3.80.25 \(\int \frac {2+e^{16-x} (1+x)+\log (x)-\log (x^2)}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+(2+2 e^{16-x}-2 x) \log (x)+\log ^2(x)+(-2-2 e^{16-x}+2 x-2 \log (x)) \log (x^2)+\log ^2(x^2)} \, dx\)

Optimal. Leaf size=24 \[ \frac {x}{1+e^{16-x}-x+\log (x)-\log \left (x^2\right )} \]

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Rubi [F]  time = 9.41, 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 {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + E^(16 - x)*(1 + x) + Log[x] - Log[x^2])/(1 + E^(32 - 2*x) + E^(16 - x)*(2 - 2*x) - 2*x + x^2 + (2 + 2
*E^(16 - x) - 2*x)*Log[x] + Log[x]^2 + (-2 - 2*E^(16 - x) + 2*x - 2*Log[x])*Log[x^2] + Log[x^2]^2),x]

[Out]

Defer[Int][(E^x*Log[x])/((1 - x + Log[x] - Log[x^2])*(E^16 + E^x - E^x*x + E^x*Log[x] - E^x*Log[x^2])), x] + D
efer[Int][E^(16 + x)/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log[x] + E^x*Log[x^2])^2), x] +
Defer[Int][(E^(16 + x)*x^2)/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log[x] + E^x*Log[x^2])^2)
, x] - Defer[Int][(E^(16 + x)*x*Log[x])/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log[x] + E^x*
Log[x^2])^2), x] + Defer[Int][(E^(16 + x)*x*Log[x^2])/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x
*Log[x] + E^x*Log[x^2])^2), x] + 2*Defer[Int][E^x/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log
[x] + E^x*Log[x^2])), x] - Defer[Int][(E^x*Log[x^2])/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*
Log[x] + E^x*Log[x^2])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (e^{16}+2 e^x+e^{16} x+e^x \log (x)-e^x \log \left (x^2\right )\right )}{\left (e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )\right )^2} \, dx\\ &=\int \left (\frac {e^x \left (2+\log (x)-\log \left (x^2\right )\right )}{\left (1-x+\log (x)-\log \left (x^2\right )\right ) \left (e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )\right )}+\frac {e^{16+x} \left (1+x^2-x \log (x)+x \log \left (x^2\right )\right )}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\int \frac {e^x \left (2+\log (x)-\log \left (x^2\right )\right )}{\left (1-x+\log (x)-\log \left (x^2\right )\right ) \left (e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )\right )} \, dx+\int \frac {e^{16+x} \left (1+x^2-x \log (x)+x \log \left (x^2\right )\right )}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )\right )^2} \, dx\\ &=\int \left (\frac {e^{16+x}}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2}+\frac {e^{16+x} x^2}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2}-\frac {e^{16+x} x \log (x)}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2}+\frac {e^{16+x} x \log \left (x^2\right )}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2}\right ) \, dx+\int \left (\frac {e^x \log (x)}{\left (1-x+\log (x)-\log \left (x^2\right )\right ) \left (e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )\right )}+\frac {2 e^x}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )}-\frac {e^x \log \left (x^2\right )}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )}\right ) \, dx\\ &=2 \int \frac {e^x}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )} \, dx+\int \frac {e^x \log (x)}{\left (1-x+\log (x)-\log \left (x^2\right )\right ) \left (e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )\right )} \, dx+\int \frac {e^{16+x}}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2} \, dx+\int \frac {e^{16+x} x^2}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{16+x} x \log (x)}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2} \, dx+\int \frac {e^{16+x} x \log \left (x^2\right )}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^x \log \left (x^2\right )}{\left (-1+x-\log (x)+\log \left (x^2\right )\right ) \left (-e^{16}-e^x+e^x x-e^x \log (x)+e^x \log \left (x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.56, size = 35, normalized size = 1.46 \begin {gather*} \frac {e^x x}{e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + E^(16 - x)*(1 + x) + Log[x] - Log[x^2])/(1 + E^(32 - 2*x) + E^(16 - x)*(2 - 2*x) - 2*x + x^2 +
(2 + 2*E^(16 - x) - 2*x)*Log[x] + Log[x]^2 + (-2 - 2*E^(16 - x) + 2*x - 2*Log[x])*Log[x^2] + Log[x^2]^2),x]

