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

Optimal. Leaf size=22 \[ x+\log \left (\frac {x}{1+e^x (-2+x-\log (3))^2}\right ) \]

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Rubi [F]  time = 1.91, 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 {1+x+e^x \left (4-x^2+4 \log (3)+\log ^2(3)\right )}{x+e^x \left (4 x-4 x^2+x^3+\left (4 x-2 x^2\right ) \log (3)+x \log ^2(3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+x+e^x \left (4-x^2+\log ^2(3)+\log (81)\right )}{x \left (1+e^x (2-x+\log (3))^2\right )} \, dx\\ &=\int \left (\frac {x^2+\log ^2(3)+\log (9)-x (2+\log (9))}{(2-x+\log (3))^2 \left (1+e^x x^2-4 e^x x \left (1+\frac {\log (3)}{2}\right )+4 e^x \left (1+\frac {1}{4} \log (3) (4+\log (3))\right )\right )}+\frac {4-x^2+\log ^2(3)+\log (81)}{x (-2+x-\log (3))^2}\right ) \, dx\\ &=\int \frac {x^2+\log ^2(3)+\log (9)-x (2+\log (9))}{(2-x+\log (3))^2 \left (1+e^x x^2-4 e^x x \left (1+\frac {\log (3)}{2}\right )+4 e^x \left (1+\frac {1}{4} \log (3) (4+\log (3))\right )\right )} \, dx+\int \frac {4-x^2+\log ^2(3)+\log (81)}{x (-2+x-\log (3))^2} \, dx\\ &=\int \frac {-x+\log (3)}{(2-x+\log (3)) \left (1+e^x x^2-4 e^x x \left (1+\frac {\log (3)}{2}\right )+4 e^x \left (1+\frac {1}{4} \log (3) (4+\log (3))\right )\right )} \, dx+\int \frac {-x+\frac {4+\log ^2(3)+\log (81)}{-2-\log (3)}}{x (-2+x-\log (3))} \, dx\\ &=\int \left (\frac {1}{x}-\frac {2}{-2+x-\log (3)}\right ) \, dx+\int \frac {-x+\log (3)}{(2-x+\log (3)) \left (1+e^x (2-x+\log (3))^2\right )} \, dx\\ &=\log (x)-2 \log (2-x+\log (3))+\int \left (\frac {1}{1+e^x x^2-4 e^x x \left (1+\frac {\log (3)}{2}\right )+4 e^x \left (1+\frac {1}{4} \log (3) (4+\log (3))\right )}+\frac {2}{(-2+x-\log (3)) \left (1+e^x x^2-4 e^x x \left (1+\frac {\log (3)}{2}\right )+4 e^x \left (1+\frac {1}{4} \log (3) (4+\log (3))\right )\right )}\right ) \, dx\\ &=\log (x)-2 \log (2-x+\log (3))+2 \int \frac {1}{(-2+x-\log (3)) \left (1+e^x x^2-4 e^x x \left (1+\frac {\log (3)}{2}\right )+4 e^x \left (1+\frac {1}{4} \log (3) (4+\log (3))\right )\right )} \, dx+\int \frac {1}{1+e^x x^2-4 e^x x \left (1+\frac {\log (3)}{2}\right )+4 e^x \left (1+\frac {1}{4} \log (3) (4+\log (3))\right )} \, dx\\ &=\log (x)-2 \log (2-x+\log (3))+2 \int \frac {1}{(-2+x-\log (3)) \left (1+e^x (2-x+\log (3))^2\right )} \, dx+\int \frac {1}{1+e^x (2-x+\log (3))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.11, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x+e^x \left (4-x^2+4 \log (3)+\log ^2(3)\right )}{x+e^x \left (4 x-4 x^2+x^3+\left (4 x-2 x^2\right ) \log (3)+x \log ^2(3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.65, size = 63, normalized size = 2.86 \begin {gather*} x - 2 \, \log \left (x - \log \relax (3) - 2\right ) + \log \relax (x) - \log \left (\frac {{\left (x^{2} - 2 \, {\left (x - 2\right )} \log \relax (3) + \log \relax (3)^{2} - 4 \, x + 4\right )} e^{x} + 1}{x^{2} - 2 \, {\left (x - 2\right )} \log \relax (3) + \log \relax (3)^{2} - 4 \, x + 4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 44, normalized size = 2.00 \begin {gather*} x - \log \left (x^{2} e^{x} - 2 \, x e^{x} \log \relax (3) + e^{x} \log \relax (3)^{2} - 4 \, x e^{x} + 4 \, e^{x} \log \relax (3) + 4 \, e^{x} + 1\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.09, size = 44, normalized size = 2.00

method	result	size
risch	\(-2 \ln \left (x -2-\ln \relax (3)\right )+\ln \relax (x )+x -\ln \left ({\mathrm e}^{x}+\frac {1}{\ln \relax (3)^{2}-2 x \ln \relax (3)+x^{2}+4 \ln \relax (3)-4 x +4}\right )\)	\(44\)
norman	\(x -\ln \left (\ln \relax (3)^{2} {\mathrm e}^{x}-2 x \ln \relax (3) {\mathrm e}^{x}+{\mathrm e}^{x} x^{2}+4 \ln \relax (3) {\mathrm e}^{x}-4 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+1\right )+\ln \relax (x )\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.54, size = 65, normalized size = 2.95 \begin {gather*} x - 2 \, \log \left (x - \log \relax (3) - 2\right ) + \log \relax (x) - \log \left (\frac {{\left (x^{2} - 2 \, x {\left (\log \relax (3) + 2\right )} + \log \relax (3)^{2} + 4 \, \log \relax (3) + 4\right )} e^{x} + 1}{x^{2} - 2 \, x {\left (\log \relax (3) + 2\right )} + \log \relax (3)^{2} + 4 \, \log \relax (3) + 4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {x+{\mathrm {e}}^x\,\left (-x^2+4\,\ln \relax (3)+{\ln \relax (3)}^2+4\right )+1}{x+{\mathrm {e}}^x\,\left (4\,x+\ln \relax (3)\,\left (4\,x-2\,x^2\right )+x\,{\ln \relax (3)}^2-4\,x^2+x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.11, size = 46, normalized size = 2.09 \begin {gather*} x + \log {\relax (x )} - \log {\left (e^{x} + \frac {1}{x^{2} - 4 x - 2 x \log {\relax (3 )} + \log {\relax (3 )}^{2} + 4 + 4 \log {\relax (3 )}} \right )} - 2 \log {\left (x - 2 - \log {\relax (3 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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