Integrand size = 108, antiderivative size = 27 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{x^2 \left (x-e^x \left (5+x-x \log \left (4 (4+x)^2\right )\right )\right )} \] Output:
exp(x^2*(x-exp(x)*(5-ln(4*(4+x)^2)*x+x)))
\[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=\int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx \] Input:
Integrate[(E^(x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*( 12*x^2 + 3*x^3 + E^x*(-40*x - 42*x^2 - 10*x^3 - x^4) + E^x*(12*x^2 + 7*x^3 + x^4)*Log[64 + 32*x + 4*x^2]))/(4 + x),x]
Output:
Integrate[(E^(x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*( 12*x^2 + 3*x^3 + E^x*(-40*x - 42*x^2 - 10*x^3 - x^4) + E^x*(12*x^2 + 7*x^3 + x^4)*Log[64 + 32*x + 4*x^2]))/(4 + x), x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (3 x^3+12 x^2+e^x \left (-x^4-10 x^3-42 x^2-40 x\right )+e^x \left (x^4+7 x^3+12 x^2\right ) \log \left (4 x^2+32 x+64\right )\right ) \exp \left (x^3+e^x \left (-x^3-5 x^2\right )+e^x x^3 \log \left (4 x^2+32 x+64\right )\right )}{x+4} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (3 x^2 \exp \left (x^3+e^x \left (-x^3-5 x^2\right )+e^x x^3 \log \left (4 x^2+32 x+64\right )\right )+\frac {x \left (-x^3+x^3 \log \left (4 (x+4)^2\right )-10 x^2+7 x^2 \log \left (4 (x+4)^2\right )-42 x+12 x \log \left (4 (x+4)^2\right )-40\right ) \exp \left (x^3+e^x \left (-x^3-5 x^2\right )+e^x x^3 \log \left (4 x^2+32 x+64\right )+x\right )}{x+4}\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (3\ 4^{e^x x^3} e^{x^3-e^x x^2 (x+5)} x^2 \left ((x+4)^2\right )^{e^x x^3}+\frac {4^{e^x x^3} e^{x^3-e^x (x+5) x^2+x} x \left ((x+4)^2\right )^{e^x x^3} \left (-x^3+x^3 \log \left (4 (x+4)^2\right )-10 x^2+7 x^2 \log \left (4 (x+4)^2\right )-42 x+12 x \log \left (4 (x+4)^2\right )-40\right )}{x+4}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (3\ 4^{e^x x^3} e^{x^3-e^x x^2 (x+5)} x^2 \left ((x+4)^2\right )^{e^x x^3}+\frac {4^{e^x x^3} e^{x^3-e^x (x+5) x^2+x} x \left ((x+4)^2\right )^{e^x x^3} \left (-x^3+x^3 \log \left (4 (x+4)^2\right )-10 x^2+7 x^2 \log \left (4 (x+4)^2\right )-42 x+12 x \log \left (4 (x+4)^2\right )-40\right )}{x+4}\right )dx\) |
Input:
Int[(E^(x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*(12*x^2 + 3*x^3 + E^x*(-40*x - 42*x^2 - 10*x^3 - x^4) + E^x*(12*x^2 + 7*x^3 + x^4 )*Log[64 + 32*x + 4*x^2]))/(4 + x),x]
Output:
$Aborted
Time = 0.93 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19
| method | result | size |
| parallelrisch | \({\mathrm e}^{x^{2} \left (x \,{\mathrm e}^{x} \ln \left (4 x^{2}+32 x +64\right )-{\mathrm e}^{x} x -5 \,{\mathrm e}^{x}+x \right )}\) | \(32\) |
| risch | \(4^{{\mathrm e}^{x} x^{3}} \left (4+x \right )^{2 \,{\mathrm e}^{x} x^{3}} {\mathrm e}^{-\frac {x^{2} \left (i {\mathrm e}^{x} x \pi \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-2 i {\mathrm e}^{x} x \pi \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (4+x \right )\right )+i {\mathrm e}^{x} x \pi \,\operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i \left (4+x \right )\right )^{2}+2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}-2 x \right )}{2}}\) | \(108\) |
Input:
int(((x^4+7*x^3+12*x^2)*exp(x)*ln(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^2-40*x) *exp(x)+3*x^3+12*x^2)*exp(x^3*exp(x)*ln(4*x^2+32*x+64)+(-x^3-5*x^2)*exp(x) +x^3)/(4+x),x,method=_RETURNVERBOSE)
