Integrand size = 19, antiderivative size = 108 \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=-\frac {x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )+\frac {(1-b (7+4 p)) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )}{b (7+4 p)}+\frac {3}{5} x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-b x^4\right ) \]
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Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1221, 1907, 251, 371} \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )+\frac {3}{5} x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-b x^4\right )-x^3 \left (1-\frac {1}{4 b p+7 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )-\frac {x^3 \left (b x^4+1\right )^{p+1}}{b (4 p+7)} \]
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Rule 251
Rule 371
Rule 1221
Rule 1907
Rubi steps \begin{align*} \text {integral}& = -\frac {x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\int \left (1+b x^4\right )^p \left (b (7+4 p)+3 (1-b (7+4 p)) x^2+3 b (7+4 p) x^4\right ) \, dx}{b (7+4 p)} \\ & = -\frac {x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\int \left (b (7+4 p) \left (1+b x^4\right )^p+3 (1-b (7+4 p)) x^2 \left (1+b x^4\right )^p+3 b (7+4 p) x^4 \left (1+b x^4\right )^p\right ) \, dx}{b (7+4 p)} \\ & = -\frac {x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+3 \int x^4 \left (1+b x^4\right )^p \, dx-\left (3 \left (1-\frac {1}{7 b+4 b p}\right )\right ) \int x^2 \left (1+b x^4\right )^p \, dx+\int \left (1+b x^4\right )^p \, dx \\ & = -\frac {x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+x \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-b x^4\right )-\left (1-\frac {1}{7 b+4 b p}\right ) x^3 \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-b x^4\right )+\frac {3}{5} x^5 \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-b x^4\right ) \\ \end{align*}
Time = 0.87 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )+\frac {3}{5} x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-b x^4\right )-\frac {1}{7} x^7 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-b x^4\right ) \]
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Time = 4.73 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69
method | result | size |
meijerg | \(-\frac {x^{7} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {7}{4},-p ;\frac {11}{4};-b \,x^{4}\right )}{7}+\frac {3 x^{5} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {5}{4},-p ;\frac {9}{4};-b \,x^{4}\right )}{5}-x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{4},-p ;\frac {7}{4};-b \,x^{4}\right )+x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},-p ;\frac {5}{4};-b \,x^{4}\right )\) | \(75\) |
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\[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )}^{3} {\left (b x^{4} + 1\right )}^{p} \,d x } \]
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Result contains complex when optimal does not.
Time = 50.67 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.19 \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=- \frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, - p \\ \frac {11}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {3 x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} - \frac {3 x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]
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\[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )}^{3} {\left (b x^{4} + 1\right )}^{p} \,d x } \]
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\[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )}^{3} {\left (b x^{4} + 1\right )}^{p} \,d x } \]
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Timed out. \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=-\int {\left (x^2-1\right )}^3\,{\left (b\,x^4+1\right )}^p \,d x \]
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