Integrand size = 19, antiderivative size = 101 \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=x \operatorname {AppellF1}\left (\frac {1}{4},3,-p,\frac {5}{4},x^4,-b x^4\right )+x^3 \operatorname {AppellF1}\left (\frac {3}{4},3,-p,\frac {7}{4},x^4,-b x^4\right )+\frac {3}{5} x^5 \operatorname {AppellF1}\left (\frac {5}{4},3,-p,\frac {9}{4},x^4,-b x^4\right )+\frac {1}{7} x^7 \operatorname {AppellF1}\left (\frac {7}{4},3,-p,\frac {11}{4},x^4,-b x^4\right ) \]
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Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1254, 440, 524} \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=x \operatorname {AppellF1}\left (\frac {1}{4},3,-p,\frac {5}{4},x^4,-b x^4\right )+\frac {1}{7} x^7 \operatorname {AppellF1}\left (\frac {7}{4},3,-p,\frac {11}{4},x^4,-b x^4\right )+\frac {3}{5} x^5 \operatorname {AppellF1}\left (\frac {5}{4},3,-p,\frac {9}{4},x^4,-b x^4\right )+x^3 \operatorname {AppellF1}\left (\frac {3}{4},3,-p,\frac {7}{4},x^4,-b x^4\right ) \]
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Rule 440
Rule 524
Rule 1254
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}-\frac {3 x^2 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}-\frac {3 x^4 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}-\frac {x^6 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3}\right ) \, dx \\ & = -\left (3 \int \frac {x^2 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx\right )-3 \int \frac {x^4 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx-\int \frac {\left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx-\int \frac {x^6 \left (1+b x^4\right )^p}{\left (-1+x^4\right )^3} \, dx \\ & = x F_1\left (\frac {1}{4};3,-p;\frac {5}{4};x^4,-b x^4\right )+x^3 F_1\left (\frac {3}{4};3,-p;\frac {7}{4};x^4,-b x^4\right )+\frac {3}{5} x^5 F_1\left (\frac {5}{4};3,-p;\frac {9}{4};x^4,-b x^4\right )+\frac {1}{7} x^7 F_1\left (\frac {7}{4};3,-p;\frac {11}{4};x^4,-b x^4\right ) \\ \end{align*}
\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=\int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx \]
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\[\int \frac {\left (b \,x^{4}+1\right )^{p}}{\left (-x^{2}+1\right )^{3}}d x\]
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\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=\int { -\frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=\int { -\frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{3}} \,d x } \]
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\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=\int { -\frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^3} \, dx=\int -\frac {{\left (b\,x^4+1\right )}^p}{{\left (x^2-1\right )}^3} \,d x \]
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