3.69.95 \(\int (-3 x^2+e^{1-4 x+\log ^2(\log (3))} (2 x-4 x^2)) \, dx\)

Optimal. Leaf size=20 \[ \left (e^{1-4 x+\log ^2(\log (3))}-x\right ) x^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} x^2 e^{-4 x+1+\log ^2(\log (3))}-x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-3*x^2 + E^(1 - 4*x + Log[Log[3]]^2)*(2*x - 4*x^2),x]

[Out]

E^(1 - 4*x + Log[Log[3]]^2)*x^2 - x^3

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-x^3+\int e^{1-4 x+\log ^2(\log (3))} \left (2 x-4 x^2\right ) \, dx\\ &=-x^3+\int e^{1-4 x+\log ^2(\log (3))} (2-4 x) x \, dx\\ &=-x^3+\int \left (2 e^{1-4 x+\log ^2(\log (3))} x-4 e^{1-4 x+\log ^2(\log (3))} x^2\right ) \, dx\\ &=-x^3+2 \int e^{1-4 x+\log ^2(\log (3))} x \, dx-4 \int e^{1-4 x+\log ^2(\log (3))} x^2 \, dx\\ &=-\frac {1}{2} e^{1-4 x+\log ^2(\log (3))} x+e^{1-4 x+\log ^2(\log (3))} x^2-x^3+\frac {1}{2} \int e^{1-4 x+\log ^2(\log (3))} \, dx-2 \int e^{1-4 x+\log ^2(\log (3))} x \, dx\\ &=-\frac {1}{8} e^{1-4 x+\log ^2(\log (3))}+e^{1-4 x+\log ^2(\log (3))} x^2-x^3-\frac {1}{2} \int e^{1-4 x+\log ^2(\log (3))} \, dx\\ &=e^{1-4 x+\log ^2(\log (3))} x^2-x^3\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 20, normalized size = 1.00 \begin {gather*} \left (e^{1-4 x+\log ^2(\log (3))}-x\right ) x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-3*x^2 + E^(1 - 4*x + Log[Log[3]]^2)*(2*x - 4*x^2),x]

[Out]

(E^(1 - 4*x + Log[Log[3]]^2) - x)*x^2

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 21, normalized size = 1.05 \begin {gather*} -x^{3} + x^{2} e^{\left (\log \left (\log \relax (3)\right )^{2} - 4 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x^2*e^(log(log(3))^2 - 4*x + 1)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 21, normalized size = 1.05 \begin {gather*} -x^{3} + x^{2} e^{\left (\log \left (\log \relax (3)\right )^{2} - 4 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x^2*e^(log(log(3))^2 - 4*x + 1)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 22, normalized size = 1.10

method	result	size
norman	\({\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} x^{2}-x^{3}\)	\(22\)
risch	\({\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} x^{2}-x^{3}\)	\(22\)
default	\(-\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right ) \ln \left (\ln \relax (3)\right )^{2}}{8}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \ln \left (\ln \relax (3)\right )^{2}}{8}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right )^{2}}{16}-\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right )}{8}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1}}{16}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \ln \left (\ln \relax (3)\right )^{4}}{16}-x^{3}\)	\(132\)
derivativedivides	\(\frac {\left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right )^{3}}{64}-\frac {3 \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right )^{2} \ln \left (\ln \relax (3)\right )^{2}}{64}-\frac {3 \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right )^{2}}{64}+\frac {3 \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right ) \ln \left (\ln \relax (3)\right )^{4}}{64}+\frac {3 \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right ) \ln \left (\ln \relax (3)\right )^{2}}{32}-\frac {3 x}{16}+\frac {1}{32}-\frac {\ln \left (\ln \relax (3)\right )^{6}}{64}-\frac {3 \ln \left (\ln \relax (3)\right )^{4}}{64}-\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right ) \ln \left (\ln \relax (3)\right )^{2}}{8}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \ln \left (\ln \relax (3)\right )^{2}}{8}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right )^{2}}{16}-\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \left (\ln \left (\ln \relax (3)\right )^{2}-4 x +1\right )}{8}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1}}{16}+\frac {{\mathrm e}^{\ln \left (\ln \relax (3)\right )^{2}-4 x +1} \ln \left (\ln \relax (3)\right )^{4}}{16}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^2+2*x)*exp(ln(ln(3))^2-4*x+1)-3*x^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.45, size = 71, normalized size = 3.55 \begin {gather*} -x^{3} + \frac {1}{8} \, {\left (8 \, x^{2} e^{\left (\log \left (\log \relax (3)\right )^{2} + 1\right )} + 4 \, x e^{\left (\log \left (\log \relax (3)\right )^{2} + 1\right )} + e^{\left (\log \left (\log \relax (3)\right )^{2} + 1\right )}\right )} e^{\left (-4 \, x\right )} - \frac {1}{8} \, {\left (4 \, x e^{\left (\log \left (\log \relax (3)\right )^{2} + 1\right )} + e^{\left (\log \left (\log \relax (3)\right )^{2} + 1\right )}\right )} e^{\left (-4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + 1/8*(8*x^2*e^(log(log(3))^2 + 1) + 4*x*e^(log(log(3))^2 + 1) + e^(log(log(3))^2 + 1))*e^(-4*x) - 1/8*(4
*x*e^(log(log(3))^2 + 1) + e^(log(log(3))^2 + 1))*e^(-4*x)

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 22, normalized size = 1.10 \begin {gather*} x^2\,{\mathrm {e}}^{-4\,x}\,\mathrm {e}\,{\mathrm {e}}^{{\ln \left (\ln \relax (3)\right )}^2}-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(log(log(3))^2 - 4*x + 1)*(2*x - 4*x^2) - 3*x^2,x)

[Out]

x^2*exp(-4*x)*exp(1)*exp(log(log(3))^2) - x^3

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 19, normalized size = 0.95 \begin {gather*} - x^{3} + x^{2} e^{- 4 x + \log {\left (\log {\relax (3 )} \right )}^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3 + x**2*exp(-4*x + log(log(3))**2 + 1)

________________________________________________________________________________________