Optimal. Leaf size=82 \[ \frac{5}{4 b^2 \sqrt{x} (a x+b)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2}}+\frac{1}{2 b \sqrt{x} (a x+b)^2}-\frac{15}{4 b^3 \sqrt{x}} \]
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Rubi [A] time = 0.0286377, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {263, 51, 63, 205} \[ \frac{5}{4 b^2 \sqrt{x} (a x+b)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2}}+\frac{1}{2 b \sqrt{x} (a x+b)^2}-\frac{15}{4 b^3 \sqrt{x}} \]
Antiderivative was successfully verified.
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Rule 263
Rule 51
Rule 63
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{\left (a+\frac{b}{x}\right )^3 x^{9/2}} \, dx &=\int \frac{1}{x^{3/2} (b+a x)^3} \, dx\\ &=\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5 \int \frac{1}{x^{3/2} (b+a x)^2} \, dx}{4 b}\\ &=\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}+\frac{15 \int \frac{1}{x^{3/2} (b+a x)} \, dx}{8 b^2}\\ &=-\frac{15}{4 b^3 \sqrt{x}}+\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}-\frac{(15 a) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 b^3}\\ &=-\frac{15}{4 b^3 \sqrt{x}}+\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}-\frac{(15 a) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 b^3}\\ &=-\frac{15}{4 b^3 \sqrt{x}}+\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2}}\\ \end{align*}
Mathematica [C] time = 0.0042387, size = 25, normalized size = 0.3 \[ -\frac{2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\frac{a x}{b}\right )}{b^3 \sqrt{x}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.012, size = 66, normalized size = 0.8 \begin{align*} -{\frac{7\,{a}^{2}}{4\,{b}^{3} \left ( ax+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{9\,a}{4\,{b}^{2} \left ( ax+b \right ) ^{2}}\sqrt{x}}-{\frac{15\,a}{4\,{b}^{3}}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-2\,{\frac{1}{{b}^{3}\sqrt{x}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.78374, size = 466, normalized size = 5.68 \begin{align*} \left [\frac{15 \,{\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{a x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - b}{a x + b}\right ) - 2 \,{\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt{x}}{8 \,{\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}, \frac{15 \,{\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt{x}}{4 \,{\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.09443, size = 80, normalized size = 0.98 \begin{align*} -\frac{15 \, a \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{3}} - \frac{2}{b^{3} \sqrt{x}} - \frac{7 \, a^{2} x^{\frac{3}{2}} + 9 \, a b \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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