Integrand size = 22, antiderivative size = 89 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {a (4 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \]
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Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {470, 327, 223, 212} \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a (4 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}}+\frac {x \sqrt {a+b x^2} (4 A b-3 a B)}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b} \]
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Rule 212
Rule 223
Rule 327
Rule 470
Rubi steps \begin{align*} \text {integral}& = \frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {(-4 A b+3 a B) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{4 b} \\ & = \frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {(a (4 A b-3 a B)) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b^2} \\ & = \frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {(a (4 A b-3 a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b^2} \\ & = \frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {a (4 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (4 A b-3 a B+2 b B x^2\right )}{8 b^2}+\frac {a (-4 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{4 b^{5/2}} \]
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Time = 2.83 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {x \left (2 b B \,x^{2}+4 A b -3 B a \right ) \sqrt {b \,x^{2}+a}}{8 b^{2}}-\frac {a \left (4 A b -3 B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\) | \(64\) |
pseudoelliptic | \(\frac {\left (-a b A +\frac {3}{4} a^{2} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\left (\left (\frac {x^{2} B}{2}+A \right ) b^{\frac {3}{2}}-\frac {3 B \sqrt {b}\, a}{4}\right ) x \sqrt {b \,x^{2}+a}}{2 b^{\frac {5}{2}}}\) | \(68\) |
default | \(B \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+A \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\) | \(106\) |
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Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.82 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\left [-\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, B b^{2} x^{3} - {\left (3 \, B a b - 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, -\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b^{2} x^{3} - {\left (3 \, B a b - 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \]
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Time = 0.33 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.22 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {a \left (A - \frac {3 B a}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {B x^{3}}{4 b} + \frac {x \left (A - \frac {3 B a}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B x^{3}}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x}{2 \, b} + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} \]
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Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{8} \, \sqrt {b x^{2} + a} {\left (\frac {2 \, B x^{2}}{b} - \frac {3 \, B a b - 4 \, A b^{2}}{b^{3}}\right )} x - \frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \]
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Timed out. \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {x^2\,\left (B\,x^2+A\right )}{\sqrt {b\,x^2+a}} \,d x \]
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