Optimal. Leaf size=37 \[ \frac {\left (-5+e^{e^{x+x^2}}\right )^2 x^2 \left (e^{e^3}+x^2\right )^2}{x-x^2} \]
________________________________________________________________________________________
Rubi [F] time = 19.53, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {25 e^{2 e^3}+125 x^4-100 x^5+e^{e^3} \left (150 x^2-100 x^3\right )+e^{2 e^{x+x^2}} \left (5 x^4-4 x^5+e^{x+x^2} \left (2 x^5+2 x^6-4 x^7\right )+e^{2 e^3} \left (1+e^{x+x^2} \left (2 x+2 x^2-4 x^3\right )\right )+e^{e^3} \left (6 x^2-4 x^3+e^{x+x^2} \left (4 x^3+4 x^4-8 x^5\right )\right )\right )+e^{e^{x+x^2}} \left (-50 x^4+40 x^5+e^{x+x^2} \left (-10 x^5-10 x^6+20 x^7\right )+e^{2 e^3} \left (-10+e^{x+x^2} \left (-10 x-10 x^2+20 x^3\right )\right )+e^{e^3} \left (-60 x^2+40 x^3+e^{x+x^2} \left (-20 x^3-20 x^4+40 x^5\right )\right )\right )}{1-2 x+x^2} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 e^{2 e^3}+125 x^4-100 x^5+e^{e^3} \left (150 x^2-100 x^3\right )+e^{2 e^{x+x^2}} \left (5 x^4-4 x^5+e^{x+x^2} \left (2 x^5+2 x^6-4 x^7\right )+e^{2 e^3} \left (1+e^{x+x^2} \left (2 x+2 x^2-4 x^3\right )\right )+e^{e^3} \left (6 x^2-4 x^3+e^{x+x^2} \left (4 x^3+4 x^4-8 x^5\right )\right )\right )+e^{e^{x+x^2}} \left (-50 x^4+40 x^5+e^{x+x^2} \left (-10 x^5-10 x^6+20 x^7\right )+e^{2 e^3} \left (-10+e^{x+x^2} \left (-10 x-10 x^2+20 x^3\right )\right )+e^{e^3} \left (-60 x^2+40 x^3+e^{x+x^2} \left (-20 x^3-20 x^4+40 x^5\right )\right )\right )}{(-1+x)^2} \, dx\\ &=\int \frac {\left (5-e^{e^{x+x^2}}\right ) \left (e^{e^3}+x^2\right ) \left (5 e^{e^3}-e^{e^3+e^{x+x^2}}+5 (5-4 x) x^2+e^{e^{x+x^2}} x^2 (-5+4 x)+2 e^{e^3+e^{x+x^2}+x+x^2} x \left (-1-x+2 x^2\right )+2 e^{e^{x+x^2}+x+x^2} x^3 \left (-1-x+2 x^2\right )\right )}{(1-x)^2} \, dx\\ &=\int \left (\frac {e^{e^{x (1+x)}} \left (5-e^{e^{x+x^2}}\right ) (5-4 x) x^2 \left (-e^{e^3}-x^2\right )}{(1-x)^2}-\frac {5 e^{e^3} \left (-5+e^{e^{x+x^2}}\right ) \left (e^{e^3}+x^2\right )}{(-1+x)^2}+\frac {e^{e^3+e^{x+x^2}} \left (-5+e^{e^{x+x^2}}\right ) \left (e^{e^3}+x^2\right )}{(-1+x)^2}+\frac {5 \left (-5+e^{e^{x+x^2}}\right ) x^2 (-5+4 x) \left (e^{e^3}+x^2\right )}{(-1+x)^2}-\frac {2 e^{e^{x+x^2}+x+x^2} \left (-5+e^{e^{x+x^2}}\right ) x (1+2 x) \left (e^{e^3}+x^2\right )^2}{-1+x}\right ) \, dx\\ &=-\left (2 \int \frac {e^{e^{x+x^2}+x+x^2} \left (-5+e^{e^{x+x^2}}\right ) x (1+2 x) \left (e^{e^3}+x^2\right )^2}{-1+x} \, dx\right )+5 \int \frac {\left (-5+e^{e^{x+x^2}}\right ) x^2 (-5+4 x) \left (e^{e^3}+x^2\right )}{(-1+x)^2} \, dx-\left (5 e^{e^3}\right ) \int \frac {\left (-5+e^{e^{x+x^2}}\right ) \left (e^{e^3}+x^2\right )}{(-1+x)^2} \, dx+\int \frac {e^{e^{x (1+x)}} \left (5-e^{e^{x+x^2}}\right ) (5-4 x) x^2 \left (-e^{e^3}-x^2\right )}{(1-x)^2} \, dx+\int \frac {e^{e^3+e^{x+x^2}} \left (-5+e^{e^{x+x^2}}\right ) \left (e^{e^3}+x^2\right )}{(-1+x)^2} \, dx\\ &=-\left (2 \int \left (\frac {e^{e^{x (1+x)}+e^{x+x^2}+x+x^2} (-1-2 x) x \left (e^{e^3}+x^2\right )^2}{1-x}-\frac {5 e^{e^{x+x^2}+x+x^2} x (1+2 x) \left (e^{e^3}+x^2\right )^2}{-1+x}\right ) \, dx\right )+5 \int \left (\frac {e^{e^{x (1+x)}} (5-4 x) x^2 \left (-e^{e^3}-x^2\right )}{(1-x)^2}-\frac {5 x^2 (-5+4 x) \left (e^{e^3}+x^2\right )}{(-1+x)^2}\right ) \, dx-\left (5 e^{e^3}\right ) \int \left (\frac {e^{e^{x (1+x)}} \left (e^{e^3}+x^2\right )}{(1-x)^2}-\frac {5 \left (e^{e^3}+x^2\right )}{(-1+x)^2}\right ) \, dx+\int \left (\frac {e^{e^3+e^{x (1+x)}+e^{x+x^2}} \left (e^{e^3}+x^2\right )}{(1-x)^2}-\frac {5 e^{e^3+e^{x+x^2}} \left (e^{e^3}+x^2\right )}{(-1+x)^2}\right ) \, dx+\int \left (\frac {e^{2 e^{x (1+x)}} (5-4 x) x^2 \left (e^{e^3}+x^2\right )}{(1-x)^2}+\frac {5 e^{e^{x (1+x)}} x^2 (-5+4 x) \left (e^{e^3}+x^2\right )}{(-1+x)^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{e^{x (1+x)}+e^{x+x^2}+x+x^2} (-1-2 x) x \left (e^{e^3}+x^2\right )^2}{1-x} \, dx\right )+5 \int \frac {e^{e^{x (1+x)}} (5-4 x) x^2 \left (-e^{e^3}-x^2\right )}{(1-x)^2} \, dx-5 \int \frac {e^{e^3+e^{x+x^2}} \left (e^{e^3}+x^2\right )}{(-1+x)^2} \, dx+5 \int \frac {e^{e^{x (1+x)}} x^2 (-5+4 x) \left (e^{e^3}+x^2\right )}{(-1+x)^2} \, dx+10 \int \frac {e^{e^{x+x^2}+x+x^2} x (1+2 x) \left (e^{e^3}+x^2\right )^2}{-1+x} \, dx-25 \int \frac {x^2 (-5+4 x) \left (e^{e^3}+x^2\right )}{(-1+x)^2} \, dx-\left (5 e^{e^3}\right ) \int \frac {e^{e^{x (1+x)}} \left (e^{e^3}+x^2\right )}{(1-x)^2} \, dx+\left (25 e^{e^3}\right ) \int \frac {e^{e^3}+x^2}{(-1+x)^2} \, dx+\int \frac {e^{e^3+e^{x (1+x)}+e^{x+x^2}} \left (e^{e^3}+x^2\right )}{(1-x)^2} \, dx+\int \frac {e^{2 e^{x (1+x)}} (5-4 x) x^2 \left (e^{e^3}+x^2\right )}{(1-x)^2} \, dx\\ &=-\left (2 \int \frac {e^{2 e^{x (1+x)}+x+x^2} (-1-2 x) x \left (e^{e^3}+x^2\right )^2}{1-x} \, dx\right )-5 \int \left (e^{e^3+e^{x+x^2}}+\frac {e^{e^3+e^{x+x^2}} \left (1+e^{e^3}\right )}{(-1+x)^2}+\frac {2 e^{e^3+e^{x+x^2}}}{-1+x}\right ) \, dx+2 \left (5 \int \left (e^{e^{x (1+x)}} \left (1+3 e^{e^3}\right )-\frac {e^{e^{x (1+x)}} \left (1+e^{e^3}\right )}{(-1+x)^2}+\frac {2 e^{e^3+e^{x (1+x)}}}{-1+x}+2 e^{e^{x (1+x)}} \left (1+2 e^{e^3}\right ) x+3 e^{e^{x (1+x)}} x^2+4 e^{e^{x (1+x)}} x^3\right ) \, dx\right )+10 \int \left (3 e^{e^{x+x^2}+x+x^2} \left (1+e^{e^3}\right )^2+\frac {3 e^{e^{x+x^2}+x+x^2} \left (1+e^{e^3}\right )^2}{-1+x}+e^{e^{x+x^2}+x+x^2} \left (3+6 e^{e^3}+2 e^{2 e^3}\right ) x+3 e^{e^{x+x^2}+x+x^2} \left (1+2 e^{e^3}\right ) x^2+e^{e^{x+x^2}+x+x^2} \left (3+4 e^{e^3}\right ) x^3+3 e^{e^{x+x^2}+x+x^2} x^4+2 