3.24.95 \(\int \frac {6-20 x-5 x^2+e^{e^x} (-3-e^x x)+(-2+e^{e^x}+5 x+x^2) \log (\frac {5}{-4 x^3+2 e^{e^x} x^3+10 x^4+2 x^5})}{-2+e^{e^x}+5 x+x^2} \, dx\)

Optimal. Leaf size=24 \[ x \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+x (5+x)\right )}\right ) \]

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Rubi [A]  time = 1.66, antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 14, number of rules used = 3, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6742, 6688, 2548} \begin {gather*} x \log \left (-\frac {5}{2 x^3 \left (-x^2-5 x-e^{e^x}+2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 - 20*x - 5*x^2 + E^E^x*(-3 - E^x*x) + (-2 + E^E^x + 5*x + x^2)*Log[5/(-4*x^3 + 2*E^E^x*x^3 + 10*x^4 + 2
*x^5)])/(-2 + E^E^x + 5*x + x^2),x]

[Out]

x*Log[-5/(2*x^3*(2 - E^E^x - 5*x - x^2))]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2}+\frac {6-3 e^{e^x}-20 x-5 x^2-2 \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )+e^{e^x} \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )+5 x \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )+x^2 \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )}{-2+e^{e^x}+5 x+x^2}\right ) \, dx\\ &=-\int \frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2} \, dx+\int \frac {6-3 e^{e^x}-20 x-5 x^2-2 \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )+e^{e^x} \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )+5 x \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )+x^2 \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )}{-2+e^{e^x}+5 x+x^2} \, dx\\ &=-\int \frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2} \, dx+\int \frac {-6+3 e^{e^x}+20 x+5 x^2-\left (-2+e^{e^x}+5 x+x^2\right ) \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )}{2-e^{e^x}-5 x-x^2} \, dx\\ &=-\int \frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2} \, dx+\int \left (-3-\frac {x (5+2 x)}{-2+e^{e^x}+5 x+x^2}+\log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right )\right ) \, dx\\ &=-3 x-\int \frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2} \, dx-\int \frac {x (5+2 x)}{-2+e^{e^x}+5 x+x^2} \, dx+\int \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right ) \, dx\\ &=-3 x+x \log \left (-\frac {5}{2 x^3 \left (2-e^{e^x}-5 x-x^2\right )}\right )-\int \frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2} \, dx-\int \frac {-6+3 e^{e^x}+20 x+e^{e^x+x} x+5 x^2}{2-e^{e^x}-5 x-x^2} \, dx-\int \left (\frac {5 x}{-2+e^{e^x}+5 x+x^2}+\frac {2 x^2}{-2+e^{e^x}+5 x+x^2}\right ) \, dx\\ &=-3 x+x \log \left (-\frac {5}{2 x^3 \left (2-e^{e^x}-5 x-x^2\right )}\right )-2 \int \frac {x^2}{-2+e^{e^x}+5 x+x^2} \, dx-5 \int \frac {x}{-2+e^{e^x}+5 x+x^2} \, dx-\int \frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2} \, dx-\int \left (-\frac {e^{e^x+x} x}{-2+e^{e^x}+5 x+x^2}-\frac {-6+3 e^{e^x}+20 x+5 x^2}{-2+e^{e^x}+5 x+x^2}\right ) \, dx\\ &=-3 x+x \log \left (-\frac {5}{2 x^3 \left (2-e^{e^x}-5 x-x^2\right )}\right )-2 \int \frac {x^2}{-2+e^{e^x}+5 x+x^2} \, dx-5 \int \frac {x}{-2+e^{e^x}+5 x+x^2} \, dx+\int \frac {-6+3 e^{e^x}+20 x+5 x^2}{-2+e^{e^x}+5 x+x^2} \, dx\\ &=-3 x+x \log \left (-\frac {5}{2 x^3 \left (2-e^{e^x}-5 x-x^2\right )}\right )-2 \int \frac {x^2}{-2+e^{e^x}+5 x+x^2} \, dx-5 \int \frac {x}{-2+e^{e^x}+5 x+x^2} \, dx+\int \left (3+\frac {x (5+2 x)}{-2+e^{e^x}+5 x+x^2}\right ) \, dx\\ &=x \log \left (-\frac {5}{2 x^3 \left (2-e^{e^x}-5 x-x^2\right )}\right )-2 \int \frac {x^2}{-2+e^{e^x}+5 x+x^2} \, dx-5 \int \frac {x}{-2+e^{e^x}+5 x+x^2} \, dx+\int \frac {x (5+2 x)}{-2+e^{e^x}+5 x+x^2} \, dx\\ &=x \log \left (-\frac {5}{2 x^3 \left (2-e^{e^x}-5 x-x^2\right )}\right )-2 \int \frac {x^2}{-2+e^{e^x}+5 x+x^2} \, dx-5 \int \frac {x}{-2+e^{e^x}+5 x+x^2} \, dx+\int \left (\frac {5 x}{-2+e^{e^x}+5 x+x^2}+\frac {2 x^2}{-2+e^{e^x}+5 x+x^2}\right ) \, dx\\ &=x \log \left (-\frac {5}{2 x^3 \left (2-e^{e^x}-5 x-x^2\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 25, normalized size = 1.04 \begin {gather*} x \log \left (\frac {5}{2 x^3 \left (-2+e^{e^x}+5 x+x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 - 20*x - 5*x^2 + E^E^x*(-3 - E^x*x) + (-2 + E^E^x + 5*x + x^2)*Log[5/(-4*x^3 + 2*E^E^x*x^3 + 10*x
^4 + 2*x^5)])/(-2 + E^E^x + 5*x + x^2),x]

