Integrand size = 27, antiderivative size = 147 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {f x}{b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x}{4 a \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c+a b^2 d-5 a^2 b e+9 a^3 f\right ) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac {\left (3 b^3 c+a b^2 d+3 a^2 b e-15 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{7/2}} \]
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Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1828, 1171, 396, 211} \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}+\frac {x \left (9 a^3 f-5 a^2 b e+a b^2 d+3 b^3 c\right )}{8 a^2 b^3 \left (a+b x^2\right )}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-15 a^3 f+3 a^2 b e+a b^2 d+3 b^3 c\right )}{8 a^{5/2} b^{7/2}}+\frac {f x}{b^3} \]
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Rule 211
Rule 396
Rule 1171
Rule 1828
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}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {-\frac {3 b^3 c+a b^2 d-a^2 b e+a^3 f}{b^3}-\frac {4 a (b e-a f) x^2}{b^2}-\frac {4 a f x^4}{b}}{\left (a+b x^2\right )^2} \, dx}{4 a} \\ & = \frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x}{4 a \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c+a b^2 d-5 a^2 b e+9 a^3 f\right ) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac {\int \frac {\frac {3 b^3 c+a b^2 d+3 a^2 b e-7 a^3 f}{b^3}+\frac {8 a^2 f x^2}{b^2}}{a+b x^2} \, dx}{8 a^2} \\ & = \frac {f x}{b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x}{4 a \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c+a b^2 d-5 a^2 b e+9 a^3 f\right ) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac {\left (3 b^3 c+a b^2 d+3 a^2 b e-15 a^3 f\right ) \int \frac {1}{a+b x^2} \, dx}{8 a^2 b^3} \\ & = \frac {f x}{b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x}{4 a \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c+a b^2 d-5 a^2 b e+9 a^3 f\right ) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac {\left (3 b^3 c+a b^2 d+3 a^2 b e-15 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{7/2}} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.96 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (15 a^4 f+3 b^4 c x^2+a b^3 \left (5 c+d x^2\right )+a^3 b \left (-3 e+25 f x^2\right )-a^2 b^2 \left (d+5 e x^2-8 f x^4\right )\right )}{8 a^2 b^3 \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c+a b^2 d+3 a^2 b e-15 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{7/2}} \]
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Time = 3.50 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {f x}{b^{3}}-\frac {\frac {-\frac {b \left (9 f \,a^{3}-5 a^{2} b e +a \,b^{2} d +3 b^{3} c \right ) x^{3}}{8 a^{2}}-\frac {\left (7 f \,a^{3}-3 a^{2} b e -a \,b^{2} d +5 b^{3} c \right ) x}{8 a}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (15 f \,a^{3}-3 a^{2} b e -a \,b^{2} d -3 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}}{b^{3}}\) | \(139\) |
risch | \(\frac {f x}{b^{3}}+\frac {\frac {b \left (9 f \,a^{3}-5 a^{2} b e +a \,b^{2} d +3 b^{3} c \right ) x^{3}}{8 a^{2}}+\frac {\left (7 f \,a^{3}-3 a^{2} b e -a \,b^{2} d +5 b^{3} c \right ) x}{8 a}}{b^{3} \left (b \,x^{2}+a \right )^{2}}-\frac {15 a \ln \left (b x -\sqrt {-a b}\right ) f}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (b x -\sqrt {-a b}\right ) e}{16 b^{2} \sqrt {-a b}}+\frac {\ln \left (b x -\sqrt {-a b}\right ) d}{16 b \sqrt {-a b}\, a}+\frac {3 \ln \left (b x -\sqrt {-a b}\right ) c}{16 \sqrt {-a b}\, a^{2}}+\frac {15 a \ln \left (-b x -\sqrt {-a b}\right ) f}{16 b^{3} \sqrt {-a b}}-\frac {3 \ln \left (-b x -\sqrt {-a b}\right ) e}{16 b^{2} \sqrt {-a b}}-\frac {\ln \left (-b x -\sqrt {-a b}\right ) d}{16 b \sqrt {-a b}\, a}-\frac {3 \ln \left (-b x -\sqrt {-a b}\right ) c}{16 \sqrt {-a b}\, a^{2}}\) | \(302\) |
