Optimal. Leaf size=26 \[ -5+x+e^{\frac {(16+x)^2}{(2-x)^2 x^6}} \log (1+x) \]
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Rubi [B] time = 5.49, antiderivative size = 128, normalized size of antiderivative = 4.92, number of steps used = 14, number of rules used = 6, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {6688, 6742, 44, 77, 88, 2288} \begin {gather*} x-\frac {3 e^{\frac {(x+16)^2}{(2-x)^2 x^6}} \left (x^4 (-\log (x+1))-37 x^3 \log (x+1)-324 x^2 \log (x+1)+224 x \log (x+1)+512 \log (x+1)\right )}{(2-x)^3 x^7 (x+1) \left (-\frac {3 (x+16)^2}{(2-x)^2 x^7}+\frac {(x+16)^2}{(2-x)^3 x^6}+\frac {x+16}{(2-x)^2 x^6}\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 44
Rule 77
Rule 88
Rule 2288
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 x^7-e^{\frac {(16+x)^2}{(-2+x)^2 x^6}} (-2+x)^3 x^7-4 x^8-6 x^9+5 x^{10}-x^{11}+6 e^{\frac {(16+x)^2}{(-2+x)^2 x^6}} \left (-512-224 x+324 x^2+37 x^3+x^4\right ) \log (1+x)}{(2-x)^3 x^7 (1+x)} \, dx\\ &=\int \left (-\frac {8}{(-2+x)^3 (1+x)}+\frac {4 x}{(-2+x)^3 (1+x)}+\frac {6 x^2}{(-2+x)^3 (1+x)}-\frac {5 x^3}{(-2+x)^3 (1+x)}+\frac {x^4}{(-2+x)^3 (1+x)}+\frac {e^{\frac {(16+x)^2}{(-2+x)^2 x^6}} \left (-8 x^7+12 x^8-6 x^9+x^{10}+3072 \log (1+x)+1344 x \log (1+x)-1944 x^2 \log (1+x)-222 x^3 \log (1+x)-6 x^4 \log (1+x)\right )}{(-2+x)^3 x^7 (1+x)}\right ) \, dx\\ &=4 \int \frac {x}{(-2+x)^3 (1+x)} \, dx-5 \int \frac {x^3}{(-2+x)^3 (1+x)} \, dx+6 \int \frac {x^2}{(-2+x)^3 (1+x)} \, dx-8 \int \frac {1}{(-2+x)^3 (1+x)} \, dx+\int \frac {x^4}{(-2+x)^3 (1+x)} \, dx+\int \frac {e^{\frac {(16+x)^2}{(-2+x)^2 x^6}} \left (-8 x^7+12 x^8-6 x^9+x^{10}+3072 \log (1+x)+1344 x \log (1+x)-1944 x^2 \log (1+x)-222 x^3 \log (1+x)-6 x^4 \log (1+x)\right )}{(-2+x)^3 x^7 (1+x)} \, dx\\ &=-\frac {3 e^{\frac {(16+x)^2}{(2-x)^2 x^6}} \left (512 \log (1+x)+224 x \log (1+x)-324 x^2 \log (1+x)-37 x^3 \log (1+x)-x^4 \log (1+x)\right )}{(2-x)^3 x^7 (1+x) \left (\frac {16+x}{(2-x)^2 x^6}-\frac {3 (16+x)^2}{(2-x)^2 x^7}+\frac {(16+x)^2}{(2-x)^3 x^6}\right )}+4 \int \left (\frac {2}{3 (-2+x)^3}+\frac {1}{9 (-2+x)^2}-\frac {1}{27 (-2+x)}+\frac {1}{27 (1+x)}\right ) \, dx-5 \int \left (\frac {8}{3 (-2+x)^3}+\frac {28}{9 (-2+x)^2}+\frac {26}{27 (-2+x)}+\frac {1}{27 (1+x)}\right ) \, dx+6 \int \left (\frac {4}{3 (-2+x)^3}+\frac {8}{9 (-2+x)^2}+\frac {1}{27 (-2+x)}-\frac {1}{27 (1+x)}\right ) \, dx-8 \int \left (\frac {1}{3 (-2+x)^3}-\frac {1}{9 (-2+x)^2}+\frac {1}{27 (-2+x)}-\frac {1}{27 (1+x)}\right ) \, dx+\int \left (1+\frac {16}{3 (-2+x)^3}+\frac {80}{9 (-2+x)^2}+\frac {136}{27 (-2+x)}-\frac {1}{27 (1+x)}\right ) \, dx\\ &=x-\frac {3 e^{\frac {(16+x)^2}{(2-x)^2 x^6}} \left (512 \log (1+x)+224 x \log (1+x)-324 x^2 \log (1+x)-37 