6*x*log(2)^2)*(log(2)*(6*x + 12*x^2 + 4) - 6*log(2)^2),x)

[Out]

2^(4*x)*2^(3*x^2)*2^(4*x^3)*exp(-6*x*log(2)^2)