3.61.30 \(\int \frac {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} (1+2 e^{2 x}+2 x+e^x (2+2 x))}{(e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} (e^{2 x}+x+2 e^x x+x^2)) \log (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x})} \, dx\)

Optimal. Leaf size=21 \[ \log \left (\log \left (\frac {x+\left (e^x+x\right )^2}{e^{e^{10}}+x}\right )\right ) \]

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Rubi [F]  time = 3.25, 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 {x^2+2 e^x x^2+e^{2 x} (-1+2 x)+e^{e^{10}} \left (1+2 e^{2 x}+2 x+e^x (2+2 x)\right )}{\left (e^{2 x} x+x^2+2 e^x x^2+x^3+e^{e^{10}} \left (e^{2 x}+x+2 e^x x+x^2\right )\right ) \log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^2 + 2*E^x*x^2 + E^(2*x)*(-1 + 2*x) + E^E^10*(1 + 2*E^(2*x) + 2*x + E^x*(2 + 2*x)))/((E^(2*x)*x + x^2 +
2*E^x*x^2 + x^3 + E^E^10*(E^(2*x) + x + 2*E^x*x + x^2))*Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^10 + x)]),x]

[Out]

2*Defer[Int][Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^10 + x)]^(-1), x] - Defer[Int][1/((E^E^10 + x)*Log[(E^(2*x
) + x + 2*E^x*x + x^2)/(E^E^10 + x)]), x] + Defer[Int][1/((E^(2*x) + x + 2*E^x*x + x^2)*Log[(E^(2*x) + x + 2*E
^x*x + x^2)/(E^E^10 + x)]), x] + 2*Defer[Int][E^x/((E^(2*x) + x + 2*E^x*x + x^2)*Log[(E^(2*x) + x + 2*E^x*x +
x^2)/(E^E^10 + x)]), x] - 2*Defer[Int][(E^x*x)/((E^(2*x) + x + 2*E^x*x + x^2)*Log[(E^(2*x) + x + 2*E^x*x + x^2
)/(E^E^10 + x)]), x] - 2*Defer[Int][x^2/((E^(2*x) + x + 2*E^x*x + x^2)*Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^
10 + x)]), x]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 28, normalized size = 1.33 \begin {gather*} \log \left (\log \left (\frac {e^{2 x}+x+2 e^x x+x^2}{e^{e^{10}}+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + 2*E^x*x^2 + E^(2*x)*(-1 + 2*x) + E^E^10*(1 + 2*E^(2*x) + 2*x + E^x*(2 + 2*x)))/((E^(2*x)*x +
x^2 + 2*E^x*x^2 + x^3 + E^E^10*(E^(2*x) + x + 2*E^x*x + x^2))*Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^10 + x)])
,x]

[Out]

Log[Log[(E^(2*x) + x + 2*E^x*x + x^2)/(E^E^10 + x)]]

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fricas [A]  time = 0.85, size = 24, normalized size = 1.14 \begin {gather*} \log \left (\log \left (\frac {x^{2} + 2 \, x e^{x} + x + e^{\left (2 \, x\right )}}{x + e^{\left (e^{10}\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log((x^2 + 2*x*e^x + x + e^(2*x))/(x + e^(e^10))))

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giac [A]  time = 0.41, size = 24, normalized size = 1.14 \begin {gather*} \log \left (\log \left (\frac {x^{2} + 2 \, x e^{x} + x + e^{\left (2 \, x\right )}}{x + e^{\left (e^{10}\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log((x^2 + 2*x*e^x + x + e^(2*x))/(x + e^(e^10))))

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maple [C]  time = 0.22, size = 219, normalized size = 10.43

method	result	size
risch	\(\ln \left (\ln \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right ) \mathrm {csgn}\left (i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x^{2}+\left (2 \,{\mathrm e}^{x}+1\right ) x +{\mathrm e}^{2 x}\right )}{{\mathrm e}^{{\mathrm e}^{10}}+x}\right )^{3}-2 i \ln \left ({\mathrm e}^{{\mathrm e}^{10}}+x \right )\right )}{2}\right )\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(x^2+(2*exp(x)+1)*x+exp(2*x))-1/2*I*(Pi*csgn(I/(exp(exp(10))+x))*csgn(I*(x^2+(2*exp(x)+1)*x+exp(2*x)))*cs
gn(I/(exp(exp(10))+x)*(x^2+(2*exp(x)+1)*x+exp(2*x)))-Pi*csgn(I/(exp(exp(10))+x))*csgn(I/(exp(exp(10))+x)*(x^2+
(2*exp(x)+1)*x+exp(2*x)))^2-Pi*csgn(I*(x^2+(2*exp(x)+1)*x+exp(2*x)))*csgn(I/(exp(exp(10))+x)*(x^2+(2*exp(x)+1)
*x+exp(2*x)))^2+Pi*csgn(I/(exp(exp(10))+x)*(x^2+(2*exp(x)+1)*x+exp(2*x)))^3-2*I*ln(exp(exp(10))+x)))

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maxima [A]  time = 0.60, size = 25, normalized size = 1.19 \begin {gather*} \log \left (\log \left (x^{2} + 2 \, x e^{x} + x + e^{\left (2 \, x\right )}\right ) - \log \left (x + e^{\left (e^{10}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x^2 + 2*x*e^x + x + e^(2*x)) - log(x + e^(e^10)))

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mupad [B]  time = 6.13, size = 24, normalized size = 1.14 \begin {gather*} \ln \left (\ln \left (\frac {x+{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^x+x^2}{x+{\mathrm {e}}^{{\mathrm {e}}^{10}}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*exp(x) + exp(exp(10))*(2*x + 2*exp(2*x) + exp(x)*(2*x + 2) + 1) + exp(2*x)*(2*x - 1) + x^2)/(log((x
+ exp(2*x) + 2*x*exp(x) + x^2)/(x + exp(exp(10))))*(x*exp(2*x) + 2*x^2*exp(x) + exp(exp(10))*(x + exp(2*x) +
2*x*exp(x) + x^2) + x^2 + x^3)),x)

[Out]

log(log((x + exp(2*x) + 2*x*exp(x) + x^2)/(x + exp(exp(10)))))

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sympy [A]  time = 1.73, size = 26, normalized size = 1.24 \begin {gather*} \log {\left (\log {\left (\frac {x^{2} + 2 x e^{x} + x + e^{2 x}}{x + e^{e^{10}}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log((x**2 + 2*x*exp(x) + x + exp(2*x))/(x + exp(exp(10)))))

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