[Out]

x*Log[5/(2*x^3*(-2 + E^E^x + 5*x + x^2))]

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fricas [A]  time = 1.09, size = 28, normalized size = 1.17 \begin {gather*} x \log \left (\frac {5}{2 \, {\left (x^{5} + 5 \, x^{4} + x^{3} e^{\left (e^{x}\right )} - 2 \, x^{3}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(x))+x^2+5*x-2)*log(5/(2*x^3*exp(exp(x))+2*x^5+10*x^4-4*x^3))+(-exp(x)*x-3)*exp(exp(x))-5*x
^2-20*x+6)/(exp(exp(x))+x^2+5*x-2),x, algorithm="fricas")

[Out]

x*log(5/2/(x^5 + 5*x^4 + x^3*e^(e^x) - 2*x^3))

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giac [A]  time = 0.25, size = 39, normalized size = 1.62 \begin {gather*} x \log \left (\frac {5 \, e^{x}}{2 \, {\left (x^{5} e^{x} + 5 \, x^{4} e^{x} + x^{3} e^{\left (x + e^{x}\right )} - 2 \, x^{3} e^{x}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(x))+x^2+5*x-2)*log(5/(2*x^3*exp(exp(x))+2*x^5+10*x^4-4*x^3))+(-exp(x)*x-3)*exp(exp(x))-5*x
^2-20*x+6)/(exp(exp(x))+x^2+5*x-2),x, algorithm="giac")

[Out]

x*log(5/2*e^x/(x^5*e^x + 5*x^4*e^x + x^3*e^(x + e^x) - 2*x^3*e^x))

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maple [C]  time = 0.48, size = 315, normalized size = 13.12

method	result	size
risch	\(-x \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+x^{2}+5 x -2\right )-3 x \ln \relax (x )-i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \left ({\mathrm e}^{{\mathrm e}^{x}}+x^{2}+5 x -2\right )}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{{\mathrm e}^{x}}+x^{2}+5 x -2}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \left ({\mathrm e}^{{\mathrm e}^{x}}+x^{2}+5 x -2\right )}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )}{2}+\frac {i \pi x \mathrm {csgn}\left (i x^{3}\right )^{3}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i}{x^{3} \left ({\mathrm e}^{{\mathrm e}^{x}}+x^{2}+5 x -2\right )}\right )^{3}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{{\mathrm e}^{x}}+x^{2}+5 x -2}\right ) \mathrm {csgn}\left (\frac {i}{x^{3} \left ({\mathrm e}^{{\mathrm e}^{x}}+x^{2}+5 x -2\right )}\right )}{2}-\frac {i \pi x \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}}{2}+x \ln \relax (5)-x \ln \relax (2)\)	\(315\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(exp(x))+x^2+5*x-2)*ln(5/(2*x^3*exp(exp(x))+2*x^5+10*x^4-4*x^3))+(-exp(x)*x-3)*exp(exp(x))-5*x^2-20*x
+6)/(exp(exp(x))+x^2+5*x-2),x,method=_RETURNVERBOSE)

