3.61.63 \(\int \frac {2 x^4-6 x^5-2 x^6+e^{2+x} (-16 x-16 x^2-8 x^3)+(8 e^{2+x} x^2+2 x^5) \log (\frac {e^{-2-x} (4 e^{2+x}+x^3)}{x^2})+(2 x^3-6 x^4-2 x^5+e^{2+x} (-16-16 x-8 x^2)+(8 e^{2+x} x+2 x^4) \log (\frac {e^{-2-x} (4 e^{2+x}+x^3)}{x^2})) \log (-1-x+\log (\frac {e^{-2-x} (4 e^{2+x}+x^3)}{x^2}))}{-x^4-x^5+e^{2+x} (-4 x-4 x^2)+(4 e^{2+x} x+x^4) \log (\frac {e^{-2-x} (4 e^{2+x}+x^3)}{x^2})} \, dx\)

Optimal. Leaf size=26 \[ \left (x+\log \left (-1-x+\log \left (\frac {4}{x^2}+e^{-2-x} x\right )\right )\right )^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.91, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 241, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6688, 12, 6686} \begin {gather*} \left (\log \left (\log \left (\frac {4}{x^2}+e^{-x-2} x\right )-x-1\right )+x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^4 - 6*x^5 - 2*x^6 + E^(2 + x)*(-16*x - 16*x^2 - 8*x^3) + (8*E^(2 + x)*x^2 + 2*x^5)*Log[(E^(-2 - x)*(4
*E^(2 + x) + x^3))/x^2] + (2*x^3 - 6*x^4 - 2*x^5 + E^(2 + x)*(-16 - 16*x - 8*x^2) + (8*E^(2 + x)*x + 2*x^4)*Lo
g[(E^(-2 - x)*(4*E^(2 + x) + x^3))/x^2])*Log[-1 - x + Log[(E^(-2 - x)*(4*E^(2 + x) + x^3))/x^2]])/(-x^4 - x^5
+ E^(2 + x)*(-4*x - 4*x^2) + (4*E^(2 + x)*x + x^4)*Log[(E^(-2 - x)*(4*E^(2 + x) + x^3))/x^2]),x]

[Out]

