Integrand size = 40, antiderivative size = 25 \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=5 \left (\frac {4+2 x}{x}+e^{-x} \left (16+x^4\right )^2\right ) \]
[Out]
Time = 0.55 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 31, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6873, 12, 6874, 2225, 2207} \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=5 e^{-x} x^8+160 e^{-x} x^4+1280 e^{-x}+\frac {20}{x} \]
[In]
[Out]
Rule 12
Rule 2207
Rule 2225
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e^{-x} \left (-4 e^x-256 x^2+128 x^5-32 x^6+8 x^9-x^{10}\right )}{x^2} \, dx \\ & = 5 \int \frac {e^{-x} \left (-4 e^x-256 x^2+128 x^5-32 x^6+8 x^9-x^{10}\right )}{x^2} \, dx \\ & = 5 \int \left (-256 e^{-x}-\frac {4}{x^2}+128 e^{-x} x^3-32 e^{-x} x^4+8 e^{-x} x^7-e^{-x} x^8\right ) \, dx \\ & = \frac {20}{x}-5 \int e^{-x} x^8 \, dx+40 \int e^{-x} x^7 \, dx-160 \int e^{-x} x^4 \, dx+640 \int e^{-x} x^3 \, dx-1280 \int e^{-x} \, dx \\ & = 1280 e^{-x}+\frac {20}{x}-640 e^{-x} x^3+160 e^{-x} x^4-40 e^{-x} x^7+5 e^{-x} x^8-40 \int e^{-x} x^7 \, dx+280 \int e^{-x} x^6 \, dx-640 \int e^{-x} x^3 \, dx+1920 \int e^{-x} x^2 \, dx \\ & = 1280 e^{-x}+\frac {20}{x}-1920 e^{-x} x^2+160 e^{-x} x^4-280 e^{-x} x^6+5 e^{-x} x^8-280 \int e^{-x} x^6 \, dx+1680 \int e^{-x} x^5 \, dx-1920 \int e^{-x} x^2 \, dx+3840 \int e^{-x} x \, dx \\ & = 1280 e^{-x}+\frac {20}{x}-3840 e^{-x} x+160 e^{-x} x^4-1680 e^{-x} x^5+5 e^{-x} x^8-1680 \int e^{-x} x^5 \, dx+3840 \int e^{-x} \, dx-3840 \int e^{-x} x \, dx+8400 \int e^{-x} x^4 \, dx \\ & = -2560 e^{-x}+\frac {20}{x}-8240 e^{-x} x^4+5 e^{-x} x^8-3840 \int e^{-x} \, dx-8400 \int e^{-x} x^4 \, dx+33600 \int e^{-x} x^3 \, dx \\ & = 1280 e^{-x}+\frac {20}{x}-33600 e^{-x} x^3+160 e^{-x} x^4+5 e^{-x} x^8-33600 \int e^{-x} x^3 \, dx+100800 \int e^{-x} x^2 \, dx \\ & = 1280 e^{-x}+\frac {20}{x}-100800 e^{-x} x^2+160 e^{-x} x^4+5 e^{-x} x^8-100800 \int e^{-x} x^2 \, dx+201600 \int e^{-x} x \, dx \\ & = 1280 e^{-x}+\frac {20}{x}-201600 e^{-x} x+160 e^{-x} x^4+5 e^{-x} x^8+201600 \int e^{-x} \, dx-201600 \int e^{-x} x \, dx \\ & = -200320 e^{-x}+\frac {20}{x}+160 e^{-x} x^4+5 e^{-x} x^8-201600 \int e^{-x} \, dx \\ & = 1280 e^{-x}+\frac {20}{x}+160 e^{-x} x^4+5 e^{-x} x^8 \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=\frac {20}{x}-5 e^{-x} \left (-256-32 x^4-x^8\right ) \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {20}{x}+\left (5 x^{8}+160 x^{4}+1280\right ) {\mathrm e}^{-x}\) | \(24\) |
norman | \(\frac {\left (1280 x +160 x^{5}+5 x^{9}+20 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}\) | \(27\) |
parallelrisch | \(-\frac {\left (-5 x^{9}-160 x^{5}-1280 x -20 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}\) | \(28\) |
default | \(\frac {20}{x}+1280 \,{\mathrm e}^{-x}+160 x^{4} {\mathrm e}^{-x}+5 \,{\mathrm e}^{-x} x^{8}\) | \(31\) |
parts | \(\frac {20}{x}+1280 \,{\mathrm e}^{-x}+160 x^{4} {\mathrm e}^{-x}+5 \,{\mathrm e}^{-x} x^{8}\) | \(31\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=\frac {5 \, {\left (x^{9} + 32 \, x^{5} + 256 \, x + 4 \, e^{x}\right )} e^{\left (-x\right )}}{x} \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=\left (5 x^{8} + 160 x^{4} + 1280\right ) e^{- x} + \frac {20}{x} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 5.52 \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=5 \, {\left (x^{8} + 8 \, x^{7} + 56 \, x^{6} + 336 \, x^{5} + 1680 \, x^{4} + 6720 \, x^{3} + 20160 \, x^{2} + 40320 \, x + 40320\right )} e^{\left (-x\right )} - 40 \, {\left (x^{7} + 7 \, x^{6} + 42 \, x^{5} + 210 \, x^{4} + 840 \, x^{3} + 2520 \, x^{2} + 5040 \, x + 5040\right )} e^{\left (-x\right )} + 160 \, {\left (x^{4} + 4 \, x^{3} + 12 \, x^{2} + 24 \, x + 24\right )} e^{\left (-x\right )} - 640 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x + 6\right )} e^{\left (-x\right )} + \frac {20}{x} + 1280 \, e^{\left (-x\right )} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=\frac {5 \, {\left (x^{9} e^{\left (-x\right )} + 32 \, x^{5} e^{\left (-x\right )} + 256 \, x e^{\left (-x\right )} + 4\right )}}{x} \]
[In]
[Out]
Time = 8.71 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-x} \left (-20 e^x-1280 x^2+640 x^5-160 x^6+40 x^9-5 x^{10}\right )}{x^2} \, dx=5\,{\mathrm {e}}^{-x}\,{\left (x^4+16\right )}^2+\frac {20}{x} \]
[In]
[Out]