Output:
exp(x^2*(x*exp(x)*ln(4*x^2+32*x+64)-exp(x)*x-5*exp(x)+x))
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{\left (x^{3} e^{x} \log \left (4 \, x^{2} + 32 \, x + 64\right ) + x^{3} - {\left (x^{3} + 5 \, x^{2}\right )} e^{x}\right )} \] Input:
integrate(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^ 2-40*x)*exp(x)+3*x^3+12*x^2)*exp(x^3*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2 )*exp(x)+x^3)/(4+x),x, algorithm="fricas")
Output:
e^(x^3*e^x*log(4*x^2 + 32*x + 64) + x^3 - (x^3 + 5*x^2)*e^x)
Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{x^{3} e^{x} \log {\left (4 x^{2} + 32 x + 64 \right )} + x^{3} + \left (- x^{3} - 5 x^{2}\right ) e^{x}} \] Input:
integrate(((x**4+7*x**3+12*x**2)*exp(x)*ln(4*x**2+32*x+64)+(-x**4-10*x**3- 42*x**2-40*x)*exp(x)+3*x**3+12*x**2)*exp(x**3*exp(x)*ln(4*x**2+32*x+64)+(- x**3-5*x**2)*exp(x)+x**3)/(4+x),x)
Output:
exp(x**3*exp(x)*log(4*x**2 + 32*x + 64) + x**3 + (-x**3 - 5*x**2)*exp(x))
Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{\left (2 \, x^{3} e^{x} \log \left (2\right ) + 2 \, x^{3} e^{x} \log \left (x + 4\right ) - x^{3} e^{x} + x^{3} - 5 \, x^{2} e^{x}\right )} \] Input:
integrate(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^ 2-40*x)*exp(x)+3*x^3+12*x^2)*exp(x^3*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2 )*exp(x)+x^3)/(4+x),x, algorithm="maxima")
Output:
e^(2*x^3*e^x*log(2) + 2*x^3*e^x*log(x + 4) - x^3*e^x + x^3 - 5*x^2*e^x)
Time = 0.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{\left (x^{3} e^{x} \log \left (4 \, x^{2} + 32 \, x + 64\right ) - x^{3} e^{x} + x^{3} - 5 \, x^{2} e^{x}\right )} \] Input:
integrate(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^ 2-40*x)*exp(x)+3*x^3+12*x^2)*exp(x^3*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2 )*exp(x)+x^3)/(4+x),x, algorithm="giac")
Output:
e^(x^3*e^x*log(4*x^2 + 32*x + 64) - x^3*e^x + x^3 - 5*x^2*e^x)
Time = 2.65 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx={\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-5\,x^2\,{\mathrm {e}}^x}\,{\left (4\,x^2+32\,x+64\right )}^{x^3\,{\mathrm {e}}^x} \] Input:
int((exp(x^3 - exp(x)*(5*x^2 + x^3) + x^3*exp(x)*log(32*x + 4*x^2 + 64))*( 12*x^2 - exp(x)*(40*x + 42*x^2 + 10*x^3 + x^4) + 3*x^3 + exp(x)*log(32*x + 4*x^2 + 64)*(12*x^2 + 7*x^3 + x^4)))/(x + 4),x)
Output:
exp(x^3)*exp(-x^3*exp(x))*exp(-5*x^2*exp(x))*(32*x + 4*x^2 + 64)^(x^3*exp( x))
\[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=\int \frac {\left (\left (x^{4}+7 x^{3}+12 x^{2}\right ) {\mathrm e}^{x} \mathrm {log}\left (4 x^{2}+32 x +64\right )+\left (-x^{4}-10 x^{3}-42 x^{2}-40 x \right ) {\mathrm e}^{x}+3 x^{3}+12 x^{2}\right ) {\mathrm e}^{x^{3} {\mathrm e}^{x} \mathrm {log}\left (4 x^{2}+32 x +64\right )+\left (-x^{3}-5 x^{2}\right ) {\mathrm e}^{x}+x^{3}}}{x +4}d x \] Input:
int(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^2-40*x )*exp(x)+3*x^3+12*x^2)*exp(x^3*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2)*exp( x)+x^3)/(4+x),x)
Output:
int(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^2-40*x )*exp(x)+3*x^3+12*x^2)*exp(x^3*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2)*exp( x)+x^3)/(4+x),x)