Integrand size = 22, antiderivative size = 155 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=-\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {a^3 (8 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}} \]
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Time = 0.06 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {470, 285, 327, 223, 212} \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {a^3 (8 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}}-\frac {a^2 x \sqrt {a+b x^2} (8 A b-5 a B)}{128 b^3}+\frac {a x^3 \sqrt {a+b x^2} (8 A b-5 a B)}{192 b^2}+\frac {x^5 \sqrt {a+b x^2} (8 A b-5 a B)}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b} \]
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Rule 212
Rule 223
Rule 285
Rule 327
Rule 470
Rubi steps \begin{align*} \text {integral}& = \frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}-\frac {(-8 A b+5 a B) \int x^4 \sqrt {a+b x^2} \, dx}{8 b} \\ & = \frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {(a (8 A b-5 a B)) \int \frac {x^4}{\sqrt {a+b x^2}} \, dx}{48 b} \\ & = \frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}-\frac {\left (a^2 (8 A b-5 a B)\right ) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{64 b^2} \\ & = -\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {\left (a^3 (8 A b-5 a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{128 b^3} \\ & = -\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {\left (a^3 (8 A b-5 a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{128 b^3} \\ & = -\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {a^3 (8 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}} \\ \end{align*}
Time = 0.49 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (-24 a^2 A b+15 a^3 B+16 a A b^2 x^2-10 a^2 b B x^2+64 A b^3 x^4+8 a b^2 B x^4+48 b^3 B x^6\right )}{384 b^3}-\frac {a^3 (-8 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{64 b^{7/2}} \]
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Time = 2.82 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.70
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \left (A \,a^{3} b -\frac {5}{8} B \,a^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{2}+x \sqrt {b \,x^{2}+a}\, \left (-\frac {3 \left (\frac {5 x^{2} B}{12}+A \right ) a^{2} b^{\frac {3}{2}}}{2}+x^{2} a \left (\frac {x^{2} B}{2}+A \right ) b^{\frac {5}{2}}+\left (3 B \,x^{6}+4 A \,x^{4}\right ) b^{\frac {7}{2}}+\frac {15 B \,a^{3} \sqrt {b}}{16}\right )}{24 b^{\frac {7}{2}}}\) | \(109\) |
risch | \(-\frac {x \left (-48 b^{3} B \,x^{6}-64 A \,b^{3} x^{4}-8 B a \,b^{2} x^{4}-16 a A \,b^{2} x^{2}+10 B \,a^{2} b \,x^{2}+24 a^{2} b A -15 a^{3} B \right ) \sqrt {b \,x^{2}+a}}{384 b^{3}}+\frac {a^{3} \left (8 A b -5 B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {7}{2}}}\) | \(112\) |
default | \(B \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 b}-\frac {5 a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )}{8 b}\right )+A \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )\) | \(192\) |
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Time = 0.32 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.66 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\left [-\frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, B b^{4} x^{7} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{5} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, B b^{4} x^{7} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{5} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{4}}\right ] \]
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Time = 0.41 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.13 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\begin {cases} \frac {3 a^{2} \left (A a - \frac {5 a \left (A b + \frac {B a}{8}\right )}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {B x^{7}}{8} - \frac {3 a x \left (A a - \frac {5 a \left (A b + \frac {B a}{8}\right )}{6 b}\right )}{8 b^{2}} + \frac {x^{5} \left (A b + \frac {B a}{8}\right )}{6 b} + \frac {x^{3} \left (A a - \frac {5 a \left (A b + \frac {B a}{8}\right )}{6 b}\right )}{4 b}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{5}}{5} + \frac {B x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.07 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{5}}{8 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x^{3}}{48 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{3}}{6 \, b} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{64 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A a^{2} x}{16 \, b^{2}} - \frac {5 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} \]
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Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B x^{2} + \frac {B a b^{5} + 8 \, A b^{6}}{b^{6}}\right )} x^{2} - \frac {5 \, B a^{2} b^{4} - 8 \, A a b^{5}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{3} b^{3} - 8 \, A a^{2} b^{4}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {7}{2}}} \]
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Timed out. \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\int x^4\,\left (B\,x^2+A\right )\,\sqrt {b\,x^2+a} \,d x \]
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