Integrand size = 24, antiderivative size = 124 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {x^3}{10 b \left (a+b x^2\right )^5}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}+\frac {x}{160 a b^2 \left (a+b x^2\right )^3}+\frac {x}{128 a^2 b^2 \left (a+b x^2\right )^2}+\frac {3 x}{256 a^3 b^2 \left (a+b x^2\right )}+\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 a^{7/2} b^{5/2}} \]
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Time = 0.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 294, 205, 211} \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 a^{7/2} b^{5/2}}+\frac {3 x}{256 a^3 b^2 \left (a+b x^2\right )}+\frac {x}{128 a^2 b^2 \left (a+b x^2\right )^2}+\frac {x}{160 a b^2 \left (a+b x^2\right )^3}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}-\frac {x^3}{10 b \left (a+b x^2\right )^5} \]
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Rule 28
Rule 205
Rule 211
Rule 294
Rubi steps \begin{align*} \text {integral}& = b^6 \int \frac {x^4}{\left (a b+b^2 x^2\right )^6} \, dx \\ & = -\frac {x^3}{10 b \left (a+b x^2\right )^5}+\frac {1}{10} \left (3 b^4\right ) \int \frac {x^2}{\left (a b+b^2 x^2\right )^5} \, dx \\ & = -\frac {x^3}{10 b \left (a+b x^2\right )^5}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}+\frac {1}{80} \left (3 b^2\right ) \int \frac {1}{\left (a b+b^2 x^2\right )^4} \, dx \\ & = -\frac {x^3}{10 b \left (a+b x^2\right )^5}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}+\frac {x}{160 a b^2 \left (a+b x^2\right )^3}+\frac {b \int \frac {1}{\left (a b+b^2 x^2\right )^3} \, dx}{32 a} \\ & = -\frac {x^3}{10 b \left (a+b x^2\right )^5}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}+\frac {x}{160 a b^2 \left (a+b x^2\right )^3}+\frac {x}{128 a^2 b^2 \left (a+b x^2\right )^2}+\frac {3 \int \frac {1}{\left (a b+b^2 x^2\right )^2} \, dx}{128 a^2} \\ & = -\frac {x^3}{10 b \left (a+b x^2\right )^5}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}+\frac {x}{160 a b^2 \left (a+b x^2\right )^3}+\frac {x}{128 a^2 b^2 \left (a+b x^2\right )^2}+\frac {3 x}{256 a^3 b^2 \left (a+b x^2\right )}+\frac {3 \int \frac {1}{a b+b^2 x^2} \, dx}{256 a^3 b} \\ & = -\frac {x^3}{10 b \left (a+b x^2\right )^5}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}+\frac {x}{160 a b^2 \left (a+b x^2\right )^3}+\frac {x}{128 a^2 b^2 \left (a+b x^2\right )^2}+\frac {3 x}{256 a^3 b^2 \left (a+b x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 a^{7/2} b^{5/2}} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {-15 a^4 x-70 a^3 b x^3+128 a^2 b^2 x^5+70 a b^3 x^7+15 b^4 x^9}{1280 a^3 b^2 \left (a+b x^2\right )^5}+\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 a^{7/2} b^{5/2}} \]
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Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63
method | result | size |
default | \(\frac {\frac {3 b^{2} x^{9}}{256 a^{3}}+\frac {7 b \,x^{7}}{128 a^{2}}+\frac {x^{5}}{10 a}-\frac {7 x^{3}}{128 b}-\frac {3 a x}{256 b^{2}}}{\left (b \,x^{2}+a \right )^{5}}+\frac {3 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 a^{3} b^{2} \sqrt {a b}}\) | \(78\) |
risch | \(\frac {\frac {3 b^{2} x^{9}}{256 a^{3}}+\frac {7 b \,x^{7}}{128 a^{2}}+\frac {x^{5}}{10 a}-\frac {7 x^{3}}{128 b}-\frac {3 a x}{256 b^{2}}}{\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right )}{512 \sqrt {-a b}\, b^{2} a^{3}}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right )}{512 \sqrt {-a b}\, b^{2} a^{3}}\) | \(127\) |
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Time = 0.27 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.15 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\left [\frac {30 \, a b^{5} x^{9} + 140 \, a^{2} b^{4} x^{7} + 256 \, a^{3} b^{3} x^{5} - 140 \, a^{4} b^{2} x^{3} - 30 \, a^{5} b x - 15 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2560 \, {\left (a^{4} b^{8} x^{10} + 5 \, a^{5} b^{7} x^{8} + 10 \, a^{6} b^{6} x^{6} + 10 \, a^{7} b^{5} x^{4} + 5 \, a^{8} b^{4} x^{2} + a^{9} b^{3}\right )}}, \frac {15 \, a b^{5} x^{9} + 70 \, a^{2} b^{4} x^{7} + 128 \, a^{3} b^{3} x^{5} - 70 \, a^{4} b^{2} x^{3} - 15 \, a^{5} b x + 15 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{1280 \, {\left (a^{4} b^{8} x^{10} + 5 \, a^{5} b^{7} x^{8} + 10 \, a^{6} b^{6} x^{6} + 10 \, a^{7} b^{5} x^{4} + 5 \, a^{8} b^{4} x^{2} + a^{9} b^{3}\right )}}\right ] \]
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Time = 0.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.58 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=- \frac {3 \sqrt {- \frac {1}{a^{7} b^{5}}} \log {\left (- a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{512} + \frac {3 \sqrt {- \frac {1}{a^{7} b^{5}}} \log {\left (a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{512} + \frac {- 15 a^{4} x - 70 a^{3} b x^{3} + 128 a^{2} b^{2} x^{5} + 70 a b^{3} x^{7} + 15 b^{4} x^{9}}{1280 a^{8} b^{2} + 6400 a^{7} b^{3} x^{2} + 12800 a^{6} b^{4} x^{4} + 12800 a^{5} b^{5} x^{6} + 6400 a^{4} b^{6} x^{8} + 1280 a^{3} b^{7} x^{10}} \]
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Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {15 \, b^{4} x^{9} + 70 \, a b^{3} x^{7} + 128 \, a^{2} b^{2} x^{5} - 70 \, a^{3} b x^{3} - 15 \, a^{4} x}{1280 \, {\left (a^{3} b^{7} x^{10} + 5 \, a^{4} b^{6} x^{8} + 10 \, a^{5} b^{5} x^{6} + 10 \, a^{6} b^{4} x^{4} + 5 \, a^{7} b^{3} x^{2} + a^{8} b^{2}\right )}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} a^{3} b^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} a^{3} b^{2}} + \frac {15 \, b^{4} x^{9} + 70 \, a b^{3} x^{7} + 128 \, a^{2} b^{2} x^{5} - 70 \, a^{3} b x^{3} - 15 \, a^{4} x}{1280 \, {\left (b x^{2} + a\right )}^{5} a^{3} b^{2}} \]
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Time = 14.00 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {x^5}{10\,a}-\frac {7\,x^3}{128\,b}+\frac {7\,b\,x^7}{128\,a^2}+\frac {3\,b^2\,x^9}{256\,a^3}-\frac {3\,a\,x}{256\,b^2}}{a^5+5\,a^4\,b\,x^2+10\,a^3\,b^2\,x^4+10\,a^2\,b^3\,x^6+5\,a\,b^4\,x^8+b^5\,x^{10}}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{256\,a^{7/2}\,b^{5/2}} \]
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