Integrand size = 22, antiderivative size = 55 \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=\frac {(a-b x) \left (a^2-b^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (1,1+2 p,1+p,\frac {a+b x}{2 a}\right )}{2 a b p} \]
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Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {692, 71} \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=-\frac {2^{p-1} \left (\frac {b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1-p,p+1,p+2,\frac {a-b x}{2 a}\right )}{a^2 b (p+1)} \]
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Rule 71
Rule 692
Rubi steps \begin{align*} \text {integral}& = \frac {\left ((a-b x)^{-1-p} \left (1+\frac {b x}{a}\right )^{-1-p} \left (a^2-b^2 x^2\right )^{1+p}\right ) \int (a-b x)^p \left (1+\frac {b x}{a}\right )^{-1+p} \, dx}{a^2} \\ & = -\frac {2^{-1+p} \left (1+\frac {b x}{a}\right )^{-1-p} \left (a^2-b^2 x^2\right )^{1+p} \, _2F_1\left (1-p,1+p;2+p;\frac {a-b x}{2 a}\right )}{a^2 b (1+p)} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=-\frac {2^{-1+p} (a-b x) \left (1+\frac {b x}{a}\right )^{-p} \left (a^2-b^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {a-b x}{2 a}\right )}{a b (1+p)} \]
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\[\int \frac {\left (-b^{2} x^{2}+a^{2}\right )^{p}}{b x +a}d x\]
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\[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=\int { \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b x + a} \,d x } \]
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Result contains complex when optimal does not.
Time = 3.13 (sec) , antiderivative size = 318, normalized size of antiderivative = 5.78 \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=\begin {cases} \frac {0^{p} \log {\left (-1 + \frac {b^{2} x^{2}}{a^{2}} \right )}}{2 b} + \frac {0^{p} \operatorname {acoth}{\left (\frac {b x}{a} \right )}}{b} + \frac {a b^{2 p - 2} p x^{2 p - 1} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (\frac {1}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, \frac {1}{2} - p \\ \frac {3}{2} - p \end {matrix}\middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{2 \Gamma \left (\frac {3}{2} - p\right ) \Gamma \left (p + 1\right )} + \frac {a^{2 p} b x^{2} \Gamma \left (p\right ) \Gamma \left (1 - p\right ) {{}_{3}F_{2}\left (\begin {matrix} 2, 1, 1 - p \\ 2, 2 \end {matrix}\middle | {\frac {b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )}}{2 a^{2} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {0^{p} \log {\left (1 - \frac {b^{2} x^{2}}{a^{2}} \right )}}{2 b} + \frac {0^{p} \operatorname {atanh}{\left (\frac {b x}{a} \right )}}{b} + \frac {a b^{2 p - 2} p x^{2 p - 1} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (\frac {1}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, \frac {1}{2} - p \\ \frac {3}{2} - p \end {matrix}\middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{2 \Gamma \left (\frac {3}{2} - p\right ) \Gamma \left (p + 1\right )} + \frac {a^{2 p} b x^{2} \Gamma \left (p\right ) \Gamma \left (1 - p\right ) {{}_{3}F_{2}\left (\begin {matrix} 2, 1, 1 - p \\ 2, 2 \end {matrix}\middle | {\frac {b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )}}{2 a^{2} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} & \text {otherwise} \end {cases} \]
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\[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=\int { \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b x + a} \,d x } \]
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\[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=\int { \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b x + a} \,d x } \]
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Timed out. \[ \int \frac {\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx=\int \frac {{\left (a^2-b^2\,x^2\right )}^p}{a+b\,x} \,d x \]
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