Integrand size = 75, antiderivative size = 21 \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx=e^{5 (4+x)^{x^2} (-2+2 x) \left (x+x^2\right )} \] Output:
exp(exp(x^2*ln(4+x))*(10*x-10)*(x^2+x))
Time = 5.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx=e^{10 x (4+x)^{x^2} \left (-1+x^2\right )} \] Input:
Integrate[E^((4 + x)^x^2*(-10*x + 10*x^3))*(4 + x)^(-1 + x^2)*(-40 - 10*x + 120*x^2 + 20*x^3 + 10*x^5 + (-80*x^2 - 20*x^3 + 80*x^4 + 20*x^5)*Log[4 + x]),x]
Output:
E^(10*x*(4 + x)^x^2*(-1 + x^2))
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.
\(\displaystyle \int e^{(x+4)^{x^2} \left (10 x^3-10 x\right )} (x+4)^{x^2-1} \left (10 x^5+20 x^3+120 x^2+\left (20 x^5+80 x^4-20 x^3-80 x^2\right ) \log (x+4)-10 x-40\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} (x+4)^{x^2-1} \left (10 x^5+20 x^3+120 x^2+\left (20 x^5+80 x^4-20 x^3-80 x^2\right ) \log (x+4)-10 x-40\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-40 e^{x (x+4)^{x^2} \left (10 x^2-10\right )} (x+4)^{x^2-1}+120 e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^2 (x+4)^{x^2-1}-10 e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x (x+4)^{x^2-1}+10 e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^5 (x+4)^{x^2-1}+20 e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^3 (x+4)^{x^2-1}+20 e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^2 \left (x^3+4 x^2-x-4\right ) (x+4)^{x^2-1} \log (x+4)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -40 \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} (x+4)^{x^2-1}dx-10 \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x (x+4)^{x^2-1}dx+120 \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^2 (x+4)^{x^2-1}dx+80 \int \frac {\int e^{10 x (x+4)^{x^2} \left (x^2-1\right )} x^2 (x+4)^{x^2-1}dx}{x+4}dx-80 \log (x+4) \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^2 (x+4)^{x^2-1}dx+10 \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^5 (x+4)^{x^2-1}dx-20 \int \frac {\int e^{10 x (x+4)^{x^2} \left (x^2-1\right )} x^5 (x+4)^{x^2-1}dx}{x+4}dx+20 \log (x+4) \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^5 (x+4)^{x^2-1}dx-80 \int \frac {\int e^{10 x (x+4)^{x^2} \left (x^2-1\right )} x^4 (x+4)^{x^2-1}dx}{x+4}dx+80 \log (x+4) \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^4 (x+4)^{x^2-1}dx+20 \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^3 (x+4)^{x^2-1}dx+20 \int \frac {\int e^{10 x (x+4)^{x^2} \left (x^2-1\right )} x^3 (x+4)^{x^2-1}dx}{x+4}dx-20 \log (x+4) \int e^{x (x+4)^{x^2} \left (10 x^2-10\right )} x^3 (x+4)^{x^2-1}dx\) |
Input:
Int[E^((4 + x)^x^2*(-10*x + 10*x^3))*(4 + x)^(-1 + x^2)*(-40 - 10*x + 120* x^2 + 20*x^3 + 10*x^5 + (-80*x^2 - 20*x^3 + 80*x^4 + 20*x^5)*Log[4 + x]),x ]
Output:
$Aborted
Time = 1.81 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
