Integrand size = 11, antiderivative size = 64 \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=\frac {a^3}{9 b^4 (a+b x)^9}-\frac {3 a^2}{8 b^4 (a+b x)^8}+\frac {3 a}{7 b^4 (a+b x)^7}-\frac {1}{6 b^4 (a+b x)^6} \]
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Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=\frac {a^3}{9 b^4 (a+b x)^9}-\frac {3 a^2}{8 b^4 (a+b x)^8}+\frac {3 a}{7 b^4 (a+b x)^7}-\frac {1}{6 b^4 (a+b x)^6} \]
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Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a^3}{b^3 (a+b x)^{10}}+\frac {3 a^2}{b^3 (a+b x)^9}-\frac {3 a}{b^3 (a+b x)^8}+\frac {1}{b^3 (a+b x)^7}\right ) \, dx \\ & = \frac {a^3}{9 b^4 (a+b x)^9}-\frac {3 a^2}{8 b^4 (a+b x)^8}+\frac {3 a}{7 b^4 (a+b x)^7}-\frac {1}{6 b^4 (a+b x)^6} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.66 \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=-\frac {a^3+9 a^2 b x+36 a b^2 x^2+84 b^3 x^3}{504 b^4 (a+b x)^9} \]
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Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(-\frac {84 b^{3} x^{3}+36 a \,b^{2} x^{2}+9 a^{2} b x +a^{3}}{504 \left (b x +a \right )^{9} b^{4}}\) | \(41\) |
norman | \(\frac {-\frac {x^{3}}{6 b}-\frac {a \,x^{2}}{14 b^{2}}-\frac {a^{2} x}{56 b^{3}}-\frac {a^{3}}{504 b^{4}}}{\left (b x +a \right )^{9}}\) | \(44\) |
risch | \(\frac {-\frac {x^{3}}{6 b}-\frac {a \,x^{2}}{14 b^{2}}-\frac {a^{2} x}{56 b^{3}}-\frac {a^{3}}{504 b^{4}}}{\left (b x +a \right )^{9}}\) | \(44\) |
parallelrisch | \(\frac {-84 b^{8} x^{3}-36 a \,b^{7} x^{2}-9 a^{2} b^{6} x -a^{3} b^{5}}{504 b^{9} \left (b x +a \right )^{9}}\) | \(48\) |
default | \(\frac {a^{3}}{9 b^{4} \left (b x +a \right )^{9}}-\frac {3 a^{2}}{8 b^{4} \left (b x +a \right )^{8}}+\frac {3 a}{7 b^{4} \left (b x +a \right )^{7}}-\frac {1}{6 b^{4} \left (b x +a \right )^{6}}\) | \(57\) |
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Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (56) = 112\).
Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.05 \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=-\frac {84 \, b^{3} x^{3} + 36 \, a b^{2} x^{2} + 9 \, a^{2} b x + a^{3}}{504 \, {\left (b^{13} x^{9} + 9 \, a b^{12} x^{8} + 36 \, a^{2} b^{11} x^{7} + 84 \, a^{3} b^{10} x^{6} + 126 \, a^{4} b^{9} x^{5} + 126 \, a^{5} b^{8} x^{4} + 84 \, a^{6} b^{7} x^{3} + 36 \, a^{7} b^{6} x^{2} + 9 \, a^{8} b^{5} x + a^{9} b^{4}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (60) = 120\).
Time = 0.37 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.17 \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=\frac {- a^{3} - 9 a^{2} b x - 36 a b^{2} x^{2} - 84 b^{3} x^{3}}{504 a^{9} b^{4} + 4536 a^{8} b^{5} x + 18144 a^{7} b^{6} x^{2} + 42336 a^{6} b^{7} x^{3} + 63504 a^{5} b^{8} x^{4} + 63504 a^{4} b^{9} x^{5} + 42336 a^{3} b^{10} x^{6} + 18144 a^{2} b^{11} x^{7} + 4536 a b^{12} x^{8} + 504 b^{13} x^{9}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (56) = 112\).
Time = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.05 \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=-\frac {84 \, b^{3} x^{3} + 36 \, a b^{2} x^{2} + 9 \, a^{2} b x + a^{3}}{504 \, {\left (b^{13} x^{9} + 9 \, a b^{12} x^{8} + 36 \, a^{2} b^{11} x^{7} + 84 \, a^{3} b^{10} x^{6} + 126 \, a^{4} b^{9} x^{5} + 126 \, a^{5} b^{8} x^{4} + 84 \, a^{6} b^{7} x^{3} + 36 \, a^{7} b^{6} x^{2} + 9 \, a^{8} b^{5} x + a^{9} b^{4}\right )}} \]
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none
Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62 \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=-\frac {84 \, b^{3} x^{3} + 36 \, a b^{2} x^{2} + 9 \, a^{2} b x + a^{3}}{504 \, {\left (b x + a\right )}^{9} b^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{(a+b x)^{10}} \, dx=\frac {\frac {3\,a}{7\,{\left (a+b\,x\right )}^7}-\frac {1}{6\,{\left (a+b\,x\right )}^6}-\frac {3\,a^2}{8\,{\left (a+b\,x\right )}^8}+\frac {a^3}{9\,{\left (a+b\,x\right )}^9}}{b^4} \]
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