x^3 \log (1+x)-x^4 \log (1+x)\right )}{(2-x)^3 x^7 (1+x) \left (\frac {16+x}{(2-x)^2 x^6}-\frac {3 (16+x)^2}{(2-x)^2 x^7}+\frac {(16+x)^2}{(2-x)^3 x^6}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 23, normalized size = 0.88 \begin {gather*} x+e^{\frac {(16+x)^2}{(-2+x)^2 x^6}} \log (1+x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 33, normalized size = 1.27 \begin {gather*} e^{\left (\frac {x^{2} + 32 \, x + 256}{x^{8} - 4 \, x^{7} + 4 \, x^{6}}\right )} \log \left (x + 1\right ) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11} - 5 \, x^{10} + 6 \, x^{9} + 4 \, x^{8} - 8 \, x^{7} - 6 \, {\left (x^{4} + 37 \, x^{3} + 324 \, x^{2} - 224 \, x - 512\right )} e^{\left (\frac {x^{2} + 32 \, x + 256}{x^{8} - 4 \, x^{7} + 4 \, x^{6}}\right )} \log \left (x + 1\right ) + {\left (x^{10} - 6 \, x^{9} + 12 \, x^{8} - 8 \, x^{7}\right )} e^{\left (\frac {x^{2} + 32 \, x + 256}{x^{8} - 4 \, x^{7} + 4 \, x^{6}}\right )}}{x^{11} - 5 \, x^{10} + 6 \, x^{9} + 4 \, x^{8} - 8 \, x^{7}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 23, normalized size = 0.88
method | result | size |
risch | \({\mathrm e}^{\frac {\left (x +16\right )^{2}}{x^{6} \left (x -2\right )^{2}}} \ln \left (x +1\right )+x\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 2.88, size = 141, normalized size = 5.42 \begin {gather*} e^{\left (\frac {81}{16 \, {\left (x^{2} - 4 \, x + 4\right )}} - \frac {117}{8 \, {\left (x - 2\right )}} + \frac {117}{8 \, x} + \frac {387}{16 \, x^{2}} + \frac {153}{4 \, x^{3}} + \frac {225}{4 \, x^{4}} + \frac {72}{x^{5}} + \frac {64}{x^{6}}\right )} \log \left (x + 1\right ) + x - \frac {8 \, {\left (10 \, x - 17\right )}}{9 \, {\left (x^{2} - 4 \, x + 4\right )}} + \frac {20 \, {\left (7 \, x - 11\right )}}{9 \, {\left (x^{2} - 4 \, x + 4\right )}} - \frac {4 \, {\left (4 \, x - 5\right )}}{3 \, {\left (x^{2} - 4 \, x + 4\right )}} - \frac {4 \, {\left (2 \, x - 7\right )}}{9 \, {\left (x^{2} - 4 \, x + 4\right )}} - \frac {4 \, {\left (x + 1\right )}}{9 \, {\left (x^{2} - 4 \, x + 4\right )}} \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}}^{\frac {x^2+32\,x+256}{x^8-4\,x^7+4\,x^6}}\,\left (-x^{10}+6\,x^9-12\,x^8+8\,x^7\right )+8\,x^7-4\,x^8-6\,x^9+5\,x^{10}-x^{11}+\ln \left (x+1\right )\,{\mathrm {e}}^{\frac {x^2+32\,x+256}{x^8-4\,x^7+4\,x^6}}\,\left (6\,x^4+222\,x^3+1944\,x^2-1344\,x-3072\right )}{x^{11}-5\,x^{10}+6\,x^9+4\,x^8-8\,x^7} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.71, size = 29, normalized size = 1.12 \begin {gather*} x + e^{\frac {x^{2} + 32 x + 256}{x^{8} - 4 x^{7} + 4 x^{6}}} \log {\left (x + 1 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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