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

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 0.18, size = 62, normalized size = 2.38 \begin {gather*} 2 \left (\frac {x^2}{2}+x \log \left (-1-x+\log \left (\frac {4}{x^2}+e^{-2-x} x\right )\right )+\frac {1}{2} \log ^2\left (-1-x+\log \left (\frac {4}{x^2}+e^{-2-x} x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^4 - 6*x^5 - 2*x^6 + E^(2 + x)*(-16*x - 16*x^2 - 8*x^3) + (8*E^(2 + x)*x^2 + 2*x^5)*Log[(E^(-2 -
x)*(4*E^(2 + x) + x^3))/x^2] + (2*x^3 - 6*x^4 - 2*x^5 + E^(2 + x)*(-16 - 16*x - 8*x^2) + (8*E^(2 + x)*x + 2*x
^4)*Log[(E^(-2 - x)*(4*E^(2 + x) + x^3))/x^2])*Log[-1 - x + Log[(E^(-2 - x)*(4*E^(2 + x) + x^3))/x^2]])/(-x^4
- x^5 + E^(2 + x)*(-4*x - 4*x^2) + (4*E^(2 + x)*x + x^4)*Log[(E^(-2 - x)*(4*E^(2 + x) + x^3))/x^2]),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.91, size = 63, normalized size = 2.42 \begin {gather*} x^{2} + 2 \, x \log \left (-x + \log \left (\frac {{\left (x^{3} + 4 \, e^{\left (x + 2\right )}\right )} e^{\left (-x - 2\right )}}{x^{2}}\right ) - 1\right ) + \log \left (-x + \log \left (\frac {{\left (x^{3} + 4 \, e^{\left (x + 2\right )}\right )} e^{\left (-x - 2\right )}}{x^{2}}\right ) - 1\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*exp(2+x)+2*x^4)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x^2-16*x-16)*exp(2+x)-2*x^5-6*x^4+2*x^
3)*log(log((4*exp(2+x)+x^3)/x^2/exp(2+x))-x-1)+(8*x^2*exp(2+x)+2*x^5)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x
^3-16*x^2-16*x)*exp(2+x)-2*x^6-6*x^5+2*x^4)/((4*x*exp(2+x)+x^4)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-4*x^2-4*x
)*exp(2+x)-x^5-x^4),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{6} + 3 \, x^{5} - x^{4} + 4 \, {\left (x^{3} + 2 \, x^{2} + 2 \, x\right )} e^{\left (x + 2\right )} + {\left (x^{5} + 3 \, x^{4} - x^{3} + 4 \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (x + 2\right )} - {\left (x^{4} + 4 \, x e^{\left (x + 2\right )}\right )} \log \left (\frac {{\left (x^{3} + 4 \, e^{\left (x + 2\right )}\right )} e^{\left (-x - 2\right )}}{x^{2}}\right )\right )} \log \left (-x + \log \left (\frac {{\left (x^{3} + 4 \, e^{\left (x + 2\right )}\right )} e^{\left (-x - 2\right )}}{x^{2}}\right ) - 1\right ) - {\left (x^{5} + 4 \, x^{2} e^{\left (x + 2\right )}\right )} \log \left (\frac {{\left (x^{3} + 4 \, e^{\left (x + 2\right )}\right )} e^{\left (-x - 2\right )}}{x^{2}}\right )\right )}}{x^{5} + x^{4} + 4 \, {\left (x^{2} + x\right )} e^{\left (x + 2\right )} - {\left (x^{4} + 4 \, x e^{\left (x + 2\right )}\right )} \log \left (\frac {{\left (x^{3} + 4 \, e^{\left (x + 2\right )}\right )} e^{\left (-x - 2\right )}}{x^{2}}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*exp(2+x)+2*x^4)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x^2-16*x-16)*exp(2+x)-2*x^5-6*x^4+2*x^
3)*log(log((4*exp(2+x)+x^3)/x^2/exp(2+x))-x-1)+(8*x^2*exp(2+x)+2*x^5)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x
^3-16*x^2-16*x)*exp(2+x)-2*x^6-6*x^5+2*x^4)/((4*x*exp(2+x)+x^4)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-4*x^2-4*x
)*exp(2+x)-x^5-x^4),x, algorithm="giac")

[Out]

integrate(2*(x^6 + 3*x^5 - x^4 + 4*(x^3 + 2*x^2 + 2*x)*e^(x + 2) + (x^5 + 3*x^4 - x^3 + 4*(x^2 + 2*x + 2)*e^(x
+ 2) - (x^4 + 4*x*e^(x + 2))*log((x^3 + 4*e^(x + 2))*e^(-x - 2)/x^2))*log(-x + log((x^3 + 4*e^(x + 2))*e^(-x
- 2)/x^2) - 1) - (x^5 + 4*x^2*e^(x + 2))*log((x^3 + 4*e^(x + 2))*e^(-x - 2)/x^2))/(x^5 + x^4 + 4*(x^2 + x)*e^(
x + 2) - (x^4 + 4*x*e^(x + 2))*log((x^3 + 4*e^(x + 2))*e^(-x - 2)/x^2)), x)

