Optimal. Leaf size=34 \[ \frac {4+e^{\left (x-\frac {\left (\frac {1}{e^3}+x\right )^2}{(-2+x)^2}\right )^2}-2 x-x^2}{x} \]
________________________________________________________________________________________
Rubi [F] time = 13.21, 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 {e^{12} \left (128-320 x+352 x^2-240 x^3+120 x^4-44 x^5+10 x^6-x^7\right )+\exp \left (\frac {1+4 e^3 x+e^6 \left (-8 x+14 x^2-2 x^3\right )+e^9 \left (-16 x^2+20 x^3-4 x^4\right )+e^{12} \left (16 x^2-40 x^3+33 x^4-10 x^5+x^6\right )}{e^{12} \left (16-32 x+24 x^2-8 x^3+x^4\right )}\right ) \left (-4 x+e^3 \left (-8 x-12 x^2\right )+e^6 \left (16 x-32 x^2-16 x^3+2 x^4\right )+e^9 \left (64 x^2-88 x^3+12 x^4\right )+e^{12} \left (32-80 x+16 x^2+168 x^3-214 x^4+99 x^5-22 x^6+2 x^7\right )\right )}{e^{12} \left (-32 x^2+80 x^3-80 x^4+40 x^5-10 x^6+x^7\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{12} \left (128-320 x+352 x^2-240 x^3+120 x^4-44 x^5+10 x^6-x^7\right )+\exp \left (\frac {1+4 e^3 x+e^6 \left (-8 x+14 x^2-2 x^3\right )+e^9 \left (-16 x^2+20 x^3-4 x^4\right )+e^{12} \left (16 x^2-40 x^3+33 x^4-10 x^5+x^6\right )}{e^{12} \left (16-32 x+24 x^2-8 x^3+x^4\right )}\right ) \left (-4 x+e^3 \left (-8 x-12 x^2\right )+e^6 \left (16 x-32 x^2-16 x^3+2 x^4\right )+e^9 \left (64 x^2-88 x^3+12 x^4\right )+e^{12} \left (32-80 x+16 x^2+168 x^3-214 x^4+99 x^5-22 x^6+2 x^7\right )\right )}{-32 x^2+80 x^3-80 x^4+40 x^5-10 x^6+x^7} \, dx}{e^{12}}\\ &=\frac {\int \frac {-e^{12} \left (4+x^2\right )+\frac {\exp \left (\frac {\left (1+2 e^3 x-e^6 x \left (4-5 x+x^2\right )\right )^2}{e^{12} (-2+x)^4}\right ) \left (-4 x-4 e^3 x (2+3 x)+4 e^9 x^2 \left (16-22 x+3 x^2\right )+2 e^6 x \left (8-16 x-8 x^2+x^3\right )+e^{12} \left (32-80 x+16 x^2+168 x^3-214 x^4+99 x^5-22 x^6+2 x^7\right )\right )}{(-2+x)^5}}{x^2} \, dx}{e^{12}}\\ &=\frac {\int \left (-\frac {e^{12} \left (4+x^2\right )}{x^2}+\frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (-32 e^{12}+4 \left (1+2 e^3 \left (1-2 e^3+10 e^9\right )\right ) x+12 e^3 \left (1-\frac {4}{3} e^3 \left (-2+4 e^3+e^6\right )\right ) x^2+16 e^6 \left (1-\frac {1}{2} e^3 \left (-11+21 e^3\right )\right ) x^3-2 e^6 \left (1+6 e^3-107 e^6\right ) x^4-99 e^{12} x^5+22 e^{12} x^6-2 e^{12} x^7\right )}{(2-x)^5 x^2}\right ) \, dx}{e^{12}}\\ &=\frac {\int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (-32 e^{12}+4 \left (1+2 e^3 \left (1-2 e^3+10 e^9\right )\right ) x+12 e^3 \left (1-\frac {4}{3} e^3 \left (-2+4 e^3+e^6\right )\right ) x^2+16 e^6 \left (1-\frac {1}{2} e^3 \left (-11+21 e^3\right )\right ) x^3-2 e^6 \left (1+6 e^3-107 e^6\right ) x^4-99 e^{12} x^5+22 e^{12} x^6-2 e^{12} x^7\right )}{(2-x)^5 x^2} \, dx}{e^{12}}-\int \frac {4+x^2}{x^2} \, dx\\ &=\frac {\int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (4 x+4 e^3 x (2+3 x)-4 e^9 x^2 \left (16-22 x+3 x^2\right )-2 e^6 x \left (8-16 x-8 x^2+x^3\right )-e^{12} \left (32-80 x+16 x^2+168 x^3-214 x^4+99 x^5-22 x^6+2 x^7\right )\right )}{(2-x)^5 x^2} \, dx}{e^{12}}-\int \left (1+\frac {4}{x^2}\right ) \, dx\\ &=\frac {4}{x}-x+\frac {\int \left (2 \exp \left (12+\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )-\frac {2 \exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (1+2 e^3\right )^4}{(-2+x)^5}-\frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (1+2 e^3\right )^3 \left (-1+4 e^3\right )}{(-2+x)^4}+\frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (1+2 e^3\right )^2 \left (-1+2 e^3+4 e^6\right )}{2 (-2+x)^3}+\frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (1+2 e^3-4 e^6+16 e^{12}\right )}{4 (-2+x)^2}+\frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (-1-2 e^3+4 e^6-16 e^{12}\right )}{8 (-2+x)}-\frac {\exp \left (12+\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{x^2}+\frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \left (1+2 e^3-4 e^6\right )}{8 x}\right ) \, dx}{e^{12}}\\ &=\frac {4}{x}-x-\frac {\int \frac {\exp \left (12+\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{x^2} \, dx}{e^{12}}+\frac {2 \int \exp \left (12+\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right ) \, dx}{e^{12}}+\frac {\left (\left (1-4 e^3\right ) \left (1+2 e^3\right )^3\right ) \int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{(-2+x)^4} \, dx}{e^{12}}-\frac {\left (2 \left (1+2 e^3\right )^4\right ) \int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{(-2+x)^5} \, dx}{e^{12}}-\frac {\left (\left (1+2 e^3\right )^2 \left (1-2 e^3-4 e^6\right )\right ) \int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{(-2+x)^3} \, dx}{2 e^{12}}+\frac {\left (1+2 e^3-4 e^6\right ) \int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{x} \, dx}{8 e^{12}}-\frac {\left (1+2 e^3-4 e^6+16 e^{12}\right ) \int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{-2+x} \, dx}{8 e^{12}}+\frac {\left (1+2 e^3-4 e^6+16 e^{12}\right ) \int \frac {\exp \left (\frac {\left (1+2 e^3 \left (1-2 e^3\right ) x+5 e^6 x^2-e^6 x^3\right )^2}{e^{12} (-2+x)^4}\right )}{(-2+x)^2} \, dx}{4 e^{12}}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.53, size = 46, normalized size = 1.35 \begin {gather*} \frac {4+e^{\frac {\left (1+2 e^3 x-e^6 x \left (4-5 x+x^2\right )\right )^2}{e^{12} (-2+x)^4}}-x^2}{x} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.56, size = 104, normalized size = 3.06 \begin {gather*} -\frac {x^{2} - e^{\left (\frac {{\left ({\left (x^{6} - 10 \, x^{5} + 33 \, x^{4} - 40 \, x^{3} + 16 \, x^{2}\right )} e^{12} - 4 \, {\left (x^{4} - 5 \, x^{3} + 4 \, x^{2}\right )} e^{9} - 2 \, {\left (x^{3} - 7 \, x^{2} + 4 \, x\right )} e^{6} + 4 \, x e^{3} + 1\right )} e^{\left (-12\right )}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16}\right )} - 4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 12.57, size = 166, normalized size = 4.88 \begin {gather*} -\frac {{\left (x^{2} e^{12} - 4 \, e^{12} - e^{\left (\frac {1}{16} \, {\left (48 \, e^{12} + 1\right )} e^{\left (-12\right )} + \frac {16 \, x^{6} e^{12} - 160 \, x^{5} e^{12} + 528 \, x^{4} e^{12} - 64 \, x^{4} e^{9} - x^{4} - 640 \, x^{3} e^{12} + 320 \, x^{3} e^{9} - 32 \, x^{3} e^{6} + 8 \, x^{3} + 256 \, x^{2} e^{12} - 256 \, x^{2} e^{9} + 224 \, x^{2} e^{6} - 24 \, x^{2} - 128 \, x e^{6} + 64 \, x e^{3} + 32 \, x}{16 \, {\left (x^{4} e^{12} - 8 \, x^{3} e^{12} + 24 \, x^{2} e^{12} - 32 \, x e^{12} + 16 \, e^{12}\right )}} + 9\right )}\right )} e^{\left (-12\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (2 x^{7}-22 x^{6}+99 x^{5}-214 x^{4}+168 x^{3}+16 x^{2}-80 x +32\right ) {\mathrm e}^{12}+\left (12 x^{4}-88 x^{3}+64 x^{2}\right ) {\mathrm e}^{9}+\left (2 x^{4}-16 x^{3}-32 x^{2}+16 x \right ) {\mathrm e}^{6}+\left (-12 x^{2}-8 x \right ) {\mathrm e}^{3}-4 x \right ) {\mathrm e}^{\frac {\left (\left (x^{6}-10 x^{5}+33 x^{4}-40 x^{3}+16 x^{2}\right ) {\mathrm e}^{12}+\left (-4 x^{4}+20 x^{3}-16 x^{2}\right ) {\mathrm e}^{9}+\left (-2 x^{3}+14 x^{2}-8 x \right ) {\mathrm e}^{6}+4 x \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-12}}{x^{4}-8 x^{3}+24 x^{2}-32 x +16}}+\left (-x^{7}+10 x^{6}-44 x^{5}+120 x^{4}-240 x^{3}+352 x^{2}-320 x +128\right ) {\mathrm e}^{12}\right ) {\mathrm e}^{-12}}{x^{7}-10 x^{6}+40 x^{5}-80 x^{4}+80 x^{3}-32 x^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 1.78, size = 697, normalized size = 20.50 \begin {gather*} -\frac {1}{3} \, {\left ({\left (3 \, x - \frac {8 \, {\left (15 \, x^{3} - 75 \, x^{2} + 130 \, x - 77\right )}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} + 30 \, \log \left (x - 2\right )\right )} e^{12} - 