3.64.24 \(\int \frac {e^{1+5 e^x} (1+x-2 x^3+e^x (5 x^2+5 x^3+5 x^4)+(-1-2 x-3 x^2+e^x (5 x+5 x^2+5 x^3)) \log (x))}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx\)
Optimal. Leaf size=25 \[ \frac {e^{1+5 e^x} (x+\log (x))}{x \left (1+x+x^2\right )} \]
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Rubi [F] time = 10.39, 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^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Int[(E^(1 + 5*E^x)*(1 + x - 2*x^3 + E^x*(5*x^2 + 5*x^3 + 5*x^4) + (-1 - 2*x - 3*x^2 + E^x*(5*x + 5*x^2 + 5*x^3
))*Log[x]))/(x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6),x]
[Out]
(4*Defer[Int][E^(1 + 5*E^x)/(-1 + I*Sqrt[3] - 2*x)^2, x])/3 - (4*(1 - I*Sqrt[3])*Defer[Int][E^(1 + 5*E^x)/(-1
+ I*Sqrt[3] - 2*x)^2, x])/3 + (4*Log[x]*Defer[Int][E^(1 + 5*E^x)/(-1 + I*Sqrt[3] - 2*x)^2, x])/3 + (2*(1 - I*S
qrt[3])*Log[x]*Defer[Int][E^(1 + 5*E^x)/(-1 + I*Sqrt[3] - 2*x)^2, x])/3 - ((2*I)*Log[x]*Defer[Int][E^(1 + 5*E^
x)/(-1 + I*Sqrt[3] - 2*x), x])/Sqrt[3] + ((10*I)*Defer[Int][E^(1 + 5*E^x + x)/(-1 + I*Sqrt[3] - 2*x), x])/Sqrt
[3] + Defer[Int][E^(1 + 5*E^x)/x^2, x] - Log[x]*Defer[Int][E^(1 + 5*E^x)/x^2, x] - Defer[Int][E^(1 + 5*E^x)/x,
x] + 5*Log[x]*Defer[Int][E^(1 + 5*E^x + x)/x, x] + 2*Defer[Int][E^(1 + 5*E^x)/(1 - I*Sqrt[3] + 2*x), x] - ((3
- I*Sqrt[3])*Defer[Int][E^(1 + 5*E^x)/(1 - I*Sqrt[3] + 2*x), x])/3 - 2*Log[x]*Defer[Int][E^(1 + 5*E^x)/(1 - I
*Sqrt[3] + 2*x), x] + (2*(3 - I*Sqrt[3])*Log[x]*Defer[Int][E^(1 + 5*E^x)/(1 - I*Sqrt[3] + 2*x), x])/3 - (5*(3
- I*Sqrt[3])*Log[x]*Defer[Int][E^(1 + 5*E^x + x)/(1 - I*Sqrt[3] + 2*x), x])/3 + (4*Defer[Int][E^(1 + 5*E^x)/(1
+ I*Sqrt[3] + 2*x)^2, x])/3 - (4*(1 + I*Sqrt[3])*Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x)^2, x])/3 + (4
*Log[x]*Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x)^2, x])/3 + (2*(1 + I*Sqrt[3])*Log[x]*Defer[Int][E^(1 +
5*E^x)/(1 + I*Sqrt[3] + 2*x)^2, x])/3 + 2*Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x), x] - ((3 + I*Sqrt[3]
)*Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x), x])/3 - 2*Log[x]*Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2
*x), x] - ((2*I)*Log[x]*Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x), x])/Sqrt[3] + (2*(3 + I*Sqrt[3])*Log[x
]*Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x), x])/3 + ((10*I)*Defer[Int][E^(1 + 5*E^x + x)/(1 + I*Sqrt[3]
+ 2*x), x])/Sqrt[3] - (5*(3 + I*Sqrt[3])*Log[x]*Defer[Int][E^(1 + 5*E^x + x)/(1 + I*Sqrt[3] + 2*x), x])/3 + ((
2*I)*Defer[Int][Defer[Int][E^(1 + 5*E^x)/(-1 + I*Sqrt[3] - 2*x), x]/x, x])/Sqrt[3] - 4*Defer[Int][Defer[Int][-
(E^(1 + 5*E^x)/(I + Sqrt[3] + (2*I)*x)^2), x]/x, x] + (2*(1 - I*Sqrt[3])*Defer[Int][Defer[Int][-(E^(1 + 5*E^x)
/(I + Sqrt[3] + (2*I)*x)^2), x]/x, x])/3 + (4*(1 + I*Sqrt[3])*Defer[Int][Defer[Int][-(E^(1 + 5*E^x)/(I + Sqrt[
