Optimal. Leaf size=126 \[ \frac{63}{8} a^2 b^2 \sqrt{a+b x^2}-\frac{63}{8} a^{5/2} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{63}{40} b^2 \left (a+b x^2\right )^{5/2}+\frac{21}{8} a b^2 \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2} \]
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Rubi [A] time = 0.0764493, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 50, 63, 208} \[ \frac{63}{8} a^2 b^2 \sqrt{a+b x^2}-\frac{63}{8} a^{5/2} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{63}{40} b^2 \left (a+b x^2\right )^{5/2}+\frac{21}{8} a b^2 \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2} \]
Antiderivative was successfully verified.
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Rule 266
Rule 47
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\left (a+b x^2\right )^{9/2}}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{9/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}+\frac{1}{8} (9 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{7/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}+\frac{1}{16} \left (63 b^2\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac{63}{40} b^2 \left (a+b x^2\right )^{5/2}-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}+\frac{1}{16} \left (63 a b^2\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{21}{8} a b^2 \left (a+b x^2\right )^{3/2}+\frac{63}{40} b^2 \left (a+b x^2\right )^{5/2}-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}+\frac{1}{16} \left (63 a^2 b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{63}{8} a^2 b^2 \sqrt{a+b x^2}+\frac{21}{8} a b^2 \left (a+b x^2\right )^{3/2}+\frac{63}{40} b^2 \left (a+b x^2\right )^{5/2}-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}+\frac{1}{16} \left (63 a^3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{63}{8} a^2 b^2 \sqrt{a+b x^2}+\frac{21}{8} a b^2 \left (a+b x^2\right )^{3/2}+\frac{63}{40} b^2 \left (a+b x^2\right )^{5/2}-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}+\frac{1}{8} \left (63 a^3 b\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{63}{8} a^2 b^2 \sqrt{a+b x^2}+\frac{21}{8} a b^2 \left (a+b x^2\right )^{3/2}+\frac{63}{40} b^2 \left (a+b x^2\right )^{5/2}-\frac{9 b \left (a+b x^2\right )^{7/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{9/2}}{4 x^4}-\frac{63}{8} a^{5/2} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}
Mathematica [C] time = 0.0120723, size = 39, normalized size = 0.31 \[ -\frac{b^2 \left (a+b x^2\right )^{11/2} \, _2F_1\left (3,\frac{11}{2};\frac{13}{2};\frac{b x^2}{a}+1\right )}{11 a^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.007, size = 148, normalized size = 1.2 \begin{align*} -{\frac{1}{4\,a{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{7\,b}{8\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}+{\frac{7\,{b}^{2}}{8\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{9}{2}}}}+{\frac{9\,{b}^{2}}{8\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{63\,{b}^{2}}{40} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{21\,a{b}^{2}}{8} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{63\,{b}^{2}}{8}{a}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{63\,{b}^{2}{a}^{2}}{8}\sqrt{b{x}^{2}+a}} \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 = 1.65255, size = 451, normalized size = 3.58 \begin{align*} \left [\frac{315 \, a^{\frac{5}{2}} b^{2} x^{4} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (8 \, b^{4} x^{8} + 56 \, a b^{3} x^{6} + 288 \, a^{2} b^{2} x^{4} - 85 \, a^{3} b x^{2} - 10 \, a^{4}\right )} \sqrt{b x^{2} + a}}{80 \, x^{4}}, \frac{315 \, \sqrt{-a} a^{2} b^{2} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (8 \, b^{4} x^{8} + 56 \, a b^{3} x^{6} + 288 \, a^{2} b^{2} x^{4} - 85 \, a^{3} b x^{2} - 10 \, a^{4}\right )} \sqrt{b x^{2} + a}}{40 \, x^{4}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 8.48189, size = 175, normalized size = 1.39 \begin{align*} - \frac{63 a^{\frac{5}{2}} b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8} - \frac{a^{5}}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{19 a^{4} \sqrt{b}}{8 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{203 a^{3} b^{\frac{3}{2}}}{40 x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{43 a^{2} b^{\frac{5}{2}} x}{5 \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{8 a b^{\frac{7}{2}} x^{3}}{5 \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{b^{\frac{9}{2}} x^{5}}{5 \sqrt{\frac{a}{b x^{2}} + 1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.82305, size = 143, normalized size = 1.13 \begin{align*} \frac{1}{40} \,{\left (\frac{315 \, a^{3} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 8 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} + 40 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a + 240 \, \sqrt{b x^{2} + a} a^{2} - \frac{5 \,{\left (17 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3} - 15 \, \sqrt{b x^{2} + a} a^{4}\right )}}{b^{2} x^{4}}\right )} b^{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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