Integrand size = 20, antiderivative size = 104 \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=-\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {a^2 A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \]
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Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {847, 794, 201, 223, 212} \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=-\frac {a^2 A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}-\frac {\left (a+b x^2\right )^{3/2} (8 a B-15 A b x)}{60 b^2}-\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b} \]
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Rule 201
Rule 212
Rule 223
Rule 794
Rule 847
Rubi steps \begin{align*} \text {integral}& = \frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {\int x (-2 a B+5 A b x) \sqrt {a+b x^2} \, dx}{5 b} \\ & = \frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {(a A) \int \sqrt {a+b x^2} \, dx}{4 b} \\ & = -\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {\left (a^2 A\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b} \\ & = -\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {\left (a^2 A\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b} \\ & = -\frac {a A x \sqrt {a+b x^2}}{8 b}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {(8 a B-15 A b x) \left (a+b x^2\right )^{3/2}}{60 b^2}-\frac {a^2 A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-16 a^2 B+6 b^2 x^3 (5 A+4 B x)+a b x (15 A+8 B x)\right )+15 a^2 A \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{120 b^2} \]
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Time = 3.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {\left (24 b^{2} B \,x^{4}+30 A \,b^{2} x^{3}+8 B a b \,x^{2}+15 a A b x -16 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{120 b^{2}}-\frac {a^{2} A \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}\) | \(80\) |
default | \(B \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\) | \(96\) |
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Time = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.68 \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=\left [\frac {15 \, A a^{2} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (24 \, B b^{2} x^{4} + 30 \, A b^{2} x^{3} + 8 \, B a b x^{2} + 15 \, A a b x - 16 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}, \frac {15 \, A a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (24 \, B b^{2} x^{4} + 30 \, A b^{2} x^{3} + 8 \, B a b x^{2} + 15 \, A a b x - 16 \, B a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, b^{2}}\right ] \]
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Time = 0.42 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.17 \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=\begin {cases} - \frac {A a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \sqrt {a + b x^{2}} \left (\frac {A a x}{8 b} + \frac {A x^{3}}{4} - \frac {2 B a^{2}}{15 b^{2}} + \frac {B a x^{2}}{15 b} + \frac {B x^{4}}{5}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.83 \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{2}}{5 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x}{4 \, b} - \frac {\sqrt {b x^{2} + a} A a x}{8 \, b} - \frac {A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a}{15 \, b^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.78 \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=\frac {A a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} + \frac {1}{120} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, B x + 5 \, A\right )} x + \frac {4 \, B a}{b}\right )} x + \frac {15 \, A a}{b}\right )} x - \frac {16 \, B a^{2}}{b^{2}}\right )} \]
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Timed out. \[ \int x^2 (A+B x) \sqrt {a+b x^2} \, dx=\int x^2\,\sqrt {b\,x^2+a}\,\left (A+B\,x\right ) \,d x \]
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