Integrand size = 19, antiderivative size = 150 \[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=\frac {e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}-\frac {\left (a e^2-b c^2 (5+4 p)\right ) x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )}{b (5+4 p)}+\frac {2}{3} c e x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1221, 1218, 252, 251, 372, 371} \[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (c^2-\frac {a e^2}{4 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+\frac {2}{3} c e x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )+\frac {e^2 x \left (a+b x^4\right )^{p+1}}{b (4 p+5)} \]
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Rule 251
Rule 252
Rule 371
Rule 372
Rule 1218
Rule 1221
Rubi steps \begin{align*} \text {integral}& = \frac {e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\frac {\int \left (-a e^2+b c^2 (5+4 p)+2 b c e (5+4 p) x^2\right ) \left (a+b x^4\right )^p \, dx}{b (5+4 p)} \\ & = \frac {e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\frac {\int \left (-a e^2 \left (1-\frac {b c^2 (5+4 p)}{a e^2}\right ) \left (a+b x^4\right )^p+2 b c e (5+4 p) x^2 \left (a+b x^4\right )^p\right ) \, dx}{b (5+4 p)} \\ & = \frac {e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+(2 c e) \int x^2 \left (a+b x^4\right )^p \, dx-\left (-c^2+\frac {a e^2}{5 b+4 b p}\right ) \int \left (a+b x^4\right )^p \, dx \\ & = \frac {e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\left (2 c e \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \int x^2 \left (1+\frac {b x^4}{a}\right )^p \, dx-\left (\left (-c^2+\frac {a e^2}{5 b+4 b p}\right ) \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^4}{a}\right )^p \, dx \\ & = \frac {e^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\left (c^2-\frac {a e^2}{5 b+4 b p}\right ) x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b x^4}{a}\right )+\frac {2}{3} c e x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b x^4}{a}\right ) \\ \end{align*}
Time = 0.56 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.71 \[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=\frac {1}{15} x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (15 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+e x^2 \left (10 c \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )+3 e x^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )\right )\right ) \]
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\[\int \left (e \,x^{2}+c \right )^{2} \left (b \,x^{4}+a \right )^{p}d x\]
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\[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=\int { {\left (e x^{2} + c\right )}^{2} {\left (b x^{4} + a\right )}^{p} \,d x } \]
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Result contains complex when optimal does not.
Time = 32.63 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.79 \[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=\frac {a^{p} c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} c e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} + \frac {a^{p} e^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \]
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\[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=\int { {\left (e x^{2} + c\right )}^{2} {\left (b x^{4} + a\right )}^{p} \,d x } \]
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\[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=\int { {\left (e x^{2} + c\right )}^{2} {\left (b x^{4} + a\right )}^{p} \,d x } \]
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Timed out. \[ \int \left (c+e x^2\right )^2 \left (a+b x^4\right )^p \, dx=\int {\left (b\,x^4+a\right )}^p\,{\left (e\,x^2+c\right )}^2 \,d x \]
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