Integrand size = 26, antiderivative size = 196 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {2 a}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^4}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a^3}{3 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a^2}{2 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Time = 0.11 (sec) , antiderivative size = 196, 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} \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {3 a^2}{2 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^4}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a^3}{3 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Rule 45
Rule 660
Rule 1125
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,x^2\right ) \\ & = \frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \frac {x^4}{\left (a b+b^2 x\right )^5} \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \text {Subst}\left (\int \left (\frac {a^4}{b^9 (a+b x)^5}-\frac {4 a^3}{b^9 (a+b x)^4}+\frac {6 a^2}{b^9 (a+b x)^3}-\frac {4 a}{b^9 (a+b x)^2}+\frac {1}{b^9 (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {2 a}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^4}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a^3}{3 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a^2}{2 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}
Time = 0.74 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.33 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\frac {b x^2 \left (-a \sqrt {\left (a+b x^2\right )^2} \left (12 a^6+30 a^5 b x^2+22 a^4 b^2 x^4+3 a^3 b^3 x^6-3 a^2 b^4 x^8+3 a b^5 x^{10}-3 b^6 x^{12}\right )+\sqrt {a^2} \left (12 a^7+42 a^6 b x^2+52 a^5 b^2 x^4+25 a^4 b^3 x^6+3 b^7 x^{14}\right )\right )}{a^4 \left (a+b x^2\right )^3 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )}+12 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-12 \log \left (b^5 \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 b^5} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.39
method | result | size |
pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )+4 a \,b^{3} x^{6}+9 a^{2} b^{2} x^{4}+\frac {22 a^{3} b \,x^{2}}{3}+\frac {25 a^{4}}{12}\right )}{2 \left (b \,x^{2}+a \right )^{4} b^{5}}\) | \(76\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {2 a \,x^{6}}{b^{2}}+\frac {9 a^{2} x^{4}}{2 b^{3}}+\frac {11 a^{3} x^{2}}{3 b^{4}}+\frac {25 a^{4}}{24 b^{5}}\right )}{\left (b \,x^{2}+a \right )^{5}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{5}}\) | \(96\) |
default | \(\frac {\left (12 \ln \left (b \,x^{2}+a \right ) x^{8} b^{4}+48 \ln \left (b \,x^{2}+a \right ) x^{6} a \,b^{3}+48 a \,b^{3} x^{6}+72 \ln \left (b \,x^{2}+a \right ) x^{4} a^{2} b^{2}+108 a^{2} b^{2} x^{4}+48 \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b +88 a^{3} b \,x^{2}+12 \ln \left (b \,x^{2}+a \right ) a^{4}+25 a^{4}\right ) \left (b \,x^{2}+a \right )}{24 b^{5} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(141\) |
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Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.69 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {48 \, a b^{3} x^{6} + 108 \, a^{2} b^{2} x^{4} + 88 \, a^{3} b x^{2} + 25 \, a^{4} + 12 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}} \]
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\[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{9}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.51 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {48 \, a b^{3} x^{6} + 108 \, a^{2} b^{2} x^{4} + 88 \, a^{3} b x^{2} + 25 \, a^{4}}{24 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{5}} \]
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Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.43 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {25 \, b^{3} x^{8} + 52 \, a b^{2} x^{6} + 42 \, a^{2} b x^{4} + 12 \, a^{3} x^{2}}{24 \, {\left (b x^{2} + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \]
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Timed out. \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^9}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]
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