3.71.94 \(\int \frac {40 x-4 x^2-4 x^3+e^{e^x} (4 x^2-4 e^x x^3)+4 x^2 \log (-e^x)}{25-10 x+x^2+e^{2 e^x} x^2+e^{e^x} (10 x-2 x^2)+(10 x-2 x^2+2 e^{e^x} x^2) \log (-e^x)+x^2 \log ^2(-e^x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {4 x}{e^{e^x}+\frac {5-x}{x}+\log \left (-e^x\right )} \]

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Rubi [F]  time = 3.17, 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 {40 x-4 x^2-4 x^3+e^{e^x} \left (4 x^2-4 e^x x^3\right )+4 x^2 \log \left (-e^x\right )}{25-10 x+x^2+e^{2 e^x} x^2+e^{e^x} \left (10 x-2 x^2\right )+\left (10 x-2 x^2+2 e^{e^x} x^2\right ) \log \left (-e^x\right )+x^2 \log ^2\left (-e^x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(40*x - 4*x^2 - 4*x^3 + E^E^x*(4*x^2 - 4*E^x*x^3) + 4*x^2*Log[-E^x])/(25 - 10*x + x^2 + E^(2*E^x)*x^2 + E^
E^x*(10*x - 2*x^2) + (10*x - 2*x^2 + 2*E^E^x*x^2)*Log[-E^x] + x^2*Log[-E^x]^2),x]

[Out]

40*Defer[Int][x/(5 - x + E^E^x*x + x*Log[-E^x])^2, x] - 4*Defer[Int][x^2/(5 - x + E^E^x*x + x*Log[-E^x])^2, x]
+ 4*Defer[Int][(E^E^x*x^2)/(5 - x + E^E^x*x + x*Log[-E^x])^2, x] - 4*Defer[Int][x^3/(5 - x + E^E^x*x + x*Log[
-E^x])^2, x] - 4*Defer[Int][(E^(E^x + x)*x^3)/(5 - x + E^E^x*x + x*Log[-E^x])^2, x] + 4*Defer[Int][(x^2*Log[-E
^x])/(5 - x + E^E^x*x + x*Log[-E^x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x \left (10-x+e^{e^x} x-\left (1+e^{e^x+x}\right ) x^2+x \log \left (-e^x\right )\right )}{\left (5+\left (-1+e^{e^x}\right ) x+x \log \left (-e^x\right )\right )^2} \, dx\\ &=4 \int \frac {x \left (10-x+e^{e^x} x-\left (1+e^{e^x+x}\right ) x^2+x \log \left (-e^x\right )\right )}{\left (5+\left (-1+e^{e^x}\right ) x+x \log \left (-e^x\right )\right )^2} \, dx\\ &=4 \int \left (\frac {10 x}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2}-\frac {x^2}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2}+\frac {e^{e^x} x^2}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2}-\frac {x^3}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2}-\frac {e^{e^x+x} x^3}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2}+\frac {x^2 \log \left (-e^x\right )}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {x^2}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2} \, dx\right )+4 \int \frac {e^{e^x} x^2}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2} \, dx-4 \int \frac {x^3}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2} \, dx-4 \int \frac {e^{e^x+x} x^3}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2} \, dx+4 \int \frac {x^2 \log \left (-e^x\right )}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2} \, dx+40 \int \frac {x}{\left (5-x+e^{e^x} x+x \log \left (-e^x\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.65, size = 26, normalized size = 1.00 \begin {gather*} \frac {4 x^2}{5+\left (-1+e^{e^x}\right ) x+x \log \left (-e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40*x - 4*x^2 - 4*x^3 + E^E^x*(4*x^2 - 4*E^x*x^3) + 4*x^2*Log[-E^x])/(25 - 10*x + x^2 + E^(2*E^x)*x^
2 + E^E^x*(10*x - 2*x^2) + (10*x - 2*x^2 + 2*E^E^x*x^2)*Log[-E^x] + x^2*Log[-E^x]^2),x]

[Out]

(4*x^2)/(5 + (-1 + E^E^x)*x + x*Log[-E^x])

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fricas [C]  time = 0.82, size = 24, normalized size = 0.92 \begin {gather*} \frac {4 \, x^{2}}{i \, \pi x + x^{2} + x e^{\left (e^{x}\right )} - x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(-exp(x))+(-4*exp(x)*x^3+4*x^2)*exp(exp(x))-4*x^3-4*x^2+40*x)/(x^2*log(-exp(x))^2+(2*exp(e
xp(x))*x^2-2*x^2+10*x)*log(-exp(x))+x^2*exp(exp(x))^2+(-2*x^2+10*x)*exp(exp(x))+x^2-10*x+25),x, algorithm="fri
cas")

[Out]

4*x^2/(I*pi*x + x^2 + x*e^(e^x) - x + 5)

