Integrand size = 21, antiderivative size = 107 \[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}-\frac {\left (3 b^2+4 a c\right ) \arctan \left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{16 c^{5/2}} \]
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Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1128, 756, 654, 635, 210} \[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {\left (4 a c+3 b^2\right ) \arctan \left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{16 c^{5/2}}-\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c} \]
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Rule 210
Rule 635
Rule 654
Rule 756
Rule 1128
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right ) \\ & = -\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}-\frac {\text {Subst}\left (\int \frac {-a-\frac {3 b x}{2}}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}+\frac {\left (3 b^2+4 a c\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x-c x^2}} \, dx,x,x^2\right )}{16 c^2} \\ & = -\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}+\frac {\left (3 b^2+4 a c\right ) \text {Subst}\left (\int \frac {1}{-4 c-x^2} \, dx,x,\frac {b-2 c x^2}{\sqrt {a+b x^2-c x^4}}\right )}{8 c^2} \\ & = -\frac {3 b \sqrt {a+b x^2-c x^4}}{8 c^2}-\frac {x^2 \sqrt {a+b x^2-c x^4}}{4 c}-\frac {\left (3 b^2+4 a c\right ) \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{16 c^{5/2}} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=\frac {\left (-3 b-2 c x^2\right ) \sqrt {a+b x^2-c x^4}}{8 c^2}+\frac {\left (3 b^2+4 a c\right ) \arctan \left (\frac {\sqrt {c} x^2}{-\sqrt {a}+\sqrt {a+b x^2-c x^4}}\right )}{8 c^{5/2}} \]
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Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {\left (2 c \,x^{2}+3 b \right ) \sqrt {-c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {\left (4 a c +3 b^{2}\right ) \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{16 c^{\frac {5}{2}}}\) | \(77\) |
pseudoelliptic | \(\frac {-4 c^{\frac {3}{2}} x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}-6 b \sqrt {-c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}-4 \arctan \left (\frac {-2 c \,x^{2}+b}{2 \sqrt {c}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}}\right ) a c -3 \arctan \left (\frac {-2 c \,x^{2}+b}{2 \sqrt {c}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}}\right ) b^{2}}{16 c^{\frac {5}{2}}}\) | \(118\) |
default | \(-\frac {x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 b \sqrt {-c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 b^{2} \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{16 c^{\frac {5}{2}}}+\frac {a \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{4 c^{\frac {3}{2}}}\) | \(120\) |
elliptic | \(-\frac {x^{2} \sqrt {-c \,x^{4}+b \,x^{2}+a}}{4 c}-\frac {3 b \sqrt {-c \,x^{4}+b \,x^{2}+a}}{8 c^{2}}+\frac {3 b^{2} \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{16 c^{\frac {5}{2}}}+\frac {a \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{4 c^{\frac {3}{2}}}\) | \(120\) |
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Time = 0.28 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.97 \[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=\left [-\frac {{\left (3 \, b^{2} + 4 \, a c\right )} \sqrt {-c} \log \left (8 \, c^{2} x^{4} - 8 \, b c x^{2} + b^{2} - 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {-c} - 4 \, a c\right ) + 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + 3 \, b c\right )}}{32 \, c^{3}}, -\frac {{\left (3 \, b^{2} + 4 \, a c\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {c}}{2 \, {\left (c^{2} x^{4} - b c x^{2} - a c\right )}}\right ) + 2 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c^{2} x^{2} + 3 \, b c\right )}}{16 \, c^{3}}\right ] \]
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\[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=\int \frac {x^{5}}{\sqrt {a + b x^{2} - c x^{4}}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98 \[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {\sqrt {-c x^{4} + b x^{2} + a} x^{2}}{4 \, c} - \frac {3 \, b^{2} \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{16 \, c^{\frac {5}{2}}} - \frac {a \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{4 \, c^{\frac {3}{2}}} - \frac {3 \, \sqrt {-c x^{4} + b x^{2} + a} b}{8 \, c^{2}} \]
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Time = 0.31 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=-\frac {1}{8} \, \sqrt {-c x^{4} + b x^{2} + a} {\left (\frac {2 \, x^{2}}{c} + \frac {3 \, b}{c^{2}}\right )} - \frac {{\left (3 \, b^{2} + 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {-c} x^{2} - \sqrt {-c x^{4} + b x^{2} + a}\right )} \sqrt {-c} + b \right |}\right )}{16 \, \sqrt {-c} c^{2}} \]
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Timed out. \[ \int \frac {x^5}{\sqrt {a+b x^2-c x^4}} \, dx=\int \frac {x^5}{\sqrt {-c\,x^4+b\,x^2+a}} \,d x \]
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