[Out]

(E^x*x)/(E^16 + E^x - E^x*x + E^x*Log[x] - E^x*Log[x^2])

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fricas [A]  time = 0.57, size = 18, normalized size = 0.75 \begin {gather*} -\frac {x}{x - e^{\left (-x + 16\right )} + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x^2)+log(x)+(x+1)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*exp(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*
exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(-2*x+2)*exp(16-x)+x^2-2*x+1),x, algorithm="fricas")

[Out]

-x/(x - e^(-x + 16) + log(x) - 1)

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giac [A]  time = 0.29, size = 18, normalized size = 0.75 \begin {gather*} -\frac {x}{x - e^{\left (-x + 16\right )} + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x^2)+log(x)+(x+1)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*exp(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*
exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(-2*x+2)*exp(16-x)+x^2-2*x+1),x, algorithm="giac")

[Out]

-x/(x - e^(-x + 16) + log(x) - 1)

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maple [C]  time = 0.08, size = 72, normalized size = 3.00

method	result	size
risch	\(-\frac {2 x}{-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x -2 \,{\mathrm e}^{16-x}+2 \ln \relax (x )-2}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x^2)+ln(x)+(x+1)*exp(16-x)+2)/(ln(x^2)^2+(-2*ln(x)-2*exp(16-x)+2*x-2)*ln(x^2)+ln(x)^2+(2*exp(16-x)-2*
x+2)*ln(x)+exp(16-x)^2+(-2*x+2)*exp(16-x)+x^2-2*x+1),x,method=_RETURNVERBOSE)

[Out]

-2*x/(-I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x^2)^3+2*x-2*exp(16-x)+2*ln(x)-
2)

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maxima [A]  time = 0.51, size = 20, normalized size = 0.83 \begin {gather*} -\frac {x e^{x}}{{\left (x + \log \relax (x) - 1\right )} e^{x} - e^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x^2)+log(x)+(x+1)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*exp(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*
exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(-2*x+2)*exp(16-x)+x^2-2*x+1),x, algorithm="maxima")

[Out]

-x*e^x/((x + log(x) - 1)*e^x - e^16)

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mupad [B]  time = 5.17, size = 38, normalized size = 1.58 \begin {gather*} \frac {x\,{\mathrm {e}}^{x-16}}{{\mathrm {e}}^{x-16}-x\,{\mathrm {e}}^{x-16}+{\mathrm {e}}^{x-16}\,\ln \relax (x)-\ln \left (x^2\right )\,{\mathrm {e}}^{x-16}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) - log(x^2) + exp(16 - x)*(x + 1) + 2)/(exp(32 - 2*x) - 2*x - log(x^2)*(2*exp(16 - x) - 2*x + 2*log
(x) + 2) - exp(16 - x)*(2*x - 2) + log(x)^2 + log(x)*(2*exp(16 - x) - 2*x + 2) + log(x^2)^2 + x^2 + 1),x)

[Out]

(x*exp(x - 16))/(exp(x - 16) - x*exp(x - 16) + exp(x - 16)*log(x) - log(x^2)*exp(x - 16) + 1)

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sympy [A]  time = 0.33, size = 12, normalized size = 0.50 \begin {gather*} \frac {x}{- x + e^{16 - x} - \log {\relax (x )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x**2)+ln(x)+(x+1)*exp(16-x)+2)/(ln(x**2)**2+(-2*ln(x)-2*exp(16-x)+2*x-2)*ln(x**2)+ln(x)**2+(2*e
xp(16-x)-2*x+2)*ln(x)+exp(16-x)**2+(-2*x+2)*exp(16-x)+x**2-2*x+1),x)

[Out]

x/(-x + exp(16 - x) - log(x) + 1)

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