3] + (2*I)*x)^2), x]/x, x])/3 + Defer[Int][Defer[Int][E^(1 + 5*E^x)/x^2, x]/x, x] - 5*Defer[Int][Defer[Int][E^
(1 + 5*E^x + x)/x, x]/x, x] + 2*Defer[Int][Defer[Int][E^(1 + 5*E^x)/(1 - I*Sqrt[3] + 2*x), x]/x, x] - (2*(3 -
I*Sqrt[3])*Defer[Int][Defer[Int][E^(1 + 5*E^x)/(1 - I*Sqrt[3] + 2*x), x]/x, x])/3 + (5*(3 - I*Sqrt[3])*Defer[I
nt][Defer[Int][E^(1 + 5*E^x + x)/(1 - I*Sqrt[3] + 2*x), x]/x, x])/3 - 4*Defer[Int][Defer[Int][E^(1 + 5*E^x)/(1
+ I*Sqrt[3] + 2*x)^2, x]/x, x] + (4*(1 - I*Sqrt[3])*Defer[Int][Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x)
^2, x]/x, x])/3 + (2*(1 + I*Sqrt[3])*Defer[Int][Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x)^2, x]/x, x])/3
+ ((4*I)*Defer[Int][Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x), x]/x, x])/Sqrt[3] + (2*(9 - I*Sqrt[3])*Def
er[Int][Defer[Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x), x]/x, x])/9 - (2*(9 + (5*I)*Sqrt[3])*Defer[Int][Defer[
Int][E^(1 + 5*E^x)/(1 + I*Sqrt[3] + 2*x), x]/x, x])/9 + (5*(3 + I*Sqrt[3])*Defer[Int][Defer[Int][E^(1 + 5*E^x
+ x)/(1 + I*Sqrt[3] + 2*x), x]/x, x])/3
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2 \left (1+x+x^2\right )^2} \, dx\\ &=\int \left (\frac {e^{1+5 e^x}}{x^2 \left (1+x+x^2\right )^2}+\frac {e^{1+5 e^x}}{x \left (1+x+x^2\right )^2}-\frac {2 e^{1+5 e^x} x}{\left (1+x+x^2\right )^2}-\frac {3 e^{1+5 e^x} \log (x)}{\left (1+x+x^2\right )^2}-\frac {e^{1+5 e^x} \log (x)}{x^2 \left (1+x+x^2\right )^2}-\frac {2 e^{1+5 e^x} \log (x)}{x \left (1+x+x^2\right )^2}+\frac {5 e^{1+5 e^x+x} (x+\log (x))}{x \left (1+x+x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{1+5 e^x} x}{\left (1+x+x^2\right )^2} \, dx\right )-2 \int \frac {e^{1+5 e^x} \log (x)}{x \left (1+x+x^2\right )^2} \, dx-3 \int \frac {e^{1+5 e^x} \log (x)}{\left (1+x+x^2\right )^2} \, dx+5 \int \frac {e^{1+5 e^x+x} (x+\log (x))}{x \left (1+x+x^2\right )} \, dx+\int \frac {e^{1+5 e^x}}{x^2 \left (1+x+x^2\right )^2} \, dx+\int \frac {e^{1+5 e^x}}{x \left (1+x+x^2\right )^2} \, dx-\int \frac {e^{1+5 e^x} \log (x)}{x^2 \left (1+x+x^2\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^{1+5 e^x} (x+\log (x))}{x \left (1+x+x^2\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(E^(1 + 5*E^x)*(1 + x - 2*x^3 + E^x*(5*x^2 + 5*x^3 + 5*x^4) + (-1 - 2*x - 3*x^2 + E^x*(5*x + 5*x^2 +
5*x^3))*Log[x]))/(x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6),x]
[Out]
(E^(1 + 5*E^x)*(x + Log[x]))/(x*(1 + x + x^2))
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fricas [A] time = 0.75, size = 22, normalized size = 0.88 \begin {gather*} \frac {{\left (x + \log \relax (x)\right )} e^{\left (5 \, e^{x} + 1\right )}}{x^{3} + x^{2} + x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*log(x)+(5*x^4+5*x^3+5*x^2)*exp(x)-2*x^3+x+1)*exp(5*exp(x)+1)
/(x^6+2*x^5+3*x^4+2*x^3+x^2),x, algorithm="fricas")
[Out]
