Integrand size = 51, antiderivative size = 14 \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=-\frac {1}{9 b (a+b x)^9} \]
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Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2083, 32} \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=-\frac {1}{9 b (a+b x)^9} \]
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Rule 32
Rule 2083
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(a+b x)^{10}} \, dx \\ & = -\frac {1}{9 b (a+b x)^9} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=-\frac {1}{9 b (a+b x)^9} \]
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Time = 0.13 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {1}{9 b \left (b x +a \right )^{9}}\) | \(13\) |
norman | \(-\frac {1}{9 b \left (b x +a \right )^{9}}\) | \(13\) |
risch | \(-\frac {1}{9 b \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )^{2} \left (b x +a \right )}\) | \(53\) |
gosper | \(-\frac {1}{9 \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right ) \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 x \,a^{4} b +a^{5}\right ) b}\) | \(97\) |
parallelrisch | \(-\frac {1}{9 \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right ) \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 x \,a^{4} b +a^{5}\right ) b}\) | \(97\) |
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Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (12) = 24\).
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 7.21 \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=-\frac {1}{9 \, {\left (b^{10} x^{9} + 9 \, a b^{9} x^{8} + 36 \, a^{2} b^{8} x^{7} + 84 \, a^{3} b^{7} x^{6} + 126 \, a^{4} b^{6} x^{5} + 126 \, a^{5} b^{5} x^{4} + 84 \, a^{6} b^{4} x^{3} + 36 \, a^{7} b^{3} x^{2} + 9 \, a^{8} b^{2} x + a^{9} b\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (12) = 24\).
Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 7.79 \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=- \frac {1}{9 a^{9} b + 81 a^{8} b^{2} x + 324 a^{7} b^{3} x^{2} + 756 a^{6} b^{4} x^{3} + 1134 a^{5} b^{5} x^{4} + 1134 a^{4} b^{6} x^{5} + 756 a^{3} b^{7} x^{6} + 324 a^{2} b^{8} x^{7} + 81 a b^{9} x^{8} + 9 b^{10} x^{9}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (12) = 24\).
Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 7.21 \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=-\frac {1}{9 \, {\left (b^{10} x^{9} + 9 \, a b^{9} x^{8} + 36 \, a^{2} b^{8} x^{7} + 84 \, a^{3} b^{7} x^{6} + 126 \, a^{4} b^{6} x^{5} + 126 \, a^{5} b^{5} x^{4} + 84 \, a^{6} b^{4} x^{3} + 36 \, a^{7} b^{3} x^{2} + 9 \, a^{8} b^{2} x + a^{9} b\right )}} \]
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none
Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=-\frac {1}{9 \, {\left (b x + a\right )}^{9} b} \]
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Time = 10.73 (sec) , antiderivative size = 103, normalized size of antiderivative = 7.36 \[ \int \frac {1}{\left (a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+5 a b^4 x^4+b^5 x^5\right )^2} \, dx=-\frac {1}{9\,a^9\,b+81\,a^8\,b^2\,x+324\,a^7\,b^3\,x^2+756\,a^6\,b^4\,x^3+1134\,a^5\,b^5\,x^4+1134\,a^4\,b^6\,x^5+756\,a^3\,b^7\,x^6+324\,a^2\,b^8\,x^7+81\,a\,b^9\,x^8+9\,b^{10}\,x^9} \]
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