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Time = 0.29 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.43 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, a^{3} b^{3} f x^{5} + 2 \, {\left (3 \, a b^{5} c + a^{2} b^{4} d - 5 \, a^{3} b^{3} e + 25 \, a^{4} b^{2} f\right )} x^{3} + {\left (3 \, a^{2} b^{3} c + a^{3} b^{2} d + 3 \, a^{4} b e - 15 \, a^{5} f + {\left (3 \, b^{5} c + a b^{4} d + 3 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c + a^{2} b^{3} d + 3 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (5 \, a^{2} b^{4} c - a^{3} b^{3} d - 3 \, a^{4} b^{2} e + 15 \, a^{5} b f\right )} x}{16 \, {\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}, \frac {8 \, a^{3} b^{3} f x^{5} + {\left (3 \, a b^{5} c + a^{2} b^{4} d - 5 \, a^{3} b^{3} e + 25 \, a^{4} b^{2} f\right )} x^{3} + {\left (3 \, a^{2} b^{3} c + a^{3} b^{2} d + 3 \, a^{4} b e - 15 \, a^{5} f + {\left (3 \, b^{5} c + a b^{4} d + 3 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c + a^{2} b^{3} d + 3 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (5 \, a^{2} b^{4} c - a^{3} b^{3} d - 3 \, a^{4} b^{2} e + 15 \, a^{5} b f\right )} x}{8 \, {\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}\right ] \]
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Time = 3.40 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.65 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {\sqrt {- \frac {1}{a^{5} b^{7}}} \cdot \left (15 a^{3} f - 3 a^{2} b e - a b^{2} d - 3 b^{3} c\right ) \log {\left (- a^{3} b^{3} \sqrt {- \frac {1}{a^{5} b^{7}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{a^{5} b^{7}}} \cdot \left (15 a^{3} f - 3 a^{2} b e - a b^{2} d - 3 b^{3} c\right ) \log {\left (a^{3} b^{3} \sqrt {- \frac {1}{a^{5} b^{7}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (9 a^{3} b f - 5 a^{2} b^{2} e + a b^{3} d + 3 b^{4} c\right ) + x \left (7 a^{4} f - 3 a^{3} b e - a^{2} b^{2} d + 5 a b^{3} c\right )}{8 a^{4} b^{3} + 16 a^{3} b^{4} x^{2} + 8 a^{2} b^{5} x^{4}} + \frac {f x}{b^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, b^{4} c + a b^{3} d - 5 \, a^{2} b^{2} e + 9 \, a^{3} b f\right )} x^{3} + {\left (5 \, a b^{3} c - a^{2} b^{2} d - 3 \, a^{3} b e + 7 \, a^{4} f\right )} x}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}} + \frac {f x}{b^{3}} + \frac {{\left (3 \, b^{3} c + a b^{2} d + 3 \, a^{2} b e - 15 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {f x}{b^{3}} + \frac {{\left (3 \, b^{3} c + a b^{2} d + 3 \, a^{2} b e - 15 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{3}} + \frac {3 \, b^{4} c x^{3} + a b^{3} d x^{3} - 5 \, a^{2} b^{2} e x^{3} + 9 \, a^{3} b f x^{3} + 5 \, a b^{3} c x - a^{2} b^{2} d x - 3 \, a^{3} b e x + 7 \, a^{4} f x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b^{3}} \]
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Time = 5.54 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {x\,\left (7\,f\,a^3-3\,e\,a^2\,b-d\,a\,b^2+5\,c\,b^3\right )}{8\,a}+\frac {x^3\,\left (9\,f\,a^3\,b-5\,e\,a^2\,b^2+d\,a\,b^3+3\,c\,b^4\right )}{8\,a^2}}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}+\frac {f\,x}{b^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-15\,f\,a^3+3\,e\,a^2\,b+d\,a\,b^2+3\,c\,b^3\right )}{8\,a^{5/2}\,b^{7/2}} \]
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