\(\int \frac {e^x (-20+12 x+7 x^2+x^4)+e^3 (-x^3+3 x^4+2 x^5)+e^x (-2+x+x^2) \log (2+x)+(e^3 (20 x^2+8 x^3+x^4+x^5)+e^3 (2 x^2+x^3) \log (2+x)) \log (10-x+x^2+\log (2+x))}{20 x^2+8 x^3+x^4+x^5+(2 x^2+x^3) \log (2+x)} \, dx\) [6637]

Optimal result

Integrand size = 139, antiderivative size = 26 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {e^x}{x}+e^3 x \log \left (10-x+x^2+\log (2+x)\right ) \]

[Out]

exp(x)/x+exp(3)*ln(ln(2+x)+x^2-x+10)*x

Rubi [F]

\[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx \]

[In]

Int[(E^x*(-20 + 12*x + 7*x^2 + x^4) + E^3*(-x^3 + 3*x^4 + 2*x^5) + E^x*(-2 + x + x^2)*Log[2 + x] + (E^3*(20*x^
2 + 8*x^3 + x^4 + x^5) + E^3*(2*x^2 + x^3)*Log[2 + x])*Log[10 - x + x^2 + Log[2 + x]])/(20*x^2 + 8*x^3 + x^4 +
x^5 + (2*x^2 + x^3)*Log[2 + x]),x]

[Out]

E^x/x + E^3*Defer[Int][(10 - x + x^2 + Log[2 + x])^(-1), x] - E^3*Defer[Int][x/(10 - x + x^2 + Log[2 + x]), x]
+ 2*E^3*Defer[Int][x^2/(10 - x + x^2 + Log[2 + x]), x] - 2*E^3*Defer[Int][1/((2 + x)*(10 - x + x^2 + Log[2 +
x])), x] + E^3*Defer[Int][Log[10 - x + x^2 + Log[2 + x]], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{x^2 (2+x) \left (10-x+x^2+\log (2+x)\right )} \, dx \\ & = \int \left (\frac {e^x (-1+x)}{x^2}+\frac {e^3 \left (-x+3 x^2+2 x^3+20 \log \left (10-x+x^2+\log (2+x)\right )+8 x \log \left (10-x+x^2+\log (2+x)\right )+x^2 \log \left (10-x+x^2+\log (2+x)\right )+x^3 \log \left (10-x+x^2+\log (2+x)\right )+2 \log (2+x) \log \left (10-x+x^2+\log (2+x)\right )+x \log (2+x) \log \left (10-x+x^2+\log (2+x)\right )\right )}{(2+x) \left (10-x+x^2+\log (2+x)\right )}\right ) \, dx \\ & = e^3 \int \frac {-x+3 x^2+2 x^3+20 \log \left (10-x+x^2+\log (2+x)\right )+8 x \log \left (10-x+x^2+\log (2+x)\right )+x^2 \log \left (10-x+x^2+\log (2+x)\right )+x^3 \log \left (10-x+x^2+\log (2+x)\right )+2 \log (2+x) \log \left (10-x+x^2+\log (2+x)\right )+x \log (2+x) \log \left (10-x+x^2+\log (2+x)\right )}{(2+x) \left (10-x+x^2+\log (2+x)\right )} \, dx+\int \frac {e^x (-1+x)}{x^2} \, dx \\ & = \frac {e^x}{x}+e^3 \int \left (\frac {x \left (-1+3 x+2 x^2\right )}{(2+x) \left (10-x+x^2+\log (2+x)\right )}+\log \left (10-x+x^2+\log (2+x)\right )\right ) \, dx \\ & = \frac {e^x}{x}+e^3 \int \frac {x \left (-1+3 x+2 x^2\right )}{(2+x) \left (10-x+x^2+\log (2+x)\right )} \, dx+e^3 \int \log \left (10-x+x^2+\log (2+x)\right ) \, dx \\ & = \frac {e^x}{x}+e^3 \int \left (\frac {1}{10-x+x^2+\log (2+x)}-\frac {x}{10-x+x^2+\log (2+x)}+\frac {2 x^2}{10-x+x^2+\log (2+x)}-\frac {2}{(2+x) \left (10-x+x^2+\log (2+x)\right )}\right ) \, dx+e^3 \int \log \left (10-x+x^2+\log (2+x)\right ) \, dx \\ & = \frac {e^x}{x}+e^3 \int \frac {1}{10-x+x^2+\log (2+x)} \, dx-e^3 \int \frac {x}{10-x+x^2+\log (2+x)} \, dx+e^3 \int \log \left (10-x+x^2+\log (2+x)\right ) \, dx+\left (2 e^3\right ) \int \frac {x^2}{10-x+x^2+\log (2+x)} \, dx-\left (2 e^3\right ) \int \frac {1}{(2+x) \left (10-x+x^2+\log (2+x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {e^x}{x}+e^3 x \log \left (10-x+x^2+\log (2+x)\right ) \]

[In]

