Integrand size = 17, antiderivative size = 40 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {\left (b+c x^2\right )^4}{10 b x^{10}}+\frac {c \left (b+c x^2\right )^4}{40 b^2 x^8} \]
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Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1598, 272, 47, 37} \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=\frac {c \left (b+c x^2\right )^4}{40 b^2 x^8}-\frac {\left (b+c x^2\right )^4}{10 b x^{10}} \]
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Rule 37
Rule 47
Rule 272
Rule 1598
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (b+c x^2\right )^3}{x^{11}} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {(b+c x)^3}{x^6} \, dx,x,x^2\right ) \\ & = -\frac {\left (b+c x^2\right )^4}{10 b x^{10}}-\frac {c \text {Subst}\left (\int \frac {(b+c x)^3}{x^5} \, dx,x,x^2\right )}{10 b} \\ & = -\frac {\left (b+c x^2\right )^4}{10 b x^{10}}+\frac {c \left (b+c x^2\right )^4}{40 b^2 x^8} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.08 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {b^3}{10 x^{10}}-\frac {3 b^2 c}{8 x^8}-\frac {b c^2}{2 x^6}-\frac {c^3}{4 x^4} \]
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Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {3 b^{2} c}{8 x^{8}}-\frac {b \,c^{2}}{2 x^{6}}-\frac {c^{3}}{4 x^{4}}-\frac {b^{3}}{10 x^{10}}\) | \(36\) |
risch | \(\frac {-\frac {1}{4} c^{3} x^{6}-\frac {1}{2} b \,c^{2} x^{4}-\frac {3}{8} b^{2} c \,x^{2}-\frac {1}{10} b^{3}}{x^{10}}\) | \(37\) |
gosper | \(-\frac {10 c^{3} x^{6}+20 b \,c^{2} x^{4}+15 b^{2} c \,x^{2}+4 b^{3}}{40 x^{10}}\) | \(38\) |
parallelrisch | \(\frac {-10 c^{3} x^{6}-20 b \,c^{2} x^{4}-15 b^{2} c \,x^{2}-4 b^{3}}{40 x^{10}}\) | \(38\) |
norman | \(\frac {-\frac {1}{10} b^{3} x^{6}-\frac {1}{4} c^{3} x^{12}-\frac {1}{2} b \,c^{2} x^{10}-\frac {3}{8} b^{2} c \,x^{8}}{x^{16}}\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {10 \, c^{3} x^{6} + 20 \, b c^{2} x^{4} + 15 \, b^{2} c x^{2} + 4 \, b^{3}}{40 \, x^{10}} \]
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Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.98 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=\frac {- 4 b^{3} - 15 b^{2} c x^{2} - 20 b c^{2} x^{4} - 10 c^{3} x^{6}}{40 x^{10}} \]
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Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {10 \, c^{3} x^{6} + 20 \, b c^{2} x^{4} + 15 \, b^{2} c x^{2} + 4 \, b^{3}}{40 \, x^{10}} \]
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Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {10 \, c^{3} x^{6} + 20 \, b c^{2} x^{4} + 15 \, b^{2} c x^{2} + 4 \, b^{3}}{40 \, x^{10}} \]
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Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {\frac {b^3}{10}+\frac {3\,b^2\,c\,x^2}{8}+\frac {b\,c^2\,x^4}{2}+\frac {c^3\,x^6}{4}}{x^{10}} \]
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