________________________________________________________________________________________

maple [C]  time = 0.31, size = 522, normalized size = 20.08

method	result	size
risch	\(x^{2}+2 x \ln \left (-2 \ln \relax (x )-\ln \left ({\mathrm e}^{2+x}\right )+\ln \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x -2}\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )+\mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )\right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right ) {\mathrm e}^{-x -2}}{x^{2}}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right ) {\mathrm e}^{-x -2}}{x^{2}}\right )+\mathrm {csgn}\left (\frac {i}{x^{2}}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right ) {\mathrm e}^{-x -2}}{x^{2}}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )\right )}{2}-x -1\right )+\ln \left (-2 \ln \relax (x )-\ln \left ({\mathrm e}^{2+x}\right )+\ln \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x -2}\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )+\mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )\right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right ) {\mathrm e}^{-x -2}}{x^{2}}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right ) {\mathrm e}^{-x -2}}{x^{2}}\right )+\mathrm {csgn}\left (\frac {i}{x^{2}}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{2+x}+x^{3}\right ) {\mathrm e}^{-x -2}}{x^{2}}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x -2} \left (4 \,{\mathrm e}^{2+x}+x^{3}\right )\right )\right )}{2}-x -1\right )^{2}\)	\(522\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x*exp(2+x)+2*x^4)*ln((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x^2-16*x-16)*exp(2+x)-2*x^5-6*x^4+2*x^3)*ln(l
n((4*exp(2+x)+x^3)/x^2/exp(2+x))-x-1)+(8*x^2*exp(2+x)+2*x^5)*ln((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x^3-16*x^2-
16*x)*exp(2+x)-2*x^6-6*x^5+2*x^4)/((4*x*exp(2+x)+x^4)*ln((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-4*x^2-4*x)*exp(2+x)-
x^5-x^4),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.43, size = 51, normalized size = 1.96 \begin {gather*} x^{2} + 2 \, x \log \left (-2 \, x + \log \left (x^{3} + 4 \, e^{\left (x + 2\right )}\right ) - 2 \, \log \relax (x) - 3\right ) + \log \left (-2 \, x + \log \left (x^{3} + 4 \, e^{\left (x + 2\right )}\right ) - 2 \, \log \relax (x) - 3\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*exp(2+x)+2*x^4)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x^2-16*x-16)*exp(2+x)-2*x^5-6*x^4+2*x^
3)*log(log((4*exp(2+x)+x^3)/x^2/exp(2+x))-x-1)+(8*x^2*exp(2+x)+2*x^5)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-8*x
^3-16*x^2-16*x)*exp(2+x)-2*x^6-6*x^5+2*x^4)/((4*x*exp(2+x)+x^4)*log((4*exp(2+x)+x^3)/x^2/exp(2+x))+(-4*x^2-4*x
)*exp(2+x)-x^5-x^4),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.84, size = 55, normalized size = 2.12 \begin {gather*} x^2+2\,x\,\ln \left (\ln \left (\frac {x^3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-2}+4}{x^2}\right )-x-1\right )+{\ln \left (\ln \left (\frac {x^3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-2}+4}{x^2}\right )-x-1\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log((exp(- x - 2)*(4*exp(x + 2) + x^3))/x^2) - x - 1)*(exp(x + 2)*(16*x + 8*x^2 + 16) - log((exp(- x
- 2)*(4*exp(x + 2) + x^3))/x^2)*(8*x*exp(x + 2) + 2*x^4) - 2*x^3 + 6*x^4 + 2*x^5) + exp(x + 2)*(16*x + 16*x^2
+ 8*x^3) - log((exp(- x - 2)*(4*exp(x + 2) + x^3))/x^2)*(8*x^2*exp(x + 2) + 2*x^5) - 2*x^4 + 6*x^5 + 2*x^6)/(e
xp(x + 2)*(4*x + 4*x^2) - log((exp(- x - 2)*(4*exp(x + 2) + x^3))/x^2)*(4*x*exp(x + 2) + x^4) + x^4 + x^5),x)

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 26.16, size = 61, normalized size = 2.35 \begin {gather*} x^{2} + 2 x \log {\left (- x + \log {\left (\frac {\left (x^{3} + 4 e^{x + 2}\right ) e^{- x - 2}}{x^{2}} \right )} - 1 \right )} + \log {\left (- x + \log {\left (\frac {\left (x^{3} + 4 e^{x + 2}\right ) e^{- x - 2}}{x^{2}} \right )} - 1 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x*exp(2+x)+2*x**4)*ln((4*exp(2+x)+x**3)/x**2/exp(2+x))+(-8*x**2-16*x-16)*exp(2+x)-2*x**5-6*x**4
+2*x**3)*ln(ln((4*exp(2+x)+x**3)/x**2/exp(2+x))-x-1)+(8*x**2*exp(2+x)+2*x**5)*ln((4*exp(2+x)+x**3)/x**2/exp(2+
x))+(-8*x**3-16*x**2-16*x)*exp(2+x)-2*x**6-6*x**5+2*x**4)/((4*x*exp(2+x)+x**4)*ln((4*exp(2+x)+x**3)/x**2/exp(2
+x))+(-4*x**2-4*x)*exp(2+x)-x**5-x**4),x)

[Out]

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

________________________________________________________________________________________