Integrand size = 26, antiderivative size = 72 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 a x^{14}}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{84 a^2 x^{14}} \]
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Time = 0.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1124} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{84 a^2 x^{14}}-\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 a x^{14}} \]
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Rule 1124
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 a x^{14}}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{84 a^2 x^{14}} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (6 a^5+35 a^4 b x^2+84 a^3 b^2 x^4+105 a^2 b^3 x^6+70 a b^4 x^8+21 b^5 x^{10}\right )}{84 x^{14} \left (a+b x^2\right )} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94
method | result | size |
pseudoelliptic | \(-\frac {\left (21 x^{10} b^{5}+70 a \,x^{8} b^{4}+105 a^{2} x^{6} b^{3}+84 a^{3} x^{4} b^{2}+35 x^{2} a^{4} b +6 a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{84 x^{14}}\) | \(68\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{14} a^{5}-\frac {5}{12} x^{2} a^{4} b -a^{3} x^{4} b^{2}-\frac {5}{4} a^{2} x^{6} b^{3}-\frac {5}{6} a \,x^{8} b^{4}-\frac {1}{4} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{14}}\) | \(79\) |
gosper | \(-\frac {\left (21 x^{10} b^{5}+70 a \,x^{8} b^{4}+105 a^{2} x^{6} b^{3}+84 a^{3} x^{4} b^{2}+35 x^{2} a^{4} b +6 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{84 x^{14} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (21 x^{10} b^{5}+70 a \,x^{8} b^{4}+105 a^{2} x^{6} b^{3}+84 a^{3} x^{4} b^{2}+35 x^{2} a^{4} b +6 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{84 x^{14} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
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none
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=-\frac {21 \, b^{5} x^{10} + 70 \, a b^{4} x^{8} + 105 \, a^{2} b^{3} x^{6} + 84 \, a^{3} b^{2} x^{4} + 35 \, a^{4} b x^{2} + 6 \, a^{5}}{84 \, x^{14}} \]
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\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{15}}\, dx \]
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none
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=-\frac {b^{5}}{4 \, x^{4}} - \frac {5 \, a b^{4}}{6 \, x^{6}} - \frac {5 \, a^{2} b^{3}}{4 \, x^{8}} - \frac {a^{3} b^{2}}{x^{10}} - \frac {5 \, a^{4} b}{12 \, x^{12}} - \frac {a^{5}}{14 \, x^{14}} \]
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none
Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=-\frac {21 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 70 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 84 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 35 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{84 \, x^{14}} \]
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Time = 13.24 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.21 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{15}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^4\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^6\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{12\,x^{12}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^8\,\left (b\,x^2+a\right )}-\frac {a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^{10}\,\left (b\,x^2+a\right )} \]
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