(x + log(x))*e^(5*e^x + 1)/(x^3 + x^2 + x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, x^{3} - 5 \, {\left (x^{4} + x^{3} + x^{2}\right )} e^{x} + {\left (3 \, x^{2} - 5 \, {\left (x^{3} + x^{2} + x\right )} e^{x} + 2 \, x + 1\right )} \log \relax (x) - x - 1\right )} e^{\left (5 \, e^{x} + 1\right )}}{x^{6} + 2 \, x^{5} + 3 \, x^{4} + 2 \, x^{3} + x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*log(x)+(5*x^4+5*x^3+5*x^2)*exp(x)-2*x^3+x+1)*exp(5*exp(x)+1)
/(x^6+2*x^5+3*x^4+2*x^3+x^2),x, algorithm="giac")
[Out]
integrate(-(2*x^3 - 5*(x^4 + x^3 + x^2)*e^x + (3*x^2 - 5*(x^3 + x^2 + x)*e^x + 2*x + 1)*log(x) - x - 1)*e^(5*e
^x + 1)/(x^6 + 2*x^5 + 3*x^4 + 2*x^3 + x^2), x)
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maple [A] time = 0.06, size = 24, normalized size = 0.96
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\(\frac {{\mathrm e}^{5 \,{\mathrm e}^{x}+1} \left (x +\ln \relax (x )\right )}{\left (x^{2}+x +1\right ) x}\) |
\(24\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*ln(x)+(5*x^4+5*x^3+5*x^2)*exp(x)-2*x^3+x+1)*exp(5*exp(x)+1)/(x^6+2
*x^5+3*x^4+2*x^3+x^2),x,method=_RETURNVERBOSE)
[Out]
exp(5*exp(x)+1)/(x^2+x+1)/x*(x+ln(x))
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maxima [A] time = 0.45, size = 26, normalized size = 1.04 \begin {gather*} \frac {{\left (x e + e \log \relax (x)\right )} e^{\left (5 \, e^{x}\right )}}{x^{3} + x^{2} + x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*log(x)+(5*x^4+5*x^3+5*x^2)*exp(x)-2*x^3+x+1)*exp(5*exp(x)+1)
/(x^6+2*x^5+3*x^4+2*x^3+x^2),x, algorithm="maxima")
[Out]
(x*e + e*log(x))*e^(5*e^x)/(x^3 + x^2 + x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{5\,{\mathrm {e}}^x+1}\,\left (x+{\mathrm {e}}^x\,\left (5\,x^4+5\,x^3+5\,x^2\right )-\ln \relax (x)\,\left (2\,x+3\,x^2-{\mathrm {e}}^x\,\left (5\,x^3+5\,x^2+5\,x\right )+1\right )-2\,x^3+1\right )}{x^6+2\,x^5+3\,x^4+2\,x^3+x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((exp(5*exp(x) + 1)*(x + exp(x)*(5*x^2 + 5*x^3 + 5*x^4) - log(x)*(2*x + 3*x^2 - exp(x)*(5*x + 5*x^2 + 5*x^3
) + 1) - 2*x^3 + 1))/(x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6),x)
[Out]
int((exp(5*exp(x) + 1)*(x + exp(x)*(5*x^2 + 5*x^3 + 5*x^4) - log(x)*(2*x + 3*x^2 - exp(x)*(5*x + 5*x^2 + 5*x^3
) + 1) - 2*x^3 + 1))/(x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6), x)
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sympy [A] time = 0.39, size = 20, normalized size = 0.80 \begin {gather*} \frac {\left (x + \log {\relax (x )}\right ) e^{5 e^{x} + 1}}{x^{3} + x^{2} + x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((((5*x**3+5*x**2+5*x)*exp(x)-3*x**2-2*x-1)*ln(x)+(5*x**4+5*x**3+5*x**2)*exp(x)-2*x**3+x+1)*exp(5*exp
(x)+1)/(x**6+2*x**5+3*x**4+2*x**3+x**2),x)
[Out]
(x + log(x))*exp(5*exp(x) + 1)/(x**3 + x**2 + x)
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