\(\int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 (2 x+4 x^2)+e^{e^x} (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x (-4 e^{10} x^2-8 e^5 x^3-4 x^4))}{e^{10} x+2 e^5 x^2+x^3} \, dx\) [3799]

Optimal result

Integrand size = 101, antiderivative size = 25 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=\left (1-4 e^{e^x}\right ) x+\frac {x^2}{e^5+x}+\log (x) \]

[Out]

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

Rubi [F]

\[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=\int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx \]

[In]

Int[(x^2 + 2*x^3 + E^10*(1 + x) + E^5*(2*x + 4*x^2) + E^E^x*(-4*E^10*x - 8*E^5*x^2 - 4*x^3 + E^x*(-4*E^10*x^2
- 8*E^5*x^3 - 4*x^4)))/(E^10*x + 2*E^5*x^2 + x^3),x]

[Out]

2*x + E^5/(E^5 + x) - (2*E^10)/(E^5 + x) - (2*E^5*(1 - 2*E^5))/(E^5 + x) + (E^5*(1 - E^5))/(E^5 + x) - 4*ExpIn
tegralEi[E^x] + Log[x] - 4*Defer[Int][E^(E^x + x)*x, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{x \left (e^{10}+2 e^5 x+x^2\right )} \, dx \\ & = \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{x \left (e^5+x\right )^2} \, dx \\ & = \int \frac {x^2+2 x^3+e^{10} (1+x)+2 e^5 x (1+2 x)-4 e^{e^x} x \left (e^5+x\right )^2 \left (1+e^x x\right )}{x \left (e^5+x\right )^2} \, dx \\ & = \int \left (-4 e^{e^x}-4 e^{e^x+x} x+\frac {x}{\left (e^5+x\right )^2}+\frac {2 x^2}{\left (e^5+x\right )^2}+\frac {e^{10} (1+x)}{x \left (e^5+x\right )^2}+\frac {2 e^5 (1+2 x)}{\left (e^5+x\right )^2}\right ) \, dx \\ & = 2 \int \frac {x^2}{\left (e^5+x\right )^2} \, dx-4 \int e^{e^x} \, dx-4 \int e^{e^x+x} x \, dx+\left (2 e^5\right ) \int \frac {1+2 x}{\left (e^5+x\right )^2} \, dx+e^{10} \int \frac {1+x}{x \left (e^5+x\right )^2} \, dx+\int \frac {x}{\left (e^5+x\right )^2} \, dx \\ & = 2 \int \left (1+\frac {e^{10}}{\left (e^5+x\right )^2}-\frac {2 e^5}{e^5+x}\right ) \, dx-4 \int e^{e^x+x} x \, dx-4 \text {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right )+\left (2 e^5\right ) \int \left (\frac {1-2 e^5}{\left (e^5+x\right )^2}+\frac {2}{e^5+x}\right ) \, dx+e^{10} \int \left (\frac {1}{e^{10} x}+\frac {-1+e^5}{e^5 \left (e^5+x\right )^2}-\frac {1}{e^{10} \left (e^5+x\right )}\right ) \, dx+\int \left (-\frac {e^5}{\left (e^5+x\right )^2}+\frac {1}{e^5+x}\right ) \, dx \\ & = 2 x+\frac {e^5}{e^5+x}-\frac {2 e^{10}}{e^5+x}-\frac {2 e^5 \left (1-2 e^5\right )}{e^5+x}+\frac {e^5 \left (1-e^5\right )}{e^5+x}-4 \text {Ei}\left (e^x\right )+\log (x)-4 \int e^{e^x+x} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=2 x-4 e^{e^x} x+\frac {e^{10}}{e^5+x}+\log (x) \]

[In]

Integrate[(x^2 + 2*x^3 + E^10*(1 + x) + E^5*(2*x + 4*x^2) + E^E^x*(-4*E^10*x - 8*E^5*x^2 - 4*x^3 + E^x*(-4*E^1
0*x^2 - 8*E^5*x^3 - 4*x^4)))/(E^10*x + 2*E^5*x^2 + x^3),x]

[Out]

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

Maple [A] (verified)

Time = 3.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
risch	\(2 x +\frac {{\mathrm e}^{10}}{{\mathrm e}^{5}+x}+\ln \left (x \right )-4 x \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(24\)
norman	\(\frac {2 x^{2}-4 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-4 x \,{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{10}}{{\mathrm e}^{5}+x}+\ln \left (x \right )\)	\(39\)
parallelrisch	\(-\frac {4 x \,{\mathrm e}^{5} {\mathrm e}^{{\mathrm e}^{x}}+4 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+{\mathrm e}^{10}-{\mathrm e}^{5} \ln \left (x \right )-x \ln \left (x \right )-2 x^{2}}{{\mathrm e}^{5}+x}\)	\(46\)

[In]

int((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2*exp(5)-4*x^3)*exp(exp(x))+(1+x)*exp(5)^2+
(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x*exp(5)^2+2*x^2*exp(5)+x^3),x,method=_RETURNVERBOSE)

