Integrand size = 11, antiderivative size = 84 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}} \]
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Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {201, 223, 212} \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2} \]
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Rule 201
Rule 212
Rule 223
Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{6} (5 a) \int \left (a+b x^2\right )^{3/2} \, dx \\ & = \frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{8} \left (5 a^2\right ) \int \sqrt {a+b x^2} \, dx \\ & = \frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{16} \left (5 a^3\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx \\ & = \frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{16} \left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right ) \\ & = \frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {1}{48} \sqrt {a+b x^2} \left (33 a^2 x+26 a b x^3+8 b^2 x^5\right )-\frac {5 a^3 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 \sqrt {b}} \]
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Time = 2.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {x \left (8 b^{2} x^{4}+26 a b \,x^{2}+33 a^{2}\right ) \sqrt {b \,x^{2}+a}}{48}+\frac {5 a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 \sqrt {b}}\) | \(59\) |
pseudoelliptic | \(\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a^{3}}{16 \sqrt {b}}+\frac {11 \left (\frac {8 b^{\frac {5}{2}} x^{4}}{33}+\frac {26 x^{2} a \,b^{\frac {3}{2}}}{33}+\sqrt {b}\, a^{2}\right ) x \sqrt {b \,x^{2}+a}}{16 \sqrt {b}}\) | \(67\) |
default | \(\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\) | \(68\) |
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Time = 0.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.74 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} x^{5} + 26 \, a b^{2} x^{3} + 33 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{96 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} x^{5} + 26 \, a b^{2} x^{3} + 33 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{48 \, b}\right ] \]
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Time = 2.67 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.15 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {11 a^{\frac {5}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{16} + \frac {13 a^{\frac {3}{2}} b x^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{24} + \frac {\sqrt {a} b^{2} x^{5} \sqrt {1 + \frac {b x^{2}}{a}}}{6} + \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} \]
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Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x + \frac {5}{16} \, \sqrt {b x^{2} + a} a^{2} x + \frac {5 \, a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} \]
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Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75 \[ \int \left (a+b x^2\right )^{5/2} \, dx=-\frac {5 \, a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, \sqrt {b}} + \frac {1}{48} \, {\left (2 \, {\left (4 \, b^{2} x^{2} + 13 \, a b\right )} x^{2} + 33 \, a^{2}\right )} \sqrt {b x^{2} + a} x \]
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Time = 4.52 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \left (a+b x^2\right )^{5/2} \, dx=\frac {x\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}} \]
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