[Out]

-x*ln(exp(exp(x))+x^2+5*x-2)-3*x*ln(x)-I*Pi*x*csgn(I*x)*csgn(I*x^2)^2-1/2*I*Pi*x*csgn(I*x)*csgn(I*x^3)^2+1/2*I
*Pi*x*csgn(I*x)^2*csgn(I*x^2)+1/2*I*Pi*x*csgn(I/x^3)*csgn(I/x^3/(exp(exp(x))+x^2+5*x-2))^2+1/2*I*Pi*x*csgn(I/(
exp(exp(x))+x^2+5*x-2))*csgn(I/x^3/(exp(exp(x))+x^2+5*x-2))^2+1/2*I*Pi*x*csgn(I*x^2)^3+1/2*I*Pi*x*csgn(I*x)*cs
gn(I*x^2)*csgn(I*x^3)+1/2*I*Pi*x*csgn(I*x^3)^3-1/2*I*Pi*x*csgn(I/x^3/(exp(exp(x))+x^2+5*x-2))^3-1/2*I*Pi*x*csg
n(I/x^3)*csgn(I/(exp(exp(x))+x^2+5*x-2))*csgn(I/x^3/(exp(exp(x))+x^2+5*x-2))-1/2*I*Pi*x*csgn(I*x^2)*csgn(I*x^3
)^2+x*ln(5)-x*ln(2)

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maxima [A]  time = 0.58, size = 30, normalized size = 1.25 \begin {gather*} x {\left (\log \relax (5) - \log \relax (2)\right )} - x \log \left (x^{2} + 5 \, x + e^{\left (e^{x}\right )} - 2\right ) - 3 \, x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(x))+x^2+5*x-2)*log(5/(2*x^3*exp(exp(x))+2*x^5+10*x^4-4*x^3))+(-exp(x)*x-3)*exp(exp(x))-5*x
^2-20*x+6)/(exp(exp(x))+x^2+5*x-2),x, algorithm="maxima")

[Out]

x*(log(5) - log(2)) - x*log(x^2 + 5*x + e^(e^x) - 2) - 3*x*log(x)

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mupad [B]  time = 1.51, size = 32, normalized size = 1.33 \begin {gather*} x\,\left (\ln \left (\frac {1}{2\,x^3\,{\mathrm {e}}^{{\mathrm {e}}^x}-4\,x^3+10\,x^4+2\,x^5}\right )+\ln \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x - log(5/(2*x^3*exp(exp(x)) - 4*x^3 + 10*x^4 + 2*x^5))*(5*x + exp(exp(x)) + x^2 - 2) + exp(exp(x))*(
x*exp(x) + 3) + 5*x^2 - 6)/(5*x + exp(exp(x)) + x^2 - 2),x)

[Out]

x*(log(1/(2*x^3*exp(exp(x)) - 4*x^3 + 10*x^4 + 2*x^5)) + log(5))

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sympy [A]  time = 1.22, size = 29, normalized size = 1.21 \begin {gather*} x \log {\left (\frac {5}{2 x^{5} + 10 x^{4} + 2 x^{3} e^{e^{x}} - 4 x^{3}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(exp(x))+x**2+5*x-2)*ln(5/(2*x**3*exp(exp(x))+2*x**5+10*x**4-4*x**3))+(-exp(x)*x-3)*exp(exp(x))
-5*x**2-20*x+6)/(exp(exp(x))+x**2+5*x-2),x)

[Out]

x*log(5/(2*x**5 + 10*x**4 + 2*x**3*exp(exp(x)) - 4*x**3))

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