Integrand size = 22, antiderivative size = 189 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac {2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {b^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^2}-\frac {d^{5/2} (7 b c-5 a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2} \]
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Time = 0.20 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {483, 597, 536, 211} \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {b^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^2}+\frac {-5 a^2 d^2+2 a b c d+2 b^2 c^2}{2 a^2 c^3 x (b c-a d)}-\frac {d^{5/2} (7 b c-5 a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2}-\frac {2 b c-5 a d}{6 a c^2 x^3 (b c-a d)}-\frac {d}{2 c x^3 \left (c+d x^2\right ) (b c-a d)} \]
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Rule 211
Rule 483
Rule 536
Rule 597
Rubi steps \begin{align*} \text {integral}& = -\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {\int \frac {2 b c-5 a d-5 b d x^2}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 c (b c-a d)} \\ & = -\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}-\frac {\int \frac {3 \left (2 b^2 c^2+2 a b c d-5 a^2 d^2\right )+3 b d (2 b c-5 a d) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{6 a c^2 (b c-a d)} \\ & = -\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac {2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {\int \frac {3 \left (2 b^3 c^3+2 a b^2 c^2 d+2 a^2 b c d^2-5 a^3 d^3\right )+3 b d \left (2 b^2 c^2+2 a b c d-5 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{6 a^2 c^3 (b c-a d)} \\ & = -\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac {2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {b^4 \int \frac {1}{a+b x^2} \, dx}{a^2 (b c-a d)^2}-\frac {\left (d^3 (7 b c-5 a d)\right ) \int \frac {1}{c+d x^2} \, dx}{2 c^3 (b c-a d)^2} \\ & = -\frac {2 b c-5 a d}{6 a c^2 (b c-a d) x^3}+\frac {2 b^2 c^2+2 a b c d-5 a^2 d^2}{2 a^2 c^3 (b c-a d) x}-\frac {d}{2 c (b c-a d) x^3 \left (c+d x^2\right )}+\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)^2}-\frac {d^{5/2} (7 b c-5 a d) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {1}{3 a c^2 x^3}+\frac {b c+2 a d}{a^2 c^3 x}-\frac {d^3 x}{2 c^3 (b c-a d) \left (c+d x^2\right )}+\frac {b^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (-b c+a d)^2}-\frac {d^{5/2} (7 b c-5 a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{7/2} (b c-a d)^2} \]
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Time = 2.73 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.67
method | result | size |
default | \(-\frac {1}{3 a \,c^{2} x^{3}}-\frac {-2 a d -b c}{x \,a^{2} c^{3}}+\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a^{2} \left (a d -b c \right )^{2} \sqrt {a b}}+\frac {d^{3} \left (\frac {\left (\frac {a d}{2}-\frac {b c}{2}\right ) x}{d \,x^{2}+c}+\frac {\left (5 a d -7 b c \right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \sqrt {c d}}\right )}{c^{3} \left (a d -b c \right )^{2}}\) | \(127\) |
risch | \(\text {Expression too large to display}\) | \(1263\) |
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Time = 0.92 (sec) , antiderivative size = 1281, normalized size of antiderivative = 6.78 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\text {Timed out} \]
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Time = 0.30 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b}} - \frac {{\left (7 \, b c d^{3} - 5 \, a d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt {c d}} - \frac {2 \, a b c^{3} - 2 \, a^{2} c^{2} d - 3 \, {\left (2 \, b^{2} c^{2} d + 2 \, a b c d^{2} - 5 \, a^{2} d^{3}\right )} x^{4} - 2 \, {\left (3 \, b^{2} c^{3} + 2 \, a b c^{2} d - 5 \, a^{2} c d^{2}\right )} x^{2}}{6 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{5} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b}} - \frac {d^{3} x}{2 \, {\left (b c^{4} - a c^{3} d\right )} {\left (d x^{2} + c\right )}} - \frac {{\left (7 \, b c d^{3} - 5 \, a d^{4}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt {c d}} + \frac {3 \, b c x^{2} + 6 \, a d x^{2} - a c}{3 \, a^{2} c^{3} x^{3}} \]
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Time = 6.20 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.48 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {\frac {1}{3\,a\,c}-\frac {x^2\,\left (5\,a\,d+3\,b\,c\right )}{3\,a^2\,c^2}+\frac {x^4\,\left (-5\,a^2\,d^3+2\,a\,b\,c\,d^2+2\,b^2\,c^2\,d\right )}{2\,a^2\,c^3\,\left (a\,d-b\,c\right )}}{d\,x^5+c\,x^3}-\frac {\mathrm {atan}\left (\frac {b\,c^7\,x\,{\left (-a^5\,b^7\right )}^{3/2}\,4{}\mathrm {i}+a^{12}\,b\,d^7\,x\,\sqrt {-a^5\,b^7}\,25{}\mathrm {i}+a^{10}\,b^3\,c^2\,d^5\,x\,\sqrt {-a^5\,b^7}\,49{}\mathrm {i}-a^{11}\,b^2\,c\,d^6\,x\,\sqrt {-a^5\,b^7}\,70{}\mathrm {i}}{-25\,a^{15}\,b^4\,d^7+70\,a^{14}\,b^5\,c\,d^6-49\,a^{13}\,b^6\,c^2\,d^5+4\,a^8\,b^{11}\,c^7}\right )\,\sqrt {-a^5\,b^7}\,1{}\mathrm {i}}{a^7\,d^2-2\,a^6\,b\,c\,d+a^5\,b^2\,c^2}-\frac {\mathrm {atan}\left (\frac {a^7\,d^3\,x\,{\left (-c^7\,d^5\right )}^{3/2}\,25{}\mathrm {i}+b^7\,c^{14}\,d\,x\,\sqrt {-c^7\,d^5}\,4{}\mathrm {i}-a^6\,b\,c\,d^2\,x\,{\left (-c^7\,d^5\right )}^{3/2}\,70{}\mathrm {i}+a^5\,b^2\,c^2\,d\,x\,{\left (-c^7\,d^5\right )}^{3/2}\,49{}\mathrm {i}}{25\,a^7\,c^{11}\,d^{10}-70\,a^6\,b\,c^{12}\,d^9+49\,a^5\,b^2\,c^{13}\,d^8-4\,b^7\,c^{18}\,d^3}\right )\,\left (5\,a\,d-7\,b\,c\right )\,\sqrt {-c^7\,d^5}\,1{}\mathrm {i}}{2\,\left (a^2\,c^7\,d^2-2\,a\,b\,c^8\,d+b^2\,c^9\right )} \]
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