e^{e^{x+x^2}+x+x^2} x^5\right ) \, dx-25 \int \left (1+3 e^{e^3}-\frac {1+e^{e^3}}{(-1+x)^2}+\frac {2 e^{e^3}}{-1+x}+2 \left (1+2 e^{e^3}\right ) x+3 x^2+4 x^3\right ) \, dx-\left (5 e^{e^3}\right ) \int \left (e^{e^{x (1+x)}}+\frac {e^{e^{x (1+x)}} \left (1+e^{e^3}\right )}{(-1+x)^2}+\frac {2 e^{e^{x (1+x)}}}{-1+x}\right ) \, dx+\left (25 e^{e^3}\right ) \int \left (1+\frac {1+e^{e^3}}{(-1+x)^2}+\frac {2}{-1+x}\right ) \, dx+\int \frac {e^{e^3+2 e^{x (1+x)}} \left (e^{e^3}+x^2\right )}{(1-x)^2} \, dx+\int \frac {e^{2 e^{x+x^2}} (5-4 x) x^2 \left (e^{e^3}+x^2\right )}{(1-x)^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [B] time = 0.23, size = 144, normalized size = 3.89 \begin {gather*} -\frac {25 e^{2 e^3}+e^{2 \left (e^3+e^{x+x^2}\right )} x-10 e^{2 e^3+e^{x+x^2}} x-20 e^{e^3+e^{x+x^2}} x^3+2 e^{e^3+2 e^{x+x^2}} x^3-10 e^{e^{x+x^2}} x^5+e^{2 e^{x+x^2}} x^5+50 e^{e^3} \left (1-x+x^3\right )+25 \left (1-x+x^5\right )}{-1+x} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.73, size = 94, normalized size = 2.54 \begin {gather*} -\frac {25 \, x^{5} + {\left (x^{5} + 2 \, x^{3} e^{\left (e^{3}\right )} + x e^{\left (2 \, e^{3}\right )}\right )} e^{\left (2 \, e^{\left (x^{2} + x\right )}\right )} + 50 \, {\left (x^{3} - x + 1\right )} e^{\left (e^{3}\right )} - 10 \, {\left (x^{5} + 2 \, x^{3} e^{\left (e^{3}\right )} + x e^{\left (2 \, e^{3}\right )}\right )} e^{\left (e^{\left (x^{2} + x\right )}\right )} - 25 \, x + 25 \, e^{\left (2 \, e^{3}\right )} + 25}{x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {100 \, x^{5} - 125 \, x^{4} + {\left (4 \, x^{5} - 5 \, x^{4} + 2 \, {\left (2 \, x^{7} - x^{6} - x^{5}\right )} e^{\left (x^{2} + x\right )} + {\left (2 \, {\left (2 \, x^{3} - x^{2} - x\right )} e^{\left (x^{2} + x\right )} - 1\right )} e^{\left (2 \, e^{3}\right )} + 2 \, {\left (2 \, x^{3} - 3 \, x^{2} + 2 \, {\left (2 \, x^{5} - x^{4} - x^{3}\right )} e^{\left (x^{2} + x\right )}\right )} e^{\left (e^{3}\right )}\right )} e^{\left (2 \, e^{\left (x^{2} + x\right )}\right )} + 50 \, {\left (2 \, x^{3} - 3 \, x^{2}\right )} e^{\left (e^{3}\right )} - 10 \, {\left (4 \, x^{5} - 5 \, x^{4} + {\left (2 \, x^{7} - x^{6} - x^{5}\right )} e^{\left (x^{2} + x\right )} + {\left ({\left (2 \, x^{3} - x^{2} - x\right )} e^{\left (x^{2} + x\right )} - 1\right )} e^{\left (2 \, e^{3}\right )} + 2 \, {\left (2 \, x^{3} - 3 \, x^{2} + {\left (2 \, x^{5} - x^{4} - x^{3}\right )} e^{\left (x^{2} + x\right )}\right )} e^{\left (e^{3}\right )}\right )} e^{\left (e^{\left (x^{2} + x\right )}\right )} - 25 \, e^{\left (2 \, e^{3}\right )}}{x^{2} - 2 \, x + 1}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.54, size = 129, normalized size = 3.49
