3.45.21 \(\int \frac {-1+x-2 x^2+x^3+e^x (-2+16 e^2 x^3)+(1-3 x+x^2+e^x (2+32 e^2 x^2)) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx\)

Optimal. Leaf size=32 \[ x \left (2 x+\frac {2+e^{-x} (1-x)}{4 e^2 (x+\log (x))}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 4.87, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + x - 2*x^2 + x^3 + E^x*(-2 + 16*E^2*x^3) + (1 - 3*x + x^2 + E^x*(2 + 32*E^2*x^2))*Log[x] + 16*E^(2 +
x)*x*Log[x]^2)/(4*E^(2 + x)*x^2 + 8*E^(2 + x)*x*Log[x] + 4*E^(2 + x)*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-2-x} \left (-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)\right )}{4 (x+\log (x))^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{-2-x} \left (-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)\right )}{(x+\log (x))^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {e^{-2-x}}{(x+\log (x))^2}+\frac {e^{-2-x} x}{(x+\log (x))^2}-\frac {2 e^{-2-x} x^2}{(x+\log (x))^2}+\frac {e^{-2-x} x^3}{(x+\log (x))^2}+\frac {e^{-2-x} \log (x)}{(x+\log (x))^2}-\frac {3 e^{-2-x} x \log (x)}{(x+\log (x))^2}+\frac {e^{-2-x} x^2 \log (x)}{(x+\log (x))^2}+\frac {2 \left (-1+8 e^2 x^3+\log (x)+16 e^2 x^2 \log (x)+8 e^2 x \log ^2(x)\right )}{e^2 (x+\log (x))^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {e^{-2-x}}{(x+\log (x))^2} \, dx\right )+\frac {1}{4} \int \frac {e^{-2-x} x}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-2-x} x^3}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-2-x} \log (x)}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-2-x} x^2 \log (x)}{(x+\log (x))^2} \, dx-\frac {1}{2} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx-\frac {3}{4} \int \frac {e^{-2-x} x \log (x)}{(x+\log (x))^2} \, dx+\frac {\int \frac {-1+8 e^2 x^3+\log (x)+16 e^2 x^2 \log (x)+8 e^2 x \log ^2(x)}{(x+\log (x))^2} \, dx}{2 e^2}\\ &=-\left (\frac {1}{4} \int \frac {e^{-2-x}}{(x+\log (x))^2} \, dx\right )+\frac {1}{4} \int \frac {e^{-2-x} x}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-2-x} x^3}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \left (-\frac {e^{-2-x} x}{(x+\log (x))^2}+\frac {e^{-2-x}}{x+\log (x)}\right ) \, dx+\frac {1}{4} \int \left (-\frac {e^{-2-x} x^3}{(x+\log (x))^2}+\frac {e^{-2-x} x^2}{x+\log (x)}\right ) \, dx-\frac {1}{2} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx-\frac {3}{4} \int \left (-\frac {e^{-2-x} x^2}{(x+\log (x))^2}+\frac {e^{-2-x} x}{x+\log (x)}\right ) \, dx+\frac {\int \left (8 e^2 x+\frac {-1-x}{(x+\log (x))^2}+\frac {1}{x+\log (x)}\right ) \, dx}{2 e^2}\\ &=2 x^2-\frac {1}{4} \int \frac {e^{-2-x}}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-2-x}}{x+\log (x)} \, dx+\frac {1}{4} \int \frac {e^{-2-x} x^2}{x+\log (x)} \, dx-\frac {1}{2} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx+\frac {3}{4} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx-\frac {3}{4} \int \frac {e^{-2-x} x}{x+\log (x)} \, dx+\frac {\int \frac {-1-x}{(x+\log (x))^2} \, dx}{2 e^2}+\frac {\int \frac {1}{x+\log (x)} \, dx}{2 e^2}\\ &=2 x^2-\frac {1}{4} \int \frac {e^{-2-x}}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-2-x}}{x+\log (x)} \, dx+\frac {1}{4} \int \frac {e^{-2-x} x^2}{x+\log (x)} \, dx-\frac {1}{2} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx+\frac {3}{4} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx-\frac {3}{4} \int \frac {e^{-2-x} x}{x+\log (x)} \, dx+\frac {\int \frac {1}{x+\log (x)} \, dx}{2 e^2}+\frac {\int \left (-\frac {1}{(x+\log (x))^2}-\frac {x}{(x+\log (x))^2}\right ) \, dx}{2 e^2}\\ &=2 x^2-\frac {1}{4} \int \frac {e^{-2-x}}{(x+\log (x))^2} \, dx+\frac {1}{4} \int \frac {e^{-2-x}}{x+\log (x)} \, dx+\frac {1}{4} \int \frac {e^{-2-x} x^2}{x+\log (x)} \, dx-\frac {1}{2} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx+\frac {3}{4} \int \frac {e^{-2-x} x^2}{(x+\log (x))^2} \, dx-\frac {3}{4} \int \frac {e^{-2-x} x}{x+\log (x)} \, dx-\frac {\int \frac {1}{(x+\log (x))^2} \, dx}{2 e^2}-\frac {\int \frac {x}{(x+\log (x))^2} \, dx}{2 e^2}+\frac {\int \frac {1}{x+\log (x)} \, dx}{2 e^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 5.10, size = 33, normalized size = 1.03 \begin {gather*} \frac {1}{4} x \left (8 x+\frac {e^{-2-x} \left (1+2 e^x-x\right )}{x+\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x - 2*x^2 + x^3 + E^x*(-2 + 16*E^2*x^3) + (1 - 3*x + x^2 + E^x*(2 + 32*E^2*x^2))*Log[x] + 16*E
^(2 + x)*x*Log[x]^2)/(4*E^(2 + x)*x^2 + 8*E^(2 + x)*x*Log[x] + 4*E^(2 + x)*Log[x]^2),x]

