\(\int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} (x+x^2-24 e^2 x^2+8 x^3)+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx\) [3214]

Optimal result

Integrand size = 89, antiderivative size = 29 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x+\log \left (e^3+\frac {1}{5} e^{4 \left (3 e^2-x\right )^2} x+\log (x)\right ) \]

[Out]

x+ln(exp(3)+1/5*exp(2*(3*exp(2)-x)*(6*exp(2)-2*x))*x+ln(x))

Rubi [F]

\[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=\int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx \]

[In]

Int[(5 + 5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*(x + x^2 - 24*E^2*x^2 + 8*x^3) + 5*x*Log[x])/(5*E^3*x + E^(36
*E^4 - 24*E^2*x + 4*x^2)*x^2 + 5*x*Log[x]),x]

[Out]

(1 - 24*E^2)*x + 4*x^2 + Log[x] + 120*Defer[Int][E^(5 + 24*E^2*x)/(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*x^2)*x +
5*E^(24*E^2*x)*Log[x]), x] + 5*(1 - E^3)*Defer[Int][E^(24*E^2*x)/(x*(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*x^2)*
x + 5*E^(24*E^2*x)*Log[x])), x] - 40*Defer[Int][(E^(3 + 24*E^2*x)*x)/(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*x^2)*
x + 5*E^(24*E^2*x)*Log[x]), x] + 120*Defer[Int][(E^(2 + 24*E^2*x)*Log[x])/(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*
x^2)*x + 5*E^(24*E^2*x)*Log[x]), x] - 5*Defer[Int][(E^(24*E^2*x)*Log[x])/(x*(5*E^(3 + 24*E^2*x) + E^(36*E^4 +
4*x^2)*x + 5*E^(24*E^2*x)*Log[x])), x] - 40*Defer[Int][(E^(24*E^2*x)*x*Log[x])/(5*E^(3 + 24*E^2*x) + E^(36*E^4
+ 4*x^2)*x + 5*E^(24*E^2*x)*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1+\left (1-24 e^2\right ) x+8 x^2}{x}-\frac {5 e^{24 e^2 x} \left (-1+e^3-24 e^5 x+8 e^3 x^2+\log (x)-24 e^2 x \log (x)+8 x^2 \log (x)\right )}{x \left (5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)\right )}\right ) \, dx \\ & = -\left (5 \int \frac {e^{24 e^2 x} \left (-1+e^3-24 e^5 x+8 e^3 x^2+\log (x)-24 e^2 x \log (x)+8 x^2 \log (x)\right )}{x \left (5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)\right )} \, dx\right )+\int \frac {1+\left (1-24 e^2\right ) x+8 x^2}{x} \, dx \\ & = -\left (5 \int \left (-\frac {24 e^{5+24 e^2 x}}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)}-\frac {e^{24 e^2 x} \left (1-e^3\right )}{x \left (5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)\right )}+\frac {8 e^{3+24 e^2 x} x}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)}-\frac {24 e^{2+24 e^2 x} \log (x)}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)}+\frac {e^{24 e^2 x} \log (x)}{x \left (5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)\right )}+\frac {8 e^{24 e^2 x} x \log (x)}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)}\right ) \, dx\right )+\int \left (1-24 e^2+\frac {1}{x}+8 x\right ) \, dx \\ & = \left (1-24 e^2\right ) x+4 x^2+\log (x)-5 \int \frac {e^{24 e^2 x} \log (x)}{x \left (5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)\right )} \, dx-40 \int \frac {e^{3+24 e^2 x} x}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)} \, dx-40 \int \frac {e^{24 e^2 x} x \log (x)}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)} \, dx+120 \int \frac {e^{5+24 e^2 x}}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)} \, dx+120 \int \frac {e^{2+24 e^2 x} \log (x)}{5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)} \, dx+\left (5 \left (1-e^3\right )\right ) \int \frac {e^{24 e^2 x}}{x \left (5 e^{3+24 e^2 x}+e^{36 e^4+4 x^2} x+5 e^{24 e^2 x} \log (x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x+\log \left (5 e^3+e^{36 e^4-24 e^2 x+4 x^2} x+5 \log (x)\right ) \]

[In]

Integrate[(5 + 5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*(x + x^2 - 24*E^2*x^2 + 8*x^3) + 5*x*Log[x])/(5*E^3*x +
E^(36*E^4 - 24*E^2*x + 4*x^2)*x^2 + 5*x*Log[x]),x]

[Out]

x + Log[5*E^3 + E^(36*E^4 - 24*E^2*x + 4*x^2)*x + 5*Log[x]]

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97

method	result	size
risch	\(x +\ln \left ({\mathrm e}^{3}+\frac {{\mathrm e}^{36 \,{\mathrm e}^{4}-24 \,{\mathrm e}^{2} x +4 x^{2}} x}{5}+\ln \left (x \right )\right )\)	\(28\)
norman	\(x +\ln \left ({\mathrm e}^{36 \,{\mathrm e}^{4}-24 \,{\mathrm e}^{2} x +4 x^{2}} x +5 \ln \left (x \right )+5 \,{\mathrm e}^{3}\right )\)	\(33\)
parallelrisch	\(x +\ln \left ({\mathrm e}^{36 \,{\mathrm e}^{4}-24 \,{\mathrm e}^{2} x +4 x^{2}} x +5 \ln \left (x \right )+5 \,{\mathrm e}^{3}\right )\)	\(33\)