method | result | size |
risch | \(-25 x^{4}-50 x^{2} {\mathrm e}^{{\mathrm e}^{3}}-25 x^{3}-50 x \,{\mathrm e}^{{\mathrm e}^{3}}-25 x^{2}-25 x -\frac {25 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{x -1}-\frac {50 \,{\mathrm e}^{{\mathrm e}^{3}}}{x -1}-\frac {25}{x -1}-\frac {x \left (x^{4}+2 x^{2} {\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{2 \,{\mathrm e}^{3}}\right ) {\mathrm e}^{2 \,{\mathrm e}^{\left (x +1\right ) x}}}{x -1}+\frac {10 x \left (x^{4}+2 x^{2} {\mathrm e}^{{\mathrm e}^{3}}+{\mathrm e}^{2 \,{\mathrm e}^{3}}\right ) {\mathrm e}^{{\mathrm e}^{\left (x +1\right ) x}}}{x -1}\) | \(129\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.40, size = 148, normalized size = 4.00 \begin {gather*} -25 \, x^{4} - 25 \, x^{3} - 25 \, x^{2} - 50 \, {\left (x^{2} + 4 \, x - \frac {2}{x - 1} + 6 \, \log \left (x - 1\right )\right )} e^{\left (e^{3}\right )} + 150 \, {\left (x - \frac {1}{x - 1} + 2 \, \log \left (x - 1\right )\right )} e^{\left (e^{3}\right )} - 25 \, x - \frac {{\left (x^{5} + 2 \, x^{3} e^{\left (e^{3}\right )} + x e^{\left (2 \, e^{3}\right )}\right )} e^{\left (2 \, e^{\left (x^{2} + x\right )}\right )} - 10 \, {\left (x^{5} + 2 \, x^{3} e^{\left (e^{3}\right )} + x e^{\left (2 \, e^{3}\right )}\right )} e^{\left (e^{\left (x^{2} + x\right )}\right )}}{x - 1} - \frac {25 \, e^{\left (2 \, e^{3}\right )}}{x - 1} - \frac {25}{x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 0.97, size = 125, normalized size = 3.38 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x}\,\left (10\,x^5+20\,{\mathrm {e}}^{{\mathrm {e}}^3}\,x^3+10\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,x\right )}{x-1}-\frac {25\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}+50\,{\mathrm {e}}^{{\mathrm {e}}^3}+25}{x-1}-x^2\,\left (50\,{\mathrm {e}}^{{\mathrm {e}}^3}+25\right )-25\,x^3-25\,x^4-x\,\left (50\,{\mathrm {e}}^{{\mathrm {e}}^3}+25\right )-\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x}\,\left (x^5+2\,{\mathrm {e}}^{{\mathrm {e}}^3}\,x^3+{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,x\right )}{x-1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 0.69, size = 177, normalized size = 4.78 \begin {gather*} - 25 x^{4} - 25 x^{3} - x^{2} \left (25 + 50 e^{e^{3}}\right ) - x \left (25 + 50 e^{e^{3}}\right ) + \frac {\left (- x^{6} + x^{5} - 2 x^{4} e^{e^{3}} + 2 x^{3} e^{e^{3}} - x^{2} e^{2 e^{3}} + x e^{2 e^{3}}\right ) e^{2 e^{x^{2} + x}} + \left (10 x^{6} - 10 x^{5} + 20 x^{4} e^{e^{3}} - 20 x^{3} e^{e^{3}} + 10 x^{2} e^{2 e^{3}} - 10 x e^{2 e^{3}}\right ) e^{e^{x^{2} + x}}}{x^{2} - 2 x + 1} - \frac {25 + 50 e^{e^{3}} + 25 e^{2 e^{3}}}{x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________