Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )} \]
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Time = 0.04 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1126, 276} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )} \]
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Rule 276
Rule 1126
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^5}{x^{20}} \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5 b^5}{x^{20}}+\frac {5 a^4 b^6}{x^{18}}+\frac {10 a^3 b^7}{x^{16}}+\frac {10 a^2 b^8}{x^{14}}+\frac {5 a b^9}{x^{12}}+\frac {b^{10}}{x^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 x^{19} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 x^{17} \left (a+b x^2\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^{15} \left (a+b x^2\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (21879 a^5+122265 a^4 b x^2+277134 a^3 b^2 x^4+319770 a^2 b^3 x^6+188955 a b^4 x^8+46189 b^5 x^{10}\right )}{415701 x^{19} \left (a+b x^2\right )} \]
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Time = 17.48 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{19} a^{5}-\frac {5}{17} x^{2} a^{4} b -\frac {2}{3} a^{3} x^{4} b^{2}-\frac {10}{13} a^{2} x^{6} b^{3}-\frac {5}{11} a \,x^{8} b^{4}-\frac {1}{9} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{19}}\) | \(79\) |
gosper | \(-\frac {\left (46189 x^{10} b^{5}+188955 a \,x^{8} b^{4}+319770 a^{2} x^{6} b^{3}+277134 a^{3} x^{4} b^{2}+122265 x^{2} a^{4} b +21879 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{415701 x^{19} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (46189 x^{10} b^{5}+188955 a \,x^{8} b^{4}+319770 a^{2} x^{6} b^{3}+277134 a^{3} x^{4} b^{2}+122265 x^{2} a^{4} b +21879 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{415701 x^{19} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
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Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=-\frac {46189 \, b^{5} x^{10} + 188955 \, a b^{4} x^{8} + 319770 \, a^{2} b^{3} x^{6} + 277134 \, a^{3} b^{2} x^{4} + 122265 \, a^{4} b x^{2} + 21879 \, a^{5}}{415701 \, x^{19}} \]
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\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{20}}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=-\frac {b^{5}}{9 \, x^{9}} - \frac {5 \, a b^{4}}{11 \, x^{11}} - \frac {10 \, a^{2} b^{3}}{13 \, x^{13}} - \frac {2 \, a^{3} b^{2}}{3 \, x^{15}} - \frac {5 \, a^{4} b}{17 \, x^{17}} - \frac {a^{5}}{19 \, x^{19}} \]
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Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=-\frac {46189 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 188955 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 319770 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 277134 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 122265 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 21879 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{415701 \, x^{19}} \]
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Time = 13.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{20}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{19\,x^{19}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^9\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{11\,x^{11}\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{17\,x^{17}\,\left (b\,x^2+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{13\,x^{13}\,\left (b\,x^2+a\right )}-\frac {2\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^{15}\,\left (b\,x^2+a\right )} \]
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