2 \, {\left (\frac {2 \, {\left (15 \, x^{4} - 105 \, x^{3} + 260 \, x^{2} - 250 \, x + 48\right )}}{x^{5} - 8 \, x^{4} + 24 \, x^{3} - 32 \, x^{2} + 16 \, x} + 15 \, \log \left (x - 2\right ) - 15 \, \log \relax (x)\right )} e^{12} + 10 \, {\left (\frac {4 \, {\left (6 \, x^{3} - 27 \, x^{2} + 44 \, x - 25\right )}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} - 3 \, \log \left (x - 2\right )\right )} e^{12} + 10 \, {\left (\frac {2 \, {\left (3 \, x^{3} - 21 \, x^{2} + 52 \, x - 50\right )}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} + 3 \, \log \left (x - 2\right ) - 3 \, \log \relax (x)\right )} e^{12} - \frac {132 \, {\left (x^{3} - 3 \, x^{2} + 4 \, x - 2\right )} e^{12}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} + \frac {60 \, {\left (3 \, x^{2} - 4 \, x + 2\right )} e^{12}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} - \frac {120 \, {\left (2 \, x - 1\right )} e^{12}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} + \frac {264 \, e^{12}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} - \frac {3 \, e^{\left (x^{2} - 2 \, x + \frac {1}{x^{4} e^{12} - 8 \, x^{3} e^{12} + 24 \, x^{2} e^{12} - 32 \, x e^{12} + 16 \, e^{12}} + \frac {8}{x^{4} e^{9} - 8 \, x^{3} e^{9} + 24 \, x^{2} e^{9} - 32 \, x e^{9} + 16 \, e^{9}} + \frac {24}{x^{4} e^{6} - 8 \, x^{3} e^{6} + 24 \, x^{2} e^{6} - 32 \, x e^{6} + 16 \, e^{6}} + \frac {32}{x^{4} e^{3} - 8 \, x^{3} e^{3} + 24 \, x^{2} e^{3} - 32 \, x e^{3} + 16 \, e^{3}} + \frac {16}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16} + \frac {4}{x^{3} e^{9} - 6 \, x^{2} e^{9} + 12 \, x e^{9} - 8 \, e^{9}} + \frac {24}{x^{3} e^{6} - 6 \, x^{2} e^{6} + 12 \, x e^{6} - 8 \, e^{6}} + \frac {48}{x^{3} e^{3} - 6 \, x^{2} e^{3} + 12 \, x e^{3} - 8 \, e^{3}} + \frac {32}{x^{3} - 6 \, x^{2} + 12 \, x - 8} + \frac {2}{x^{2} e^{6} - 4 \, x e^{6} + 4 \, e^{6}} + \frac {8}{x^{2} e^{3} - 4 \, x e^{3} + 4 \, e^{3}} + \frac {8}{x^{2} - 4 \, x + 4} - \frac {2}{x e^{6} - 2 \, e^{6}} - \frac {12}{x e^{3} - 2 \, e^{3}} - \frac {16}{x - 2} - 4 \, e^{\left (-3\right )} + 5\right )}}{x}\right )} e^{\left (-12\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 9.43, size = 358, normalized size = 10.53 \begin {gather*} \frac {4}{x}-x+\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{-12}}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{\frac {x^6}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{-\frac {10\,x^5}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{\frac {16\,x^2}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{\frac {33\,x^4}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{-\frac {40\,x^3}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{\frac {4\,x\,{\mathrm {e}}^{-9}}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{-\frac {8\,x\,{\mathrm {e}}^{-6}}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{-\frac {2\,x^3\,{\mathrm {e}}^{-6}}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{-\frac {4\,x^4\,{\mathrm {e}}^{-3}}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{-\frac {16\,x^2\,{\mathrm {e}}^{-3}}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{\frac {14\,x^2\,{\mathrm {e}}^{-6}}{x^4-8\,x^3+24\,x^2-32\,x+16}}\,{\mathrm {e}}^{\frac {20\,x^3\,{\mathrm {e}}^{-3}}{x^4-8\,x^3+24\,x^2-32\,x+16}}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 1.67, size = 100, normalized size = 2.94 \begin {gather*} - x + \frac {e^{\frac {4 x e^{3} + \left (- 2 x^{3} + 14 x^{2} - 8 x\right ) e^{6} + \left (- 4 x^{4} + 20 x^{3} - 16 x^{2}\right ) e^{9} + \left (x^{6} - 10 x^{5} + 33 x^{4} - 40 x^{3} + 16 x^{2}\right ) e^{12} + 1}{\left (x^{4} - 8 x^{3} + 24 x^{2} - 32 x + 16\right ) e^{12}}}}{x} + \frac {4}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________