Integrand size = 30, antiderivative size = 167 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {(b e-3 a f) x}{b^4}+\frac {f x^3}{3 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{9/2}} \]
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Time = 0.18 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1818, 1599, 1271, 1167, 211} \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {x \left (11 a^3 f-7 a^2 b e+3 a b^2 d+b^3 c\right )}{8 a b^4 \left (a+b x^2\right )}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (35 a^3 f-15 a^2 b e+3 a b^2 d+b^3 c\right )}{8 a^{3/2} b^{9/2}}+\frac {x (b e-3 a f)}{b^4}+\frac {f x^3}{3 b^3} \]
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Rule 211
Rule 1167
Rule 1271
Rule 1599
Rule 1818
Rubi steps \begin{align*} \text {integral}& = \frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x \left (-\left (\left (b c+3 a d-\frac {3 a^2 e}{b}+\frac {3 a^3 f}{b^2}\right ) x\right )-4 a \left (e-\frac {a f}{b}\right ) x^3-4 a f x^5\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b} \\ & = \frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x^2 \left (-b c-3 a d+\frac {3 a^2 e}{b}-\frac {3 a^3 f}{b^2}-4 a \left (e-\frac {a f}{b}\right ) x^2-4 a f x^4\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b} \\ & = \frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\int \frac {b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f+8 a b (b e-2 a f) x^2+8 a b^2 f x^4}{a+b x^2} \, dx}{8 a b^4} \\ & = \frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\int \left (8 a (b e-3 a f)+8 a b f x^2+\frac {b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f}{a+b x^2}\right ) \, dx}{8 a b^4} \\ & = \frac {(b e-3 a f) x}{b^4}+\frac {f x^3}{3 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \int \frac {1}{a+b x^2} \, dx}{8 a b^4} \\ & = \frac {(b e-3 a f) x}{b^4}+\frac {f x^3}{3 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{9/2}} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (-105 a^4 f+3 b^4 c x^2+5 a^3 b \left (9 e-35 f x^2\right )+a^2 b^2 \left (-9 d+75 e x^2-56 f x^4\right )+a b^3 \left (-3 c-15 d x^2+24 e x^4+8 f x^6\right )\right )}{24 a b^4 \left (a+b x^2\right )^2}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{9/2}} \]
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Time = 3.49 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {-\frac {1}{3} f \,x^{3} b +3 a f x -b e x}{b^{4}}+\frac {\frac {-\frac {b \left (13 f \,a^{3}-9 a^{2} b e +5 a \,b^{2} d -b^{3} c \right ) x^{3}}{8 a}+\left (-\frac {11}{8} f \,a^{3}+\frac {7}{8} a^{2} b e -\frac {3}{8} a \,b^{2} d -\frac {1}{8} b^{3} c \right ) x}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (35 f \,a^{3}-15 a^{2} b e +3 a \,b^{2} d +b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}}{b^{4}}\) | \(151\) |
risch | \(\frac {f \,x^{3}}{3 b^{3}}-\frac {3 a f x}{b^{4}}+\frac {e x}{b^{3}}+\frac {-\frac {b \left (13 f \,a^{3}-9 a^{2} b e +5 a \,b^{2} d -b^{3} c \right ) x^{3}}{8 a}+\left (-\frac {11}{8} f \,a^{3}+\frac {7}{8} a^{2} b e -\frac {3}{8} a \,b^{2} d -\frac {1}{8} b^{3} c \right ) x}{b^{4} \left (b \,x^{2}+a \right )^{2}}-\frac {35 a^{2} \ln \left (b x +\sqrt {-a b}\right ) f}{16 b^{4} \sqrt {-a b}}+\frac {15 a \ln \left (b x +\sqrt {-a b}\right ) e}{16 b^{3} \sqrt {-a b}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) d}{16 b^{2} \sqrt {-a b}}-\frac {\ln \left (b x +\sqrt {-a b}\right ) c}{16 b \sqrt {-a b}\, a}+\frac {35 a^{2} \ln \left (-b x +\sqrt {-a b}\right ) f}{16 b^{4} \sqrt {-a b}}-\frac {15 a \ln \left (-b x +\sqrt {-a b}\right ) e}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) d}{16 b^{2} \sqrt {-a b}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) c}{16 b \sqrt {-a b}\, a}\) | \(306\) |
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Time = 0.26 (sec) , antiderivative size = 555, normalized size of antiderivative = 3.32 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, a^{2} b^{4} f x^{7} + 16 \, {\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} + 2 \, {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} - 3 \, {\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f + {\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} b f\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 6 \, {\left (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{48 \, {\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, \frac {8 \, a^{2} b^{4} f x^{7} + 8 \, {\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} + {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} + 3 \, {\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f + {\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 3 \, {\left (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{24 \, {\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \]
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Time = 5.71 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.56 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x \left (- \frac {3 a f}{b^{4}} + \frac {e}{b^{3}}\right ) - \frac {\sqrt {- \frac {1}{a^{3} b^{9}}} \cdot \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log {\left (- a^{2} b^{4} \sqrt {- \frac {1}{a^{3} b^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{3} b^{9}}} \cdot \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log {\left (a^{2} b^{4} \sqrt {- \frac {1}{a^{3} b^{9}}} + x \right )}}{16} + \frac {x^{3} \left (- 13 a^{3} b f + 9 a^{2} b^{2} e - 5 a b^{3} d + b^{4} c\right ) + x \left (- 11 a^{4} f + 7 a^{3} b e - 3 a^{2} b^{2} d - a b^{3} c\right )}{8 a^{3} b^{4} + 16 a^{2} b^{5} x^{2} + 8 a b^{6} x^{4}} + \frac {f x^{3}}{3 b^{3}} \]
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Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (b^{4} c - 5 \, a b^{3} d + 9 \, a^{2} b^{2} e - 13 \, a^{3} b f\right )} x^{3} - {\left (a b^{3} c + 3 \, a^{2} b^{2} d - 7 \, a^{3} b e + 11 \, a^{4} f\right )} x}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} + \frac {b f x^{3} + 3 \, {\left (b e - 3 \, a f\right )} x}{3 \, b^{4}} + \frac {{\left (b^{3} c + 3 \, a b^{2} d - 15 \, a^{2} b e + 35 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (b^{3} c + 3 \, a b^{2} d - 15 \, a^{2} b e + 35 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{4}} + \frac {b^{4} c x^{3} - 5 \, a b^{3} d x^{3} + 9 \, a^{2} b^{2} e x^{3} - 13 \, a^{3} b f x^{3} - a b^{3} c x - 3 \, a^{2} b^{2} d x + 7 \, a^{3} b e x - 11 \, a^{4} f x}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{4}} + \frac {b^{6} f x^{3} + 3 \, b^{6} e x - 9 \, a b^{5} f x}{3 \, b^{9}} \]
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Time = 5.51 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )-\frac {x\,\left (\frac {11\,f\,a^3}{8}-\frac {7\,e\,a^2\,b}{8}+\frac {3\,d\,a\,b^2}{8}+\frac {c\,b^3}{8}\right )-\frac {x^3\,\left (-13\,f\,a^3\,b+9\,e\,a^2\,b^2-5\,d\,a\,b^3+c\,b^4\right )}{8\,a}}{a^2\,b^4+2\,a\,b^5\,x^2+b^6\,x^4}+\frac {f\,x^3}{3\,b^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (35\,f\,a^3-15\,e\,a^2\,b+3\,d\,a\,b^2+c\,b^3\right )}{8\,a^{3/2}\,b^{9/2}} \]
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