[Out]

2*x+exp(5)^2/(exp(5)+x)+ln(x)-4*x*exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=\frac {2 \, x^{2} + 2 \, x e^{5} - 4 \, {\left (x^{2} + x e^{5}\right )} e^{\left (e^{x}\right )} + {\left (x + e^{5}\right )} \log \left (x\right ) + e^{10}}{x + e^{5}} \]

[In]

integrate((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2*exp(5)-4*x^3)*exp(exp(x))+(1+x)*exp
(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x*exp(5)^2+2*x^2*exp(5)+x^3),x, algorithm="fricas")

[Out]

(2*x^2 + 2*x*e^5 - 4*(x^2 + x*e^5)*e^(e^x) + (x + e^5)*log(x) + e^10)/(x + e^5)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=- 4 x e^{e^{x}} + 2 x + \log {\left (x \right )} + \frac {e^{10}}{x + e^{5}} \]

[In]

integrate((((-4*x**2*exp(5)**2-8*x**3*exp(5)-4*x**4)*exp(x)-4*x*exp(5)**2-8*x**2*exp(5)-4*x**3)*exp(exp(x))+(1
+x)*exp(5)**2+(4*x**2+2*x)*exp(5)+2*x**3+x**2)/(x*exp(5)**2+2*x**2*exp(5)+x**3),x)

[Out]

-4*x*exp(exp(x)) + 2*x + log(x) + exp(10)/(x + exp(5))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (23) = 46\).

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.72 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=-{\left (e^{\left (-10\right )} \log \left (x + e^{5}\right ) - e^{\left (-10\right )} \log \left (x\right ) - \frac {1}{x e^{5} + e^{10}}\right )} e^{10} + 4 \, {\left (\frac {e^{5}}{x + e^{5}} + \log \left (x + e^{5}\right )\right )} e^{5} - 4 \, x e^{\left (e^{x}\right )} - 4 \, e^{5} \log \left (x + e^{5}\right ) + 2 \, x - \frac {3 \, e^{10}}{x + e^{5}} - \frac {e^{5}}{x + e^{5}} + \log \left (x + e^{5}\right ) \]

[In]

integrate((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2*exp(5)-4*x^3)*exp(exp(x))+(1+x)*exp
(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x*exp(5)^2+2*x^2*exp(5)+x^3),x, algorithm="maxima")

[Out]

-(e^(-10)*log(x + e^5) - e^(-10)*log(x) - 1/(x*e^5 + e^10))*e^10 + 4*(e^5/(x + e^5) + log(x + e^5))*e^5 - 4*x*
e^(e^x) - 4*e^5*log(x + e^5) + 2*x - 3*e^10/(x + e^5) - e^5/(x + e^5) + log(x + e^5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (23) = 46\).

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.72 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=-\frac {4 \, x^{2} e^{\left (x + e^{x}\right )} - 2 \, x^{2} e^{x} - x e^{x} \log \left (x\right ) + 4 \, x e^{\left (x + e^{x} + 5\right )} - 2 \, x e^{\left (x + 5\right )} - e^{\left (x + 5\right )} \log \left (x\right ) - e^{\left (x + 10\right )}}{x e^{x} + e^{\left (x + 5\right )}} \]

[In]

integrate((((-4*x^2*exp(5)^2-8*x^3*exp(5)-4*x^4)*exp(x)-4*x*exp(5)^2-8*x^2*exp(5)-4*x^3)*exp(exp(x))+(1+x)*exp
(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)/(x*exp(5)^2+2*x^2*exp(5)+x^3),x, algorithm="giac")

[Out]

-(4*x^2*e^(x + e^x) - 2*x^2*e^x - x*e^x*log(x) + 4*x*e^(x + e^x + 5) - 2*x*e^(x + 5) - e^(x + 5)*log(x) - e^(x
+ 10))/(x*e^x + e^(x + 5))

Mupad [B] (verification not implemented)

Time = 10.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {x^2+2 x^3+e^{10} (1+x)+e^5 \left (2 x+4 x^2\right )+e^{e^x} \left (-4 e^{10} x-8 e^5 x^2-4 x^3+e^x \left (-4 e^{10} x^2-8 e^5 x^3-4 x^4\right )\right )}{e^{10} x+2 e^5 x^2+x^3} \, dx=2\,x+\ln \left (x\right )-4\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+\frac {{\mathrm {e}}^{10}}{x+{\mathrm {e}}^5} \]

[In]

int((exp(5)*(2*x + 4*x^2) - exp(exp(x))*(exp(x)*(8*x^3*exp(5) + 4*x^2*exp(10) + 4*x^4) + 4*x*exp(10) + 8*x^2*e
xp(5) + 4*x^3) + exp(10)*(x + 1) + x^2 + 2*x^3)/(x*exp(10) + 2*x^2*exp(5) + x^3),x)

[Out]

2*x + log(x) - 4*x*exp(exp(x)) + exp(10)/(x + exp(5))