risch | \({\mathrm e}^{10 x \left (-1+x \right ) \left (1+x \right ) \left (4+x \right )^{x^{2}}}\) | \(18\) |
parallelrisch | \({\mathrm e}^{\left (10 x^{3}-10 x \right ) {\mathrm e}^{x^{2} \ln \left (4+x \right )}}\) | \(21\) |
Input:
int(((20*x^5+80*x^4-20*x^3-80*x^2)*ln(4+x)+10*x^5+20*x^3+120*x^2-10*x-40)* exp(x^2*ln(4+x))*exp((10*x^3-10*x)*exp(x^2*ln(4+x)))/(4+x),x,method=_RETUR NVERBOSE)
Output:
exp(10*x*(-1+x)*(1+x)*(4+x)^(x^2))
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx=e^{\left (10 \, {\left (x^{3} - x\right )} {\left (x + 4\right )}^{\left (x^{2}\right )}\right )} \] Input:
integrate(((20*x^5+80*x^4-20*x^3-80*x^2)*log(4+x)+10*x^5+20*x^3+120*x^2-10 *x-40)*exp(x^2*log(4+x))*exp((10*x^3-10*x)*exp(x^2*log(4+x)))/(4+x),x, alg orithm="fricas")
Output:
e^(10*(x^3 - x)*(x + 4)^(x^2))
Time = 0.93 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx=e^{\left (10 x^{3} - 10 x\right ) e^{x^{2} \log {\left (x + 4 \right )}}} \] Input:
integrate(((20*x**5+80*x**4-20*x**3-80*x**2)*ln(4+x)+10*x**5+20*x**3+120*x **2-10*x-40)*exp(x**2*ln(4+x))*exp((10*x**3-10*x)*exp(x**2*ln(4+x)))/(4+x) ,x)
Output:
exp((10*x**3 - 10*x)*exp(x**2*log(x + 4)))
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx=e^{\left (10 \, {\left (x + 4\right )}^{\left (x^{2}\right )} x^{3} - 10 \, {\left (x + 4\right )}^{\left (x^{2}\right )} x\right )} \] Input:
integrate(((20*x^5+80*x^4-20*x^3-80*x^2)*log(4+x)+10*x^5+20*x^3+120*x^2-10 *x-40)*exp(x^2*log(4+x))*exp((10*x^3-10*x)*exp(x^2*log(4+x)))/(4+x),x, alg orithm="maxima")
Output:
e^(10*(x + 4)^(x^2)*x^3 - 10*(x + 4)^(x^2)*x)
Timed out. \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx=\text {Timed out} \] Input:
integrate(((20*x^5+80*x^4-20*x^3-80*x^2)*log(4+x)+10*x^5+20*x^3+120*x^2-10 *x-40)*exp(x^2*log(4+x))*exp((10*x^3-10*x)*exp(x^2*log(4+x)))/(4+x),x, alg orithm="giac")
Output:
Timed out
Time = 3.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx={\mathrm {e}}^{-10\,x\,{\left (x+4\right )}^{x^2}}\,{\mathrm {e}}^{10\,x^3\,{\left (x+4\right )}^{x^2}} \] Input:
int(-(exp(-exp(x^2*log(x + 4))*(10*x - 10*x^3))*exp(x^2*log(x + 4))*(10*x - 120*x^2 - 20*x^3 - 10*x^5 + log(x + 4)*(80*x^2 + 20*x^3 - 80*x^4 - 20*x^ 5) + 40))/(x + 4),x)
Output:
exp(-10*x*(x + 4)^(x^2))*exp(10*x^3*(x + 4)^(x^2))
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int e^{(4+x)^{x^2} \left (-10 x+10 x^3\right )} (4+x)^{-1+x^2} \left (-40-10 x+120 x^2+20 x^3+10 x^5+\left (-80 x^2-20 x^3+80 x^4+20 x^5\right ) \log (4+x)\right ) \, dx=\frac {e^{10 \left (x +4\right )^{x^{2}} x^{3}}}{e^{10 \left (x +4\right )^{x^{2}} x}} \] Input:
int(((20*x^5+80*x^4-20*x^3-80*x^2)*log(4+x)+10*x^5+20*x^3+120*x^2-10*x-40) *exp(x^2*log(4+x))*exp((10*x^3-10*x)*exp(x^2*log(4+x)))/(4+x),x)
Output:
e**(10*(x + 4)**(x**2)*x**3)/e**(10*(x + 4)**(x**2)*x)