[In]

int((5*x*ln(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*ln(x)+x^2*ex
p(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x,method=_RETURNVERBOSE)

[Out]

x+ln(exp(3)+1/5*exp(36*exp(4)-24*exp(2)*x+4*x^2)*x+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x + \log \left (x e^{\left (4 \, x^{2} - 24 \, x e^{2} + 36 \, e^{4}\right )} + 5 \, e^{3} + 5 \, \log \left (x\right )\right ) \]

[In]

integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x
)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x, algorithm="fricas")

[Out]

x + log(x*e^(4*x^2 - 24*x*e^2 + 36*e^4) + 5*e^3 + 5*log(x))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=x + \log {\left (x \right )} + \log {\left (e^{4 x^{2} - 24 x e^{2} + 36 e^{4}} + \frac {5 \log {\left (x \right )} + 5 e^{3}}{x} \right )} \]

[In]

integrate((5*x*ln(x)+(-24*x**2*exp(2)+8*x**3+x**2+x)*exp(36*exp(2)**2-24*exp(2)*x+4*x**2)+5*x*exp(3)+5)/(5*x*l
n(x)+x**2*exp(36*exp(2)**2-24*exp(2)*x+4*x**2)+5*x*exp(3)),x)

[Out]

x + log(x) + log(exp(4*x**2 - 24*x*exp(2) + 36*exp(4)) + (5*log(x) + 5*exp(3))/x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).

Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=-x {\left (24 \, e^{2} - 1\right )} + \log \left (\frac {x e^{\left (4 \, x^{2} + 36 \, e^{4}\right )} + 5 \, {\left (e^{3} + \log \left (x\right )\right )} e^{\left (24 \, x e^{2}\right )}}{5 \, {\left (e^{3} + \log \left (x\right )\right )}}\right ) + \log \left (e^{3} + \log \left (x\right )\right ) \]

[In]

integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x
)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x, algorithm="maxima")

[Out]

-x*(24*e^2 - 1) + log(1/5*(x*e^(4*x^2 + 36*e^4) + 5*(e^3 + log(x))*e^(24*x*e^2))/(e^3 + log(x))) + log(e^3 + l
og(x))

Giac [F]

\[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=\int { \frac {5 \, x e^{3} + {\left (8 \, x^{3} - 24 \, x^{2} e^{2} + x^{2} + x\right )} e^{\left (4 \, x^{2} - 24 \, x e^{2} + 36 \, e^{4}\right )} + 5 \, x \log \left (x\right ) + 5}{x^{2} e^{\left (4 \, x^{2} - 24 \, x e^{2} + 36 \, e^{4}\right )} + 5 \, x e^{3} + 5 \, x \log \left (x\right )} \,d x } \]

[In]

integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x
)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x, algorithm="giac")

[Out]

integrate((5*x*e^3 + (8*x^3 - 24*x^2*e^2 + x^2 + x)*e^(4*x^2 - 24*x*e^2 + 36*e^4) + 5*x*log(x) + 5)/(x^2*e^(4*
x^2 - 24*x*e^2 + 36*e^4) + 5*x*e^3 + 5*x*log(x)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {5+5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} \left (x+x^2-24 e^2 x^2+8 x^3\right )+5 x \log (x)}{5 e^3 x+e^{36 e^4-24 e^2 x+4 x^2} x^2+5 x \log (x)} \, dx=\int \frac {5\,x\,{\mathrm {e}}^3+5\,x\,\ln \left (x\right )+{\mathrm {e}}^{4\,x^2-24\,{\mathrm {e}}^2\,x+36\,{\mathrm {e}}^4}\,\left (x-24\,x^2\,{\mathrm {e}}^2+x^2+8\,x^3\right )+5}{x^2\,{\mathrm {e}}^{4\,x^2-24\,{\mathrm {e}}^2\,x+36\,{\mathrm {e}}^4}+5\,x\,{\mathrm {e}}^3+5\,x\,\ln \left (x\right )} \,d x \]

[In]

int((5*x*exp(3) + 5*x*log(x) + exp(36*exp(4) - 24*x*exp(2) + 4*x^2)*(x - 24*x^2*exp(2) + x^2 + 8*x^3) + 5)/(x^
2*exp(36*exp(4) - 24*x*exp(2) + 4*x^2) + 5*x*exp(3) + 5*x*log(x)),x)

[Out]

int((5*x*exp(3) + 5*x*log(x) + exp(36*exp(4) - 24*x*exp(2) + 4*x^2)*(x - 24*x^2*exp(2) + x^2 + 8*x^3) + 5)/(x^
2*exp(36*exp(4) - 24*x*exp(2) + 4*x^2) + 5*x*exp(3) + 5*x*log(x)), x)