Optimal. Leaf size=55 \[ \frac{(a-b x) \left (a^2-b^2 x^2\right )^p \, _2F_1\left (1,2 p+1;p+1;\frac{a+b x}{2 a}\right )}{2 a b p} \]
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Rubi [A] time = 0.0299029, antiderivative size = 73, normalized size of antiderivative = 1.33, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {678, 69} \[ -\frac{2^{p-1} \left (\frac{b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (1-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{a^2 b (p+1)} \]
Antiderivative was successfully verified.
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Rule 678
Rule 69
Rubi steps
\begin{align*} \int \frac{\left (a^2-b^2 x^2\right )^p}{a+b x} \, dx &=\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*}
Mathematica [A] time = 0.0516805, size = 75, normalized size = 1.36 \[ -\frac{2^{p-1} (a-b x) \left (\frac{b x}{a}+1\right )^{-p} \left (a^2-b^2 x^2\right )^p \, _2F_1\left (1-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{a b (p+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.532, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}}{bx+a}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b x + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b x + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] time = 5.05242, size = 323, normalized size = 5.87 \begin{align*} \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} p x^{2 p} 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 b^{2} x \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}\: \frac{\left |{b^{2} x^{2}}\right |}{\left |{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} p x^{2 p} 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 b^{2} x \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b x + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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