\(\int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 (2 x-8 x^2)}{-1+4 x} \, dx\) [6052]

Optimal result

Integrand size = 50, antiderivative size = 34 \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx=-3+x \left (x \left (-e^2+x\right )-\frac {-5+e^{e^{2 x}}+\log \left (-\frac {1}{4}+x\right )}{x}\right ) \]

[Out]

x*(x*(x-exp(2))-(exp(exp(x)^2)-5+ln(x-1/4))/x)-3

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6874, 2320, 2225, 1864} \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx=x^3-e^2 x^2-e^{e^{2 x}}-\log (1-4 x) \]

[In]

Int[(-4 + E^(E^(2*x) + 2*x)*(2 - 8*x) - 3*x^2 + 12*x^3 + E^2*(2*x - 8*x^2))/(-1 + 4*x),x]

[Out]

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

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 2225

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-2 e^{e^{2 x}+2 x}+\frac {4-2 e^2 x+\left (3+8 e^2\right ) x^2-12 x^3}{1-4 x}\right ) \, dx \\ & = -\left (2 \int e^{e^{2 x}+2 x} \, dx\right )+\int \frac {4-2 e^2 x+\left (3+8 e^2\right ) x^2-12 x^3}{1-4 x} \, dx \\ & = \int \left (-2 e^2 x+3 x^2-\frac {4}{-1+4 x}\right ) \, dx-\text {Subst}\left (\int e^x \, dx,x,e^{2 x}\right ) \\ & = -e^{e^{2 x}}-e^2 x^2+x^3-\log (1-4 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx=-e^{e^{2 x}}-e^2 x^2+x^3-\log (1-4 x) \]

[In]

Integrate[(-4 + E^(E^(2*x) + 2*x)*(2 - 8*x) - 3*x^2 + 12*x^3 + E^2*(2*x - 8*x^2))/(-1 + 4*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(-x^{2} {\mathrm e}^{2}+x^{3}-{\mathrm e}^{{\mathrm e}^{2 x}}-\ln \left (x -\frac {1}{4}\right )\)	\(25\)
norman	\(x^{3}-x^{2} {\mathrm e}^{2}-{\mathrm e}^{{\mathrm e}^{2 x}}-\ln \left (-1+4 x \right )\)	\(27\)
risch	\(x^{3}-x^{2} {\mathrm e}^{2}-{\mathrm e}^{{\mathrm e}^{2 x}}-\ln \left (-1+4 x \right )\)	\(27\)
parts	\(x^{3}-x^{2} {\mathrm e}^{2}-{\mathrm e}^{{\mathrm e}^{2 x}}-\ln \left (-1+4 x \right )\)	\(27\)

[In]

int(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(-1+4*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32 \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx={\left ({\left (x^{3} - x^{2} e^{2}\right )} e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x - 1\right ) - e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )} \]

[In]

integrate(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(-1+4*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx=x^{3} - x^{2} e^{2} - e^{e^{2 x}} - \log {\left (4 x - 1 \right )} \]

[In]

integrate(((-8*x+2)*exp(x)**2*exp(exp(x)**2)+(-8*x**2+2*x)*exp(2)+12*x**3-3*x**2-4)/(-1+4*x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx=x^{3} - \frac {1}{8} \, {\left (8 \, x^{2} + 4 \, x + \log \left (4 \, x - 1\right )\right )} e^{2} + \frac {1}{8} \, {\left (4 \, x + \log \left (4 \, x - 1\right )\right )} e^{2} - e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (4 \, x - 1\right ) \]

[In]

integrate(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(-1+4*x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx={\left (x^{3} e^{\left (2 \, x\right )} - x^{2} e^{\left (2 \, x + 2\right )} - e^{\left (2 \, x\right )} \log \left (4 \, x - 1\right ) - e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )} \]

[In]

integrate(((-8*x+2)*exp(x)^2*exp(exp(x)^2)+(-8*x^2+2*x)*exp(2)+12*x^3-3*x^2-4)/(-1+4*x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.92 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-4+e^{e^{2 x}+2 x} (2-8 x)-3 x^2+12 x^3+e^2 \left (2 x-8 x^2\right )}{-1+4 x} \, dx=x^3-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x^2\,{\mathrm {e}}^2-\ln \left (x-\frac {1}{4}\right ) \]

[In]

int(-(3*x^2 - exp(2)*(2*x - 8*x^2) - 12*x^3 + exp(2*x)*exp(exp(2*x))*(8*x - 2) + 4)/(4*x - 1),x)

[Out]

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