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giac [B]  time = 0.26, size = 81, normalized size = 3.12 \begin {gather*} \frac {4 \, {\left (x^{4} + x^{3} e^{\left (e^{x}\right )} - x^{3} + 5 \, x^{2}\right )}}{\pi ^{2} x^{2} + x^{4} + 2 \, x^{3} e^{\left (e^{x}\right )} - 2 \, x^{3} + x^{2} e^{\left (2 \, e^{x}\right )} - 2 \, x^{2} e^{\left (e^{x}\right )} + 11 \, x^{2} + 10 \, x e^{\left (e^{x}\right )} - 10 \, x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(-exp(x))+(-4*exp(x)*x^3+4*x^2)*exp(exp(x))-4*x^3-4*x^2+40*x)/(x^2*log(-exp(x))^2+(2*exp(e
xp(x))*x^2-2*x^2+10*x)*log(-exp(x))+x^2*exp(exp(x))^2+(-2*x^2+10*x)*exp(exp(x))+x^2-10*x+25),x, algorithm="gia
c")

[Out]

4*(x^4 + x^3*e^(e^x) - x^3 + 5*x^2)/(pi^2*x^2 + x^4 + 2*x^3*e^(e^x) - 2*x^3 + x^2*e^(2*e^x) - 2*x^2*e^(e^x) +
11*x^2 + 10*x*e^(e^x) - 10*x + 25)

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maple [B]  time = 1.01, size = 50, normalized size = 1.92

method	result	size
default	\(\frac {\left (-4 \ln \left (-{\mathrm e}^{x}\right )+4 x +4\right ) x -4 x \,{\mathrm e}^{{\mathrm e}^{x}}-20}{x^{2}+x \,{\mathrm e}^{{\mathrm e}^{x}}+x \left (\ln \left (-{\mathrm e}^{x}\right )-x \right )-x +5}\)	\(50\)
risch	\(\frac {8 i x^{2}}{2 \pi x \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}-2 \pi x \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3}-2 \pi x +2 i x \,{\mathrm e}^{{\mathrm e}^{x}}+2 i x \ln \left ({\mathrm e}^{x}\right )-2 i x +10 i}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2*ln(-exp(x))+(-4*exp(x)*x^3+4*x^2)*exp(exp(x))-4*x^3-4*x^2+40*x)/(x^2*ln(-exp(x))^2+(2*exp(exp(x))*x
^2-2*x^2+10*x)*ln(-exp(x))+x^2*exp(exp(x))^2+(-2*x^2+10*x)*exp(exp(x))+x^2-10*x+25),x,method=_RETURNVERBOSE)

[Out]

((-4*ln(-exp(x))+4*x+4)*x-4*x*exp(exp(x))-20)/(x^2+x*exp(exp(x))+x*(ln(-exp(x))-x)-x+5)

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maxima [A]  time = 0.43, size = 24, normalized size = 0.92 \begin {gather*} \frac {4 \, x^{2}}{x e^{\left (e^{x}\right )} + x \log \left (-e^{x}\right ) - x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(-exp(x))+(-4*exp(x)*x^3+4*x^2)*exp(exp(x))-4*x^3-4*x^2+40*x)/(x^2*log(-exp(x))^2+(2*exp(e
xp(x))*x^2-2*x^2+10*x)*log(-exp(x))+x^2*exp(exp(x))^2+(-2*x^2+10*x)*exp(exp(x))+x^2-10*x+25),x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 4.43, size = 98, normalized size = 3.77 \begin {gather*} \frac {4\,\left (5\,x^3\,{\mathrm {e}}^x-x^4\,{\mathrm {e}}^x+x^5\,{\mathrm {e}}^x+5\,x^2-x^4+\pi \,x^4\,{\mathrm {e}}^x\,1{}\mathrm {i}\right )}{\left (x\,{\mathrm {e}}^{{\mathrm {e}}^x}-x+x^2+5+\pi \,x\,1{}\mathrm {i}\right )\,\left (x^3\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x+5\,x\,{\mathrm {e}}^x-x^2+5+\pi \,x^2\,{\mathrm {e}}^x\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x))*(4*x^3*exp(x) - 4*x^2) - 40*x + 4*x^2 + 4*x^3 - 4*x^2*log(-exp(x)))/(exp(exp(x))*(10*x - 2*x
^2) - 10*x + x^2*log(-exp(x))^2 + x^2 + x^2*exp(2*exp(x)) + log(-exp(x))*(10*x + 2*x^2*exp(exp(x)) - 2*x^2) +
25),x)

[Out]

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

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sympy [C]  time = 0.67, size = 24, normalized size = 0.92 \begin {gather*} - \frac {4 x^{2}}{- x^{2} - x e^{e^{x}} + x - i \pi x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2*ln(-exp(x))+(-4*exp(x)*x**3+4*x**2)*exp(exp(x))-4*x**3-4*x**2+40*x)/(x**2*ln(-exp(x))**2+(2*
exp(exp(x))*x**2-2*x**2+10*x)*ln(-exp(x))+x**2*exp(exp(x))**2+(-2*x**2+10*x)*exp(exp(x))+x**2-10*x+25),x)

[Out]

-4*x**2/(-x**2 - x*exp(exp(x)) + x - I*pi*x - 5)

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