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

Optimal. Leaf size=24 \[ \log \left (\left (e^{2+x}+e^{4+\frac {1}{1-x}+x^2}\right ) \log (x)\right ) \]

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Rubi [F]  time = 6.78, 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 {e^{2+x} \left (1-2 x+x^2\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (1-2 x+x^2\right )+\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x+2 x^2-4 x^3+2 x^4\right )\right ) \log (x)}{\left (e^{2+x} \left (x-2 x^2+x^3\right )+e^{\frac {-5+4 x-x^2+x^3}{-1+x}} \left (x-2 x^2+x^3\right )\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2 + x)*(1 - 2*x + x^2) + E^((-5 + 4*x - x^2 + x^3)/(-1 + x))*(1 - 2*x + x^2) + (E^(2 + x)*(x - 2*x^2 +
x^3) + E^((-5 + 4*x - x^2 + x^3)/(-1 + x))*(x + 2*x^2 - 4*x^3 + 2*x^4))*Log[x])/((E^(2 + x)*(x - 2*x^2 + x^3)
+ E^((-5 + 4*x - x^2 + x^3)/(-1 + x))*(x - 2*x^2 + x^3))*Log[x]),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 2.40, size = 29, normalized size = 1.21 \begin {gather*} -\frac {1}{-1+x}+\log \left (e^{\frac {1}{-1+x}+x}+e^{2+x^2}\right )+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 30, normalized size = 1.25 \begin {gather*} \log \left (e^{\left (x + 2\right )} + e^{\left (\frac {x^{3} - x^{2} + 4 \, x - 5}{x - 1}\right )}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.98, size = 29, normalized size = 1.21 \begin {gather*} \log \left (e^{x} + e^{\left (\frac {x^{3} - x^{2} - x}{x - 1} + 3\right )}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 32, normalized size = 1.33

method	result	size
risch	\(\ln \left ({\mathrm e}^{\frac {x^{3}-x^{2}+4 x -5}{x -1}}+{\mathrm e}^{2+x}\right )-4+\ln \left (\ln \relax (x )\right )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.02, size = 39, normalized size = 1.62 \begin {gather*} \frac {x^{2} - x - 1}{x - 1} + \log \left ({\left (e^{\left (x^{2} + 2\right )} + e^{\left (x + \frac {1}{x - 1}\right )}\right )} e^{\left (-x\right )}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.22, size = 50, normalized size = 2.08 \begin {gather*} \ln \left (\ln \relax (x)\right )+\ln \left ({\mathrm {e}}^2\,{\mathrm {e}}^x+{\mathrm {e}}^{\frac {4\,x}{x-1}}\,{\mathrm {e}}^{\frac {x^3}{x-1}}\,{\mathrm {e}}^{-\frac {x^2}{x-1}}\,{\mathrm {e}}^{-\frac {5}{x-1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.88, size = 27, normalized size = 1.12 \begin {gather*} \log {\left (e^{\frac {x^{3} - x^{2} + 4 x - 5}{x - 1}} + e^{x + 2} \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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