Integrate[(E^x*(-20 + 12*x + 7*x^2 + x^4) + E^3*(-x^3 + 3*x^4 + 2*x^5) + E^x*(-2 + x + x^2)*Log[2 + x] + (E^3*
(20*x^2 + 8*x^3 + x^4 + x^5) + E^3*(2*x^2 + x^3)*Log[2 + x])*Log[10 - x + x^2 + Log[2 + x]])/(20*x^2 + 8*x^3 +
x^4 + x^5 + (2*x^2 + x^3)*Log[2 + x]),x]

[Out]

E^x/x + E^3*x*Log[10 - x + x^2 + Log[2 + x]]

Maple [A] (verified)

Time = 32.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {{\mathrm e}^{x}}{x}+{\mathrm e}^{3} \ln \left (\ln \left (2+x \right )+x^{2}-x +10\right ) x\)	\(25\)
parallelrisch	\(\frac {\ln \left (\ln \left (2+x \right )+x^{2}-x +10\right ) {\mathrm e}^{3} x^{2}+{\mathrm e}^{x}}{x}\)	\(27\)

[In]

int((((x^3+2*x^2)*exp(3)*ln(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*ln(ln(2+x)+x^2-x+10)+(x^2+x-2)*exp(x)*ln(2+x)+
(x^4+7*x^2+12*x-20)*exp(x)+(2*x^5+3*x^4-x^3)*exp(3))/((x^3+2*x^2)*ln(2+x)+x^5+x^4+8*x^3+20*x^2),x,method=_RETU
RNVERBOSE)

[Out]

exp(x)/x+exp(3)*ln(ln(2+x)+x^2-x+10)*x

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {x^{2} e^{3} \log \left (x^{2} - x + \log \left (x + 2\right ) + 10\right ) + e^{x}}{x} \]

[In]

integrate((((x^3+2*x^2)*exp(3)*log(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*log(log(2+x)+x^2-x+10)+(x^2+x-2)*exp(x)
*log(2+x)+(x^4+7*x^2+12*x-20)*exp(x)+(2*x^5+3*x^4-x^3)*exp(3))/((x^3+2*x^2)*log(2+x)+x^5+x^4+8*x^3+20*x^2),x,
algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\left (x e^{3} + e^{3}\right ) \log {\left (x^{2} - x + \log {\left (x + 2 \right )} + 10 \right )} - e^{3} \log {\left (x^{2} - x + \log {\left (x + 2 \right )} + 10 \right )} + \frac {e^{x}}{x} \]

[In]

integrate((((x**3+2*x**2)*exp(3)*ln(2+x)+(x**5+x**4+8*x**3+20*x**2)*exp(3))*ln(ln(2+x)+x**2-x+10)+(x**2+x-2)*e
xp(x)*ln(2+x)+(x**4+7*x**2+12*x-20)*exp(x)+(2*x**5+3*x**4-x**3)*exp(3))/((x**3+2*x**2)*ln(2+x)+x**5+x**4+8*x**
3+20*x**2),x)

[Out]

(x*exp(3) + exp(3))*log(x**2 - x + log(x + 2) + 10) - exp(3)*log(x**2 - x + log(x + 2) + 10) + exp(x)/x

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {x^{2} e^{3} \log \left (x^{2} - x + \log \left (x + 2\right ) + 10\right ) + e^{x}}{x} \]

[In]

integrate((((x^3+2*x^2)*exp(3)*log(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*log(log(2+x)+x^2-x+10)+(x^2+x-2)*exp(x)
*log(2+x)+(x^4+7*x^2+12*x-20)*exp(x)+(2*x^5+3*x^4-x^3)*exp(3))/((x^3+2*x^2)*log(2+x)+x^5+x^4+8*x^3+20*x^2),x,
algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {x^{2} e^{3} \log \left (x^{2} - x + \log \left (x + 2\right ) + 10\right ) + e^{x}}{x} \]

[In]

integrate((((x^3+2*x^2)*exp(3)*log(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*log(log(2+x)+x^2-x+10)+(x^2+x-2)*exp(x)
*log(2+x)+(x^4+7*x^2+12*x-20)*exp(x)+(2*x^5+3*x^4-x^3)*exp(3))/((x^3+2*x^2)*log(2+x)+x^5+x^4+8*x^3+20*x^2),x,
algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {{\mathrm {e}}^x}{x}+x\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+2\right )-x+x^2+10\right ) \]

[In]

int((log(log(x + 2) - x + x^2 + 10)*(exp(3)*(20*x^2 + 8*x^3 + x^4 + x^5) + log(x + 2)*exp(3)*(2*x^2 + x^3)) +
exp(x)*(12*x + 7*x^2 + x^4 - 20) + exp(3)*(3*x^4 - x^3 + 2*x^5) + log(x + 2)*exp(x)*(x + x^2 - 2))/(log(x + 2)
*(2*x^2 + x^3) + 20*x^2 + 8*x^3 + x^4 + x^5),x)

[Out]

exp(x)/x + x*exp(3)*log(log(x + 2) - x + x^2 + 10)