[Out]

(x*(8*x + (E^(-2 - x)*(1 + 2*E^x - x))/(x + Log[x])))/4

________________________________________________________________________________________

fricas [B]  time = 0.66, size = 56, normalized size = 1.75 \begin {gather*} \frac {8 \, x^{2} e^{\left (x + 4\right )} \log \relax (x) - {\left (x^{2} - x\right )} e^{2} + 2 \, {\left (4 \, x^{3} e^{2} + x\right )} e^{\left (x + 2\right )}}{4 \, {\left (x e^{\left (x + 4\right )} + e^{\left (x + 4\right )} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1)*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^
3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2+8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.18, size = 49, normalized size = 1.53 \begin {gather*} \frac {8 \, x^{3} e^{2} + 8 \, x^{2} e^{2} \log \relax (x) - x^{2} e^{\left (-x\right )} + x e^{\left (-x\right )} + 2 \, x}{4 \, {\left (x e^{2} + e^{2} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1)*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^
3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2+8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.22, size = 29, normalized size = 0.91

method	result	size
risch	\(2 x^{2}-\frac {x \left (x -2 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-x -2}}{4 \left (x +\ln \relax (x )\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x*exp(2)*exp(x)*ln(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1)*ln(x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+
x-1)/(4*exp(2)*exp(x)*ln(x)^2+8*x*exp(2)*exp(x)*ln(x)+4*x^2*exp(2)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 47, normalized size = 1.47 \begin {gather*} \frac {8 \, x^{3} e^{2} + 8 \, x^{2} e^{2} \log \relax (x) - {\left (x^{2} - x\right )} e^{\left (-x\right )} + 2 \, x}{4 \, {\left (x e^{2} + e^{2} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1)*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^
3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2+8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.62, size = 122, normalized size = 3.81 \begin {gather*} \frac {\frac {x\,{\mathrm {e}}^{-x-2}\,\left (2\,{\mathrm {e}}^x-x+2\,x^2-x^3+1\right )}{4\,\left (x+1\right )}-\frac {x\,{\mathrm {e}}^{-x-2}\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^x-3\,x+x^2+1\right )}{4\,\left (x+1\right )}}{x+\ln \relax (x)}-\frac {1}{2\,{\mathrm {e}}^2+2\,x\,{\mathrm {e}}^2}+2\,x^2+\frac {{\mathrm {e}}^{-x}\,\left (\frac {{\mathrm {e}}^{-2}\,x^3}{4}-\frac {3\,{\mathrm {e}}^{-2}\,x^2}{4}+\frac {{\mathrm {e}}^{-2}\,x}{4}\right )}{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + exp(x)*(16*x^3*exp(2) - 2) + log(x)*(exp(x)*(32*x^2*exp(2) + 2) - 3*x + x^2 + 1) - 2*x^2 + x^3 + 16*x
*exp(2)*exp(x)*log(x)^2 - 1)/(4*x^2*exp(2)*exp(x) + 4*exp(2)*exp(x)*log(x)^2 + 8*x*exp(2)*exp(x)*log(x)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.38, size = 44, normalized size = 1.38 \begin {gather*} 2 x^{2} + \frac {x}{2 x e^{2} + 2 e^{2} \log {\relax (x )}} + \frac {\left (- x^{2} + x\right ) e^{- x}}{4 x e^{2} + 4 e^{2} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*exp(2)*exp(x)*ln(x)**2+((32*x**2*exp(2)+2)*exp(x)+x**2-3*x+1)*ln(x)+(16*x**3*exp(2)-2)*exp(x)+
x**3-2*x**2+x-1)/(4*exp(2)*exp(x)*ln(x)**2+8*x*exp(2)*exp(x)*ln(x)+4*x**2*exp(2)*exp(x)),x)

[Out]

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

________________________________________________________________________________________