Integrand size = 103, antiderivative size = 27 \begin {dmath*} \int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx=e^{2+\frac {\left (4+\frac {1}{x^3}\right ) \left (5-\frac {1}{5 x}\right ) x}{2+e^x}} \end {dmath*}
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \begin {dmath*} \int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx=e^{2+\frac {-1+25 x-4 x^3+100 x^4}{5 \left (2+e^x\right ) x^3}} \end {dmath*}
Integrate[(E^((-1 + 25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(10*x^3 + 5*E^x* x^3))*(6 - 100*x + 200*x^4 + E^x*(3 - 49*x - 25*x^2 + 104*x^4 - 100*x^5))) /(20*x^4 + 20*E^x*x^4 + 5*E^(2*x)*x^4),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.
\(\displaystyle \int \frac {\left (200 x^4+e^x \left (-100 x^5+104 x^4-25 x^2-49 x+3\right )-100 x+6\right ) \exp \left (\frac {100 x^4+10 e^x x^3+16 x^3+25 x-1}{5 e^x x^3+10 x^3}\right )}{20 e^x x^4+5 e^{2 x} x^4+20 x^4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (200 x^4+e^x \left (-100 x^5+104 x^4-25 x^2-49 x+3\right )-100 x+6\right ) \exp \left (\frac {100 x^4+10 e^x x^3+16 x^3+25 x-1}{5 \left (e^x+2\right ) x^3}\right )}{5 \left (e^x+2\right )^2 x^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right ) \left (200 x^4-100 x+e^x \left (-100 x^5+104 x^4-25 x^2-49 x+3\right )+6\right )}{\left (2+e^x\right )^2 x^4}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {2 \exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right ) \left (100 x^4-4 x^3+25 x-1\right )}{\left (2+e^x\right )^2 x^3}-\frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right ) \left (100 x^5-104 x^4+25 x^2+49 x-3\right )}{\left (2+e^x\right ) x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-8 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2}dx+104 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right )}{2+e^x}dx+3 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^4}dx-2 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2 x^3}dx-49 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^3}dx+200 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right ) x}{\left (2+e^x\right )^2}dx-100 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right ) x}{2+e^x}dx+50 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2 x^2}dx-25 \int \frac {\exp \left (-\frac {-100 x^4-10 e^x x^3-16 x^3-25 x+1}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^2}dx\right )\) |
Int[(E^((-1 + 25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(10*x^3 + 5*E^x*x^3))* (6 - 100*x + 200*x^4 + E^x*(3 - 49*x - 25*x^2 + 104*x^4 - 100*x^5)))/(20*x ^4 + 20*E^x*x^4 + 5*E^(2*x)*x^4),x]
3.10.91.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 2.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30
method | result | size |
risch | \({\mathrm e}^{\frac {10 \,{\mathrm e}^{x} x^{3}+100 x^{4}+16 x^{3}+25 x -1}{5 x^{3} \left ({\mathrm e}^{x}+2\right )}}\) | \(35\) |
parallelrisch | \({\mathrm e}^{\frac {10 \,{\mathrm e}^{x} x^{3}+100 x^{4}+16 x^{3}+25 x -1}{5 x^{3} \left ({\mathrm e}^{x}+2\right )}}\) | \(35\) |
int(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((10*exp( x)*x^3+100*x^4+16*x^3+25*x-1)/(5*exp(x)*x^3+10*x^3))/(5*exp(x)^2*x^4+20*ex p(x)*x^4+20*x^4),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \begin {dmath*} \int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx=e^{\left (\frac {100 \, x^{4} + 10 \, x^{3} e^{x} + 16 \, x^{3} + 25 \, x - 1}{5 \, {\left (x^{3} e^{x} + 2 \, x^{3}\right )}}\right )} \end {dmath*}
integrate(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((1 0*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/(5*exp(x)*x^3+10*x^3))/(5*exp(x)^2*x^4 +20*exp(x)*x^4+20*x^4),x, algorithm=\
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \begin {dmath*} \int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx=e^{\frac {100 x^{4} + 10 x^{3} e^{x} + 16 x^{3} + 25 x - 1}{5 x^{3} e^{x} + 10 x^{3}}} \end {dmath*}
integrate(((-100*x**5+104*x**4-25*x**2-49*x+3)*exp(x)+200*x**4-100*x+6)*ex p((10*exp(x)*x**3+100*x**4+16*x**3+25*x-1)/(5*exp(x)*x**3+10*x**3))/(5*exp (x)**2*x**4+20*exp(x)*x**4+20*x**4),x)
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (22) = 44\).
Time = 0.41 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26 \begin {dmath*} \int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx=e^{\left (\frac {20 \, x}{e^{x} + 2} + \frac {2 \, e^{x}}{e^{x} + 2} - \frac {1}{5 \, {\left (x^{3} e^{x} + 2 \, x^{3}\right )}} + \frac {5}{x^{2} e^{x} + 2 \, x^{2}} + \frac {16}{5 \, {\left (e^{x} + 2\right )}}\right )} \end {dmath*}
integrate(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((1 0*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/(5*exp(x)*x^3+10*x^3))/(5*exp(x)^2*x^4 +20*exp(x)*x^4+20*x^4),x, algorithm=\
e^(20*x/(e^x + 2) + 2*e^x/(e^x + 2) - 1/5/(x^3*e^x + 2*x^3) + 5/(x^2*e^x + 2*x^2) + 16/5/(e^x + 2))
Exception generated. \begin {dmath*} \int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx=\text {Exception raised: TypeError} \end {dmath*}
integrate(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((1 0*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/(5*exp(x)*x^3+10*x^3))/(5*exp(x)^2*x^4 +20*exp(x)*x^4+20*x^4),x, algorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1000000000000,[1,29]%%%}+%%%{-2200000000000,[1,28]%%%}+%%% {13360000
Time = 15.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52 \begin {dmath*} \int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx={\mathrm {e}}^{\frac {2\,{\mathrm {e}}^x}{{\mathrm {e}}^x+2}}\,{\mathrm {e}}^{\frac {16}{5\,{\mathrm {e}}^x+10}}\,{\mathrm {e}}^{\frac {5}{x^2\,{\mathrm {e}}^x+2\,x^2}}\,{\mathrm {e}}^{-\frac {1}{5\,x^3\,{\mathrm {e}}^x+10\,x^3}}\,{\mathrm {e}}^{\frac {20\,x}{{\mathrm {e}}^x+2}} \end {dmath*}
int(-(exp((25*x + 10*x^3*exp(x) + 16*x^3 + 100*x^4 - 1)/(5*x^3*exp(x) + 10 *x^3))*(100*x + exp(x)*(49*x + 25*x^2 - 104*x^4 + 100*x^5 - 3) - 200*x^4 - 6))/(20*x^4*exp(x) + 5*x^4*exp(2*x) + 20*x^4),x)