3.3.12 \(\int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx\)

Optimal. Leaf size=21 \[ -5+2 x+\log \left (\frac {5+e^2}{x^2}+x-\log (x)\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 0.59, 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 {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10 + E^2*(2 - 2*x) - 10*x + x^2 - x^3 - 2*x^4 + 2*x^3*Log[x])/(-5*x - E^2*x - x^4 + x^3*Log[x]),x]

[Out]

2*x - 2*(5 + E^2)*Defer[Int][1/(x*(5*(1 + E^2/5) + x^3 - x^2*Log[x])), x] + Defer[Int][x^2/(5*(1 + E^2/5) + x^
3 - x^2*Log[x]), x] + Defer[Int][x/(-5*(1 + E^2/5) - x^3 + x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{\left (-5-e^2\right ) x-x^4+x^3 \log (x)} \, dx\\ &=\int \left (2+\frac {-2 \left (5+e^2\right )-x^2+x^3}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )}\right ) \, dx\\ &=2 x+\int \frac {-2 \left (5+e^2\right )-x^2+x^3}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )} \, dx\\ &=2 x+\int \left (\frac {2 \left (-5-e^2\right )}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )}+\frac {x^2}{5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)}+\frac {x}{-5 \left (1+\frac {e^2}{5}\right )-x^3+x^2 \log (x)}\right ) \, dx\\ &=2 x-\left (2 \left (5+e^2\right )\right ) \int \frac {1}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )} \, dx+\int \frac {x^2}{5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)} \, dx+\int \frac {x}{-5 \left (1+\frac {e^2}{5}\right )-x^3+x^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 24, normalized size = 1.14 \begin {gather*} 2 x-2 \log (x)+\log \left (5+e^2+x^3-x^2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + E^2*(2 - 2*x) - 10*x + x^2 - x^3 - 2*x^4 + 2*x^3*Log[x])/(-5*x - E^2*x - x^4 + x^3*Log[x]),x]

[Out]

2*x - 2*Log[x] + Log[5 + E^2 + x^3 - x^2*Log[x]]

________________________________________________________________________________________

fricas [A]  time = 1.12, size = 24, normalized size = 1.14 \begin {gather*} 2 \, x + \log \left (-\frac {x^{3} - x^{2} \log \relax (x) + e^{2} + 5}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*log(x)+(-2*x+2)*exp(1)^2-2*x^4-x^3+x^2-10*x+10)/(x^3*log(x)-x*exp(1)^2-x^4-5*x),x, algorithm=
"fricas")

[Out]

2*x + log(-(x^3 - x^2*log(x) + e^2 + 5)/x^2)

________________________________________________________________________________________

giac [A]  time = 0.24, size = 26, normalized size = 1.24 \begin {gather*} 2 \, x + \log \left (-x^{3} + x^{2} \log \relax (x) - e^{2} - 5\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*log(x)+(-2*x+2)*exp(1)^2-2*x^4-x^3+x^2-10*x+10)/(x^3*log(x)-x*exp(1)^2-x^4-5*x),x, algorithm=
"giac")

[Out]

2*x + log(-x^3 + x^2*log(x) - e^2 - 5) - 2*log(x)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 21, normalized size = 1.00

method	result	size
risch	\(2 x +\ln \left (\ln \relax (x )-\frac {x^{3}+{\mathrm e}^{2}+5}{x^{2}}\right )\)	\(21\)
norman	\(-2 \ln \relax (x )+2 x +\ln \left (x^{3}-x^{2} \ln \relax (x )+{\mathrm e}^{2}+5\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3*ln(x)+(-2*x+2)*exp(1)^2-2*x^4-x^3+x^2-10*x+10)/(x^3*ln(x)-x*exp(1)^2-x^4-5*x),x,method=_RETURNVERBO
SE)

[Out]

2*x+ln(ln(x)-(x^3+exp(2)+5)/x^2)

________________________________________________________________________________________

maxima [A]  time = 0.46, size = 24, normalized size = 1.14 \begin {gather*} 2 \, x + \log \left (-\frac {x^{3} - x^{2} \log \relax (x) + e^{2} + 5}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3*log(x)+(-2*x+2)*exp(1)^2-2*x^4-x^3+x^2-10*x+10)/(x^3*log(x)-x*exp(1)^2-x^4-5*x),x, algorithm=
"maxima")

[Out]

2*x + log(-(x^3 - x^2*log(x) + e^2 + 5)/x^2)

________________________________________________________________________________________

mupad [B]  time = 0.53, size = 34, normalized size = 1.62 \begin {gather*} \ln \left ({\mathrm {e}}^2-x^2\,\ln \relax (x)+x^3+5\right )-\frac {2\,x^2\,\ln \relax (x)-2\,x^3}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x - 2*x^3*log(x) - x^2 + x^3 + 2*x^4 + exp(2)*(2*x - 2) - 10)/(5*x - x^3*log(x) + x*exp(2) + x^4),x)

[Out]

log(exp(2) - x^2*log(x) + x^3 + 5) - (2*x^2*log(x) - 2*x^3)/x^2

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 20, normalized size = 0.95 \begin {gather*} 2 x + \log {\left (\log {\relax (x )} + \frac {- x^{3} - e^{2} - 5}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3*ln(x)+(-2*x+2)*exp(1)**2-2*x**4-x**3+x**2-10*x+10)/(x**3*ln(x)-x*exp(1)**2-x**4-5*x),x)

[Out]

2*x + log(log(x) + (-x**3 - exp(2) - 5)/x**2)

________________________________________________________________________________________