Integrand size = 22, antiderivative size = 135 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {x \left (a+b x^2\right )}{7 a \left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}}+\frac {6 x}{35 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/4}}+\frac {8 x \left (a+b x^2\right )}{35 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/4}}+\frac {16 x}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \]
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Time = 0.02 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1103, 198, 197} \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {16 x}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {8 x}{35 a^3 \left (a+b x^2\right ) \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \]
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Rule 197
Rule 198
Rule 1103
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+\frac {b x^2}{a}} \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{9/2}} \, dx}{a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {\left (6 \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{7/2}} \, dx}{7 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {\left (24 \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/2}} \, dx}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {8 x}{35 a^3 \left (a+b x^2\right ) \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {\left (16 \sqrt {1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/2}} \, dx}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {16 x}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {8 x}{35 a^3 \left (a+b x^2\right ) \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {\left (a+b x^2\right ) \left (35 a^3 x+70 a^2 b x^3+56 a b^2 x^5+16 b^3 x^7\right )}{35 a^4 \left (\left (a+b x^2\right )^2\right )^{9/4}} \]
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Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.49
method | result | size |
gosper | \(\frac {\left (b \,x^{2}+a \right ) x \left (16 b^{3} x^{6}+56 b^{2} x^{4} a +70 a^{2} b \,x^{2}+35 a^{3}\right )}{35 a^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {9}{4}}}\) | \(66\) |
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Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {{\left (16 \, b^{3} x^{7} + 56 \, a b^{2} x^{5} + 70 \, a^{2} b x^{3} + 35 \, a^{3} x\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}}}{35 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} \]
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\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {9}{4}}}\, dx \]
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\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {9}{4}}} \,d x } \]
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\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {9}{4}}} \,d x } \]
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Time = 12.96 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx=\frac {x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{7\,a\,{\left (b\,x^2+a\right )}^5}+\frac {6\,x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{35\,a^2\,{\left (b\,x^2+a\right )}^4}+\frac {8\,x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{35\,a^3\,{\left (b\,x^2+a\right )}^3}+\frac {16\,x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{35\,a^4\,{\left (b\,x^2+a\right )}^2} \]
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