Integrand size = 26, antiderivative size = 209 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Time = 0.05 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1126, 296, 331, 211} \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Rule 211
Rule 296
Rule 331
Rule 1126
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^4 \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (7 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^4 \left (a b+b^2 x^2\right )^2} \, dx}{4 a \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (35 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^4 \left (a b+b^2 x^2\right )} \, dx}{8 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (35 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (35 b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = \frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}
Time = 1.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-8 a^3+56 a^2 b x^2+175 a b^2 x^4+105 b^3 x^6\right )+105 b^{3/2} x^3 \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{24 a^{9/2} x^3 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]
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Time = 1.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.67
method | result | size |
default | \(-\frac {\left (-105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{4} x^{7}-105 \sqrt {a b}\, b^{3} x^{6}-210 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{3} x^{5}-175 \sqrt {a b}\, a \,b^{2} x^{4}-105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{2} x^{3}-56 \sqrt {a b}\, a^{2} b \,x^{2}+8 \sqrt {a b}\, a^{3}\right ) \left (b \,x^{2}+a \right )}{24 \sqrt {a b}\, x^{3} a^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(139\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {35 b^{3} x^{6}}{8 a^{4}}+\frac {175 b^{2} x^{4}}{24 a^{3}}+\frac {7 b \,x^{2}}{3 a^{2}}-\frac {1}{3 a}\right )}{\left (b \,x^{2}+a \right )^{3} x^{3}}+\frac {35 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) a^{5}}-\frac {35 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) a^{5}}\) | \(153\) |
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Time = 0.26 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [\frac {210 \, b^{3} x^{6} + 350 \, a b^{2} x^{4} + 112 \, a^{2} b x^{2} - 16 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{48 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}, \frac {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}\right ] \]
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\[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{4} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3}}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {35 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {11 \, b^{3} x^{3} + 13 \, a b^{2} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {9 \, b x^{2} - a}{3 \, a^{4} x^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \]
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Timed out. \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]
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