Optimal. Leaf size=26 \[ -79-x+\frac {x}{5+\left (e^x-x\right )^2 (-5+x)^2} \]
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Rubi [F] time = 44.21, 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 {-20-275 x^2+120 x^3-638 x^4+500 x^5-150 x^6+20 x^7-x^8+e^{4 x} \left (-625+500 x-150 x^2+20 x^3-x^4\right )+e^{3 x} \left (2500 x-2000 x^2+600 x^3-80 x^4+4 x^5\right )+e^{2 x} \left (-225+50 x-3741 x^2+2998 x^3-900 x^4+120 x^5-6 x^6\right )+e^x \left (500 x-170 x^2+2504 x^3-1998 x^4+600 x^5-80 x^6+4 x^7\right )}{25+250 x^2-100 x^3+635 x^4-500 x^5+150 x^6-20 x^7+x^8+e^{4 x} \left (625-500 x+150 x^2-20 x^3+x^4\right )+e^{3 x} \left (-2500 x+2000 x^2-600 x^3+80 x^4-4 x^5\right )+e^{2 x} \left (250-100 x+3760 x^2-3000 x^3+900 x^4-120 x^5+6 x^6\right )+e^x \left (-500 x+200 x^2-2520 x^3+2000 x^4-600 x^5+80 x^6-4 x^7\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20-e^{4 x} (-5+x)^4+4 e^{3 x} (-5+x)^4 x-275 x^2+120 x^3-638 x^4+500 x^5-150 x^6+20 x^7-x^8+e^{2 x} \left (-225+50 x-3741 x^2+2998 x^3-900 x^4+120 x^5-6 x^6\right )+2 e^x x \left (250-85 x+1252 x^2-999 x^3+300 x^4-40 x^5+2 x^6\right )}{\left (5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4\right )^2} \, dx\\ &=\int \left (-1-\frac {5-9 x+2 x^2}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )}+\frac {2 x \left (-20-125 e^x+130 x+200 e^x x-200 x^2-90 e^x x^2+90 x^3+16 e^x x^3-16 x^4-e^x x^4+x^5\right )}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )^2}\right ) \, dx\\ &=-x+2 \int \frac {x \left (-20-125 e^x+130 x+200 e^x x-200 x^2-90 e^x x^2+90 x^3+16 e^x x^3-16 x^4-e^x x^4+x^5\right )}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )^2} \, dx-\int \frac {5-9 x+2 x^2}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )} \, dx\\ &=-x+2 \int \frac {x \left (20+e^x (-5+x)^3 (-1+x)-130 x+200 x^2-90 x^3+16 x^4-x^5\right )}{(5-x) \left (5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4\right )^2} \, dx-\int \frac {-5+9 x-2 x^2}{(5-x) \left (5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4\right )} \, dx\\ &=-x+2 \int \left (-\frac {20+125 e^x-130 x-200 e^x x+200 x^2+90 e^x x^2-90 x^3-16 e^x x^3+16 x^4+e^x x^4-x^5}{\left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )^2}+\frac {5 \left (-20-125 e^x+130 x+200 e^x x-200 x^2-90 e^x x^2+90 x^3+16 e^x x^3-16 x^4-e^x x^4+x^5\right )}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )^2}\right ) \, dx-\int \left (\frac {1}{5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4}+\frac {10}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )}+\frac {2 x}{5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4}\right ) \, dx\\ &=-x-2 \int \frac {x}{5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4} \, dx-2 \int \frac {20+125 e^x-130 x-200 e^x x+200 x^2+90 e^x x^2-90 x^3-16 e^x x^3+16 x^4+e^x x^4-x^5}{\left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )^2} \, dx-10 \int \frac {1}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )} \, dx+10 \int \frac {-20-125 e^x+130 x+200 e^x x-200 x^2-90 e^x x^2+90 x^3+16 e^x x^3-16 x^4-e^x x^4+x^5}{(-5+x) \left (5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4\right )^2} \, dx-\int \frac {1}{5+25 e^{2 x}-50 e^x x-10 e^{2 x} x+25 x^2+20 e^x x^2+e^{2 x} x^2-10 x^3-2 e^x x^3+x^4} \, dx\\ &=-x-2 \int \frac {x}{5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4} \, dx-2 \int \frac {20+e^x (-5+x)^3 (-1+x)-130 x+200 x^2-90 x^3+16 x^4-x^5}{\left (5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4\right )^2} \, dx-10 \int \frac {1}{(-5+x) \left (5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4\right )} \, dx+10 \int \frac {20+e^x (-5+x)^3 (-1+x)-130 x+200 x^2-90 x^3+16 x^4-x^5}{(5-x) \left (5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4\right )^2} \, dx-\int \frac {1}{5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 43, normalized size = 1.65 \begin {gather*} x \left (-1+\frac {1}{5+e^{2 x} (-5+x)^2-2 e^x (-5+x)^2 x+25 x^2-10 x^3+x^4}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.77, size = 100, normalized size = 3.85 \begin {gather*} -\frac {x^{5} - 10 \, x^{4} + 25 \, x^{3} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} e^{x} + 4 \, x}{x^{4} - 10 \, x^{3} + 25 \, x^{2} + {\left (x^{2} - 10 \, x + 25\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{x} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.21, size = 64, normalized size = 2.46
method | result | size |
risch | \(-x +\frac {x}{{\mathrm e}^{2 x} x^{2}-2 \,{\mathrm e}^{x} x^{3}+x^{4}-10 x \,{\mathrm e}^{2 x}+20 \,{\mathrm e}^{x} x^{2}-10 x^{3}+25 \,{\mathrm e}^{2 x}-50 \,{\mathrm e}^{x} x +25 x^{2}+5}\) | \(64\) |
norman | \(\frac {-25 x^{3}+10 x^{4}-25 x \,{\mathrm e}^{2 x}-20 \,{\mathrm e}^{x} x^{3}+10 \,{\mathrm e}^{2 x} x^{2}+50 \,{\mathrm e}^{x} x^{2}-4 x -x^{5}+2 \,{\mathrm e}^{x} x^{4}-{\mathrm e}^{2 x} x^{3}}{{\mathrm e}^{2 x} x^{2}-2 \,{\mathrm e}^{x} x^{3}+x^{4}-10 x \,{\mathrm e}^{2 x}+20 \,{\mathrm e}^{x} x^{2}-10 x^{3}+25 \,{\mathrm e}^{2 x}-50 \,{\mathrm e}^{x} x +25 x^{2}+5}\) | \(124\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.87, size = 100, normalized size = 3.85 \begin {gather*} -\frac {x^{5} - 10 \, x^{4} + 25 \, x^{3} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} e^{x} + 4 \, x}{x^{4} - 10 \, x^{3} + 25 \, x^{2} + {\left (x^{2} - 10 \, x + 25\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{x} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{4\,x}\,\left (x^4-20\,x^3+150\,x^2-500\,x+625\right )-{\mathrm {e}}^{3\,x}\,\left (4\,x^5-80\,x^4+600\,x^3-2000\,x^2+2500\,x\right )-{\mathrm {e}}^x\,\left (4\,x^7-80\,x^6+600\,x^5-1998\,x^4+2504\,x^3-170\,x^2+500\,x\right )+275\,x^2-120\,x^3+638\,x^4-500\,x^5+150\,x^6-20\,x^7+x^8+{\mathrm {e}}^{2\,x}\,\left (6\,x^6-120\,x^5+900\,x^4-2998\,x^3+3741\,x^2-50\,x+225\right )+20}{{\mathrm {e}}^{4\,x}\,\left (x^4-20\,x^3+150\,x^2-500\,x+625\right )-{\mathrm {e}}^{3\,x}\,\left (4\,x^5-80\,x^4+600\,x^3-2000\,x^2+2500\,x\right )-{\mathrm {e}}^x\,\left (4\,x^7-80\,x^6+600\,x^5-2000\,x^4+2520\,x^3-200\,x^2+500\,x\right )+250\,x^2-100\,x^3+635\,x^4-500\,x^5+150\,x^6-20\,x^7+x^8+{\mathrm {e}}^{2\,x}\,\left (6\,x^6-120\,x^5+900\,x^4-3000\,x^3+3760\,x^2-100\,x+250\right )+25} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.69, size = 48, normalized size = 1.85 \begin {gather*} - x + \frac {x}{x^{4} - 10 x^{3} + 25 x^{2} + \left (x^{2} - 10 x + 25\right ) e^{2 x} + \left (- 2 x^{3} + 20 x^{2} - 50 x\right ) e^{x} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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