Integrand size = 29, antiderivative size = 319 \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {a d (f x)^{1+m} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {3}{2},-\frac {3}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m) \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}+\frac {a e (f x)^{3+m} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {3}{2},-\frac {3}{2},\frac {5+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f^3 (3+m) \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \]
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Time = 0.24 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1349, 1155, 524} \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {a d (f x)^{m+1} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {m+1}{2},-\frac {3}{2},-\frac {3}{2},\frac {m+3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}+\frac {a e (f x)^{m+3} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {m+3}{2},-\frac {3}{2},-\frac {3}{2},\frac {m+5}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f^3 (m+3) \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \]
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Rule 524
Rule 1155
Rule 1349
Rubi steps \begin{align*} \text {integral}& = \int \left (d (f x)^m \left (a+b x^2+c x^4\right )^{3/2}+\frac {e (f x)^{2+m} \left (a+b x^2+c x^4\right )^{3/2}}{f^2}\right ) \, dx \\ & = d \int (f x)^m \left (a+b x^2+c x^4\right )^{3/2} \, dx+\frac {e \int (f x)^{2+m} \left (a+b x^2+c x^4\right )^{3/2} \, dx}{f^2} \\ & = \frac {\left (a d \sqrt {a+b x^2+c x^4}\right ) \int (f x)^m \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2} \, dx}{\sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}+\frac {\left (a e \sqrt {a+b x^2+c x^4}\right ) \int (f x)^{2+m} \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2} \, dx}{f^2 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \\ & = \frac {a d (f x)^{1+m} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1+m}{2};-\frac {3}{2},-\frac {3}{2};\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m) \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}+\frac {a e (f x)^{3+m} \sqrt {a+b x^2+c x^4} F_1\left (\frac {3+m}{2};-\frac {3}{2},-\frac {3}{2};\frac {5+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f^3 (3+m) \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \\ \end{align*}
Time = 2.90 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.46 \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\frac {x (f x)^m \sqrt {a+b x^2+c x^4} \left (a d \left (105+71 m+15 m^2+m^3\right ) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},-\frac {1}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) x^2 \left ((b d+a e) \left (35+12 m+m^2\right ) \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},-\frac {1}{2},\frac {5+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+(3+m) x^2 \left ((c d+b e) (7+m) \operatorname {AppellF1}\left (\frac {5+m}{2},-\frac {1}{2},-\frac {1}{2},\frac {7+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+c e (5+m) x^2 \operatorname {AppellF1}\left (\frac {7+m}{2},-\frac {1}{2},-\frac {1}{2},\frac {9+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )\right )}{(1+m) (3+m) (5+m) (7+m) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}}} \]
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\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}d x\]
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\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} {\left (e x^{2} + d\right )} \left (f x\right )^{m} \,d x } \]
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\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int \left (f x\right )^{m} \left (d + e x^{2}\right ) \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \]
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\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} {\left (e x^{2} + d\right )} \left (f x\right )^{m} \,d x } \]
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\[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} {\left (e x^{2} + d\right )} \left (f x\right )^{m} \,d x } \]
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Timed out. \[ \int (f x)^m \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2} \, dx=\int {\left (f\,x\right )}^m\,\left (e\,x^2+d\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2} \,d x \]
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