Integrand size = 24, antiderivative size = 191 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {c \left (16 a^2 d^2+b c (5 b c-16 a d)\right ) x \sqrt {c+d x^2}}{128 d^3}+\frac {\left (16 a^2 d^2+b c (5 b c-16 a d)\right ) x^3 \sqrt {c+d x^2}}{64 d^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {c^2 \left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{7/2}} \]
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Time = 0.13 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {475, 470, 285, 327, 223, 212} \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=-\frac {c^2 \left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{7/2}}+\frac {1}{64} x^3 \sqrt {c+d x^2} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )+\frac {c x \sqrt {c+d x^2} \left (16 a^2 d^2+b c (5 b c-16 a d)\right )}{128 d^3}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d} \]
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Rule 212
Rule 223
Rule 285
Rule 327
Rule 470
Rule 475
Rubi steps \begin{align*} \text {integral}& = \frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}+\frac {\int x^2 \sqrt {c+d x^2} \left (8 a^2 d-b (5 b c-16 a d) x^2\right ) \, dx}{8 d} \\ & = -\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}+\frac {1}{16} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) \int x^2 \sqrt {c+d x^2} \, dx \\ & = \frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}+\frac {1}{64} \left (c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx \\ & = \frac {c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d}+\frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {\left (c^2 \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{128 d} \\ & = \frac {c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d}+\frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {\left (c^2 \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right )\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{128 d} \\ & = \frac {c \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x \sqrt {c+d x^2}}{128 d}+\frac {1}{64} \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) x^3 \sqrt {c+d x^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {c^2 \left (16 a^2+\frac {b c (5 b c-16 a d)}{d^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{3/2}} \\ \end{align*}
Time = 0.59 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.86 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (48 a^2 d^2 \left (c+2 d x^2\right )+16 a b d \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )+b^2 \left (15 c^3-10 c^2 d x^2+8 c d^2 x^4+48 d^3 x^6\right )\right )+6 c^2 \left (5 b^2 c^2-16 a b c d+16 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+d x^2}}\right )}{384 d^{7/2}} \]
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Time = 3.01 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.74
method | result | size |
pseudoelliptic | \(\frac {\left (-a^{2} c^{2} d^{2}+a b \,c^{3} d -\frac {5}{16} b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x \sqrt {d \,x^{2}+c}\, \left (c \left (\frac {1}{6} b^{2} x^{4}+\frac {2}{3} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\left (b^{2} x^{6}+\frac {8}{3} a b \,x^{4}+2 a^{2} x^{2}\right ) d^{\frac {7}{2}}-\left (\left (\frac {5 b \,x^{2}}{24}+a \right ) d^{\frac {3}{2}}-\frac {5 b \sqrt {d}\, c}{16}\right ) b \,c^{2}\right )}{8 d^{\frac {7}{2}}}\) | \(141\) |
risch | \(\frac {x \left (48 b^{2} d^{3} x^{6}+128 a b \,d^{3} x^{4}+8 b^{2} c \,d^{2} x^{4}+96 a^{2} d^{3} x^{2}+32 a b c \,d^{2} x^{2}-10 b^{2} c^{2} d \,x^{2}+48 c \,a^{2} d^{2}-48 a b \,c^{2} d +15 b^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{384 d^{3}}-\frac {c^{2} \left (16 a^{2} d^{2}-16 a b c d +5 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {7}{2}}}\) | \(157\) |
default | \(b^{2} \left (\frac {x^{5} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{8 d}-\frac {5 c \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )}{2 d}\right )}{8 d}\right )+a^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )+2 a b \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )}{2 d}\right )\) | \(257\) |
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Time = 0.30 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.79 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} - 2 \, {\left (5 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} - 48 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{768 \, d^{4}}, \frac {3 \, {\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} - 2 \, {\left (5 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} - 48 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{384 \, d^{4}}\right ] \]
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Time = 0.38 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.31 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\begin {cases} - \frac {c \left (a^{2} c - \frac {3 c \left (a^{2} d + 2 a b c - \frac {5 c \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d}\right )}{4 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2 d} + \sqrt {c + d x^{2}} \left (\frac {b^{2} x^{7}}{8} + \frac {x^{5} \cdot \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d} + \frac {x^{3} \left (a^{2} d + 2 a b c - \frac {5 c \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d}\right )}{4 d} + \frac {x \left (a^{2} c - \frac {3 c \left (a^{2} d + 2 a b c - \frac {5 c \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d}\right )}{4 d}\right )}{2 d}\right ) & \text {for}\: d \neq 0 \\\sqrt {c} \left (\frac {a^{2} x^{3}}{3} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.24 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{5}}{8 \, d} - \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x^{3}}{48 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x^{3}}{3 \, d} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x}{64 \, d^{3}} - \frac {5 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x}{4 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b c^{2} x}{8 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} x}{4 \, d} - \frac {\sqrt {d x^{2} + c} a^{2} c x}{8 \, d} - \frac {5 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {7}{2}}} + \frac {a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {5}{2}}} - \frac {a^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {3}{2}}} \]
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Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} x^{2} + \frac {b^{2} c d^{5} + 16 \, a b d^{6}}{d^{6}}\right )} x^{2} - \frac {5 \, b^{2} c^{2} d^{4} - 16 \, a b c d^{5} - 48 \, a^{2} d^{6}}{d^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{3} d^{3} - 16 \, a b c^{2} d^{4} + 16 \, a^{2} c d^{5}\right )}}{d^{6}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{128 \, d^{\frac {7}{2}}} \]
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Timed out. \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\int x^2\,{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c} \,d x \]
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