\(\int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} (16+16 x+28 x^2+19 x^3+4 x^4+(8+4 x+6 x^2+3 x^3) \log (2+x)+\log ^2(2+x))}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx\) [5746]

Optimal result

Integrand size = 92, antiderivative size = 21 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=e^{\frac {x^3}{4+2 x+\log (2+x)}} (2+x) \]

[Out]

(2+x)*exp(x^3/(ln(2+x)+2*x+4))

Rubi [F]

\[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=\int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx \]

[In]

Int[(E^(x^3/(4 + 2*x + Log[2 + x]))*(16 + 16*x + 28*x^2 + 19*x^3 + 4*x^4 + (8 + 4*x + 6*x^2 + 3*x^3)*Log[2 + x
] + Log[2 + x]^2))/(16 + 16*x + 4*x^2 + (8 + 4*x)*Log[2 + x] + Log[2 + x]^2),x]

[Out]

Defer[Int][E^(x^3/(4 + 2*x + Log[2 + x])), x] - 5*Defer[Int][(E^(x^3/(4 + 2*x + Log[2 + x]))*x^3)/(4 + 2*x + L
og[2 + x])^2, x] - 2*Defer[Int][(E^(x^3/(4 + 2*x + Log[2 + x]))*x^4)/(4 + 2*x + Log[2 + x])^2, x] + 6*Defer[In
t][(E^(x^3/(4 + 2*x + Log[2 + x]))*x^2)/(4 + 2*x + Log[2 + x]), x] + 3*Defer[Int][(E^(x^3/(4 + 2*x + Log[2 + x
]))*x^3)/(4 + 2*x + Log[2 + x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{(4+2 x+\log (2+x))^2} \, dx \\ & = \int \left (e^{\frac {x^3}{4+2 x+\log (2+x)}}-\frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3 (5+2 x)}{(4+2 x+\log (2+x))^2}+\frac {3 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2 (2+x)}{4+2 x+\log (2+x)}\right ) \, dx \\ & = 3 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2 (2+x)}{4+2 x+\log (2+x)} \, dx+\int e^{\frac {x^3}{4+2 x+\log (2+x)}} \, dx-\int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3 (5+2 x)}{(4+2 x+\log (2+x))^2} \, dx \\ & = 3 \int \left (\frac {2 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2}{4+2 x+\log (2+x)}+\frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{4+2 x+\log (2+x)}\right ) \, dx+\int e^{\frac {x^3}{4+2 x+\log (2+x)}} \, dx-\int \left (\frac {5 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{(4+2 x+\log (2+x))^2}+\frac {2 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^4}{(4+2 x+\log (2+x))^2}\right ) \, dx \\ & = -\left (2 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^4}{(4+2 x+\log (2+x))^2} \, dx\right )+3 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{4+2 x+\log (2+x)} \, dx-5 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{(4+2 x+\log (2+x))^2} \, dx+6 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2}{4+2 x+\log (2+x)} \, dx+\int e^{\frac {x^3}{4+2 x+\log (2+x)}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=e^{\frac {x^3}{4+2 x+\log (2+x)}} (2+x) \]

[In]

Integrate[(E^(x^3/(4 + 2*x + Log[2 + x]))*(16 + 16*x + 28*x^2 + 19*x^3 + 4*x^4 + (8 + 4*x + 6*x^2 + 3*x^3)*Log
[2 + x] + Log[2 + x]^2))/(16 + 16*x + 4*x^2 + (8 + 4*x)*Log[2 + x] + Log[2 + x]^2),x]

[Out]

E^(x^3/(4 + 2*x + Log[2 + x]))*(2 + x)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(\left (2+x \right ) {\mathrm e}^{\frac {x^{3}}{\ln \left (2+x \right )+2 x +4}}\)	\(21\)
parallelrisch	\({\mathrm e}^{\frac {x^{3}}{\ln \left (2+x \right )+2 x +4}} x +2 \,{\mathrm e}^{\frac {x^{3}}{\ln \left (2+x \right )+2 x +4}}\)	\(38\)

[In]

int((ln(2+x)^2+(3*x^3+6*x^2+4*x+8)*ln(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(ln(2+x)+2*x+4))/(ln(2+x)^2+(4
*x+8)*ln(2+x)+4*x^2+16*x+16),x,method=_RETURNVERBOSE)

[Out]

(2+x)*exp(x^3/(ln(2+x)+2*x+4))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx={\left (x + 2\right )} e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} \]

[In]

integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(log(2+x)+2*x+4))/(log
(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16),x, algorithm="fricas")

[Out]

(x + 2)*e^(x^3/(2*x + log(x + 2) + 4))

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((ln(2+x)**2+(3*x**3+6*x**2+4*x+8)*ln(2+x)+4*x**4+19*x**3+28*x**2+16*x+16)*exp(x**3/(ln(2+x)+2*x+4))/
(ln(2+x)**2+(4*x+8)*ln(2+x)+4*x**2+16*x+16),x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [F]

\[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=\int { \frac {{\left (4 \, x^{4} + 19 \, x^{3} + 28 \, x^{2} + {\left (3 \, x^{3} + 6 \, x^{2} + 4 \, x + 8\right )} \log \left (x + 2\right ) + \log \left (x + 2\right )^{2} + 16 \, x + 16\right )} e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )}}{4 \, x^{2} + 4 \, {\left (x + 2\right )} \log \left (x + 2\right ) + \log \left (x + 2\right )^{2} + 16 \, x + 16} \,d x } \]

[In]

integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(log(2+x)+2*x+4))/(log
(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16),x, algorithm="maxima")

[Out]

integrate((4*x^4 + 19*x^3 + 28*x^2 + (3*x^3 + 6*x^2 + 4*x + 8)*log(x + 2) + log(x + 2)^2 + 16*x + 16)*e^(x^3/(
2*x + log(x + 2) + 4))/(4*x^2 + 4*(x + 2)*log(x + 2) + log(x + 2)^2 + 16*x + 16), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx=x e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} + 2 \, e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} \]

[In]

integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(log(2+x)+2*x+4))/(log
(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16),x, algorithm="giac")

[Out]

x*e^(x^3/(2*x + log(x + 2) + 4)) + 2*e^(x^3/(2*x + log(x + 2) + 4))

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx={\mathrm {e}}^{\frac {x^3}{2\,x+\ln \left (x+2\right )+4}}\,\left (x+2\right ) \]

[In]

int((exp(x^3/(2*x + log(x + 2) + 4))*(16*x + log(x + 2)*(4*x + 6*x^2 + 3*x^3 + 8) + log(x + 2)^2 + 28*x^2 + 19
*x^3 + 4*x^4 + 16))/(16*x + log(x + 2)^2 + 4*x^2 + log(x + 2)*(4*x + 8) + 16),x)

[Out]

exp(x^3/(2*x + log(x + 2) + 4))*(x + 2)