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

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 = {} $$\begin{array}{c}\int \frac{1+x+e^x(4-x^2+4\log (3)+{\log }^2(3))}{x+e^x(4x-4x^2+x^3+(4x-2x^2)\log (3)+x{\log }^2(3))}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{1+x+e^x(4-x^2+{\log }^2(3)+\log (81))}{x(1+e^x(2-x+\log (3){)}^2)}dx\\ & =\int (\frac{x^2+{\log }^2(3)+\log (9)-x(2+\log (9))}{(2-x+\log (3){)}^2(1+e^x{x}^2-4e^{x}x(1+\frac{\log (3)}{2})+4e^x(1+\frac{1}{4}\log (3)(4+\log (3))))}+\frac{4-x^2+{\log }^2(3)+\log (81)}{x(-2+x-\log (3){)}^2})dx\\ & =\int \frac{x^2+{\log }^2(3)+\log (9)-x(2+\log (9))}{(2-x+\log (3){)}^2(1+e^x{x}^2-4e^{x}x(1+\frac{\log (3)}{2})+4e^x(1+\frac{1}{4}\log (3)(4+\log (3))))}dx+\int \frac{4-x^2+{\log }^2(3)+\log (81)}{x(-2+x-\log (3){)}^2}dx\\ & =\int \frac{-x+\log (3)}{(2-x+\log (3))(1+e^x{x}^2-4e^{x}x(1+\frac{\log (3)}{2})+4e^x(1+\frac{1}{4}\log (3)(4+\log (3))))}dx+\int \frac{-x+\frac{4+{\log }^2(3)+\log (81)}{-2-\log (3)}}{x(-2+x-\log (3))}dx\\ & =\int (\frac{1}{x}-\frac{2}{-2+x-\log (3)})dx+\int \frac{-x+\log (3)}{(2-x+\log (3))(1+e^x(2-x+\log (3){)}^2)}dx\\ & =\log (x)-2\log (2-x+\log (3))+\int (\frac{1}{1+e^x{x}^2-4e^{x}x(1+\frac{\log (3)}{2})+4e^x(1+\frac{1}{4}\log (3)(4+\log (3)))}+\frac{2}{(-2+x-\log (3))(1+e^x{x}^2-4e^{x}x(1+\frac{\log (3)}{2})+4e^x(1+\frac{1}{4}\log (3)(4+\log (3))))})dx\\ & =\log (x)-2\log (2-x+\log (3))+2\int \frac{1}{(-2+x-\log (3))(1+e^x{x}^2-4e^{x}x(1+\frac{\log (3)}{2})+4e^x(1+\frac{1}{4}\log (3)(4+\log (3))))}dx+\int \frac{1}{1+e^x{x}^2-4e^{x}x(1+\frac{\log (3)}{2})+4e^x(1+\frac{1}{4}\log (3)(4+\log (3)))}dx\\ & =\log (x)-2\log (2-x+\log (3))+2\int \frac{1}{(-2+x-\log (3))(1+e^x(2-x+\log (3){)}^2)}dx+\int \frac{1}{1+e^x(2-x+\log (3){)}^2}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}x-2\log (x-\log (3)-2)+\log (x)-\log \left(\frac{(x^2-2(x-2)\log (3)+\log (3{)}^2-4x+4)e^x+1}{x^2-2(x-2)\log (3)+\log (3{)}^2-4x+4}\right)\end{array}$$

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 $$\begin{array}{c}x-\log (x^2{e}^x-2x{e}^x\log (3)+e^x\log (3{)}^2-4x{e}^x+4e^x\log (3)+4e^x+1)+\log (x)\end{array}$$

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	$-2\ln (x-2-\ln (3))+\ln (x)+x-\ln ({\mathrm{e}}^x+\frac{1}{\ln (3{)}^2-2x\ln (3)+x^2+4\ln (3)-4x+4})$	$44$
norman	$x-\ln (\ln (3{)}^2{\mathrm{e}}^x-2x\ln (3){\mathrm{e}}^x+{\mathrm{e}}^x{x}^2+4\ln (3){\mathrm{e}}^x-4{\mathrm{e}}^{x}x+4{\mathrm{e}}^x+1)+\ln (x)$	$45$

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 $$\begin{array}{c}x-2\log (x-\log (3)-2)+\log (x)-\log \left(\frac{(x^2-2x(\log (3)+2)+\log (3{)}^2+4\log (3)+4)e^x+1}{x^2-2x(\log (3)+2)+\log (3{)}^2+4\log (3)+4}\right)\end{array}$$

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 $$\begin{array}{c}\int \frac{x+{\mathrm{e}}^x(-x^2+4\ln (3)+{\ln (3)}^2+4)+1}{x+{\mathrm{e}}^x(4x+\ln (3)(4x-2x^2)+x{\ln (3)}^2-4x^2+x^3)}dx\end{array}$$

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 $$\begin{array}{c}x+\log (x)-\log (e^x+\frac{1}{x^2-4x-2x\log (3)+\log {(3)}^2+4+4\log (3)})-2\log (x-2-\log (3))\end{array}$$

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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