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

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

________________________________________________________________________________________

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} $$\begin{array}{c}{x}^2{e}^{-4x+1+{\log }^2(\log (3))}-x^3\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-x^3+\int e^{1-4x+{\log }^2(\log (3))}(2x-4x^2)dx\\ & =-x^3+\int e^{1-4x+{\log }^2(\log (3))}(2-4x)xdx\\ & =-x^3+\int (2e^{1-4x+{\log }^2(\log (3))}x-4e^{1-4x+{\log }^2(\log (3))}{x}^2)dx\\ & =-x^3+2\int e^{1-4x+{\log }^2(\log (3))}xdx-4\int e^{1-4x+{\log }^2(\log (3))}{x}^{2}dx\\ & =-\frac{1}{2}{e}^{1-4x+{\log }^2(\log (3))}x+e^{1-4x+{\log }^2(\log (3))}{x}^2-x^3+\frac{1}{2}\int e^{1-4x+{\log }^2(\log (3))}dx-2\int e^{1-4x+{\log }^2(\log (3))}xdx\\ & =-\frac{1}{8}{e}^{1-4x+{\log }^2(\log (3))}+e^{1-4x+{\log }^2(\log (3))}{x}^2-x^3-\frac{1}{2}\int e^{1-4x+{\log }^2(\log (3))}dx\\ & =e^{1-4x+{\log }^2(\log (3))}{x}^2-x^3\end{array}\end{array}$$

________________________________________________________________________________________

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

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 $$\begin{array}{c}-x^3+x^2{e}^{(\log {(\log (3))}^2-4x+1)}\end{array}$$

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 $$\begin{array}{c}-x^3+x^2{e}^{(\log {(\log (3))}^2-4x+1)}\end{array}$$

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	${\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}{x}^2-x^3$	$22$
risch	${\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}{x}^2-x^3$	$22$
default	$-\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}(\ln {(\ln (3))}^2-4x+1)\ln {(\ln (3))}^2}{8}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}\ln {(\ln (3))}^2}{8}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}{(\ln {(\ln (3))}^2-4x+1)}^2}{16}-\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}(\ln {(\ln (3))}^2-4x+1)}{8}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}}{16}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}\ln {(\ln (3))}^4}{16}-x^3$	$132$
derivativedivides	$\frac{{(\ln {(\ln (3))}^2-4x+1)}^3}{64}-\frac{3{(\ln {(\ln (3))}^2-4x+1)}^2\ln {(\ln (3))}^2}{64}-\frac{3{(\ln {(\ln (3))}^2-4x+1)}^2}{64}+\frac{3(\ln {(\ln (3))}^2-4x+1)\ln {(\ln (3))}^4}{64}+\frac{3(\ln {(\ln (3))}^2-4x+1)\ln {(\ln (3))}^2}{32}-\frac{3x}{16}+\frac{1}{32}-\frac{\ln {(\ln (3))}^6}{64}-\frac{3\ln {(\ln (3))}^4}{64}-\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}(\ln {(\ln (3))}^2-4x+1)\ln {(\ln (3))}^2}{8}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}\ln {(\ln (3))}^2}{8}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}{(\ln {(\ln (3))}^2-4x+1)}^2}{16}-\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}(\ln {(\ln (3))}^2-4x+1)}{8}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}}{16}+\frac{{\mathrm{e}}^{\ln {(\ln (3))}^2-4x+1}\ln {(\ln (3))}^4}{16}$	$226$

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 $$\begin{array}{c}-x^3+\frac{1}{8}(8x^2{e}^{(\log {(\log (3))}^2+1)}+4x{e}^{(\log {(\log (3))}^2+1)}+e^{(\log {(\log (3))}^2+1)})e^{(-4x)}-\frac{1}{8}(4x{e}^{(\log {(\log (3))}^2+1)}+e^{(\log {(\log (3))}^2+1)})e^{(-4x)}\end{array}$$

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 $$\begin{array}{c}{x}^2{\mathrm{e}}^{-4x}\mathrm{e}{\mathrm{e}}^{{\ln (\ln (3))}^2}-x^3\end{array}$$

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 $$\begin{array}{c}-x^3+x^2{e}^{-4x+\log {(\log (3))}^2+1}\end{array}$$

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)

________________________________________________________________________________________