Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 x^{22} \left (a+b x^2\right )}-\frac {a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{20} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )} \]
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Time = 0.11 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 x^{22} \left (a+b x^2\right )}-\frac {a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{20} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )} \]
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Rule 45
Rule 660
Rule 1125
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{12}} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^5}{x^{12}} \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (\frac {a^5 b^5}{x^{12}}+\frac {5 a^4 b^6}{x^{11}}+\frac {10 a^3 b^7}{x^{10}}+\frac {10 a^2 b^8}{x^9}+\frac {5 a b^9}{x^8}+\frac {b^{10}}{x^7}\right ) \, dx,x,x^2\right )}{2 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}}{22 x^{22} \left (a+b x^2\right )}-\frac {a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{20} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^{16} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 x^{12} \left (a+b x^2\right )} \\ \end{align*}
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (252 a^5+1386 a^4 b x^2+3080 a^3 b^2 x^4+3465 a^2 b^3 x^6+1980 a b^4 x^8+462 b^5 x^{10}\right )}{5544 x^{22} \left (a+b x^2\right )} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 2.76 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\frac {11}{6} x^{10} b^{5}+\frac {55}{7} a \,x^{8} b^{4}+\frac {55}{4} a^{2} x^{6} b^{3}+\frac {110}{9} a^{3} x^{4} b^{2}+\frac {11}{2} x^{2} a^{4} b +a^{5}\right )}{22 x^{22}}\) | \(66\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{22} a^{5}-\frac {1}{4} x^{2} a^{4} b -\frac {5}{9} a^{3} x^{4} b^{2}-\frac {5}{8} a^{2} x^{6} b^{3}-\frac {5}{14} a \,x^{8} b^{4}-\frac {1}{12} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{22}}\) | \(79\) |
gosper | \(-\frac {\left (462 x^{10} b^{5}+1980 a \,x^{8} b^{4}+3465 a^{2} x^{6} b^{3}+3080 a^{3} x^{4} b^{2}+1386 x^{2} a^{4} b +252 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 x^{22} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (462 x^{10} b^{5}+1980 a \,x^{8} b^{4}+3465 a^{2} x^{6} b^{3}+3080 a^{3} x^{4} b^{2}+1386 x^{2} a^{4} b +252 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 x^{22} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
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Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {462 \, b^{5} x^{10} + 1980 \, a b^{4} x^{8} + 3465 \, a^{2} b^{3} x^{6} + 3080 \, a^{3} b^{2} x^{4} + 1386 \, a^{4} b x^{2} + 252 \, a^{5}}{5544 \, x^{22}} \]
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\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{23}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {b^{5}}{12 \, x^{12}} - \frac {5 \, a b^{4}}{14 \, x^{14}} - \frac {5 \, a^{2} b^{3}}{8 \, x^{16}} - \frac {5 \, a^{3} b^{2}}{9 \, x^{18}} - \frac {a^{4} b}{4 \, x^{20}} - \frac {a^{5}}{22 \, x^{22}} \]
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Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {462 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 1980 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 3465 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 3080 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 1386 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 252 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{5544 \, x^{22}} \]
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Time = 13.61 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{23}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{22\,x^{22}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{12\,x^{12}\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^{20}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^{16}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^{18}\,\left (b\,x^2+a\right )} \]
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