Integrand size = 24, antiderivative size = 133 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=-\frac {\left (b^2 c^2-8 a d (b c+a d)\right ) x \sqrt {c+d x^2}}{8 c d}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {\left (b^2 c^2-8 a d (b c+a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2}} \]
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Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {473, 396, 201, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}-\frac {\left (b^2 c^2-8 a d (a d+b c)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2}}-\frac {1}{8} x \sqrt {c+d x^2} \left (\frac {b^2 c}{d}-\frac {8 a (a d+b c)}{c}\right )+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d} \]
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Rule 201
Rule 212
Rule 223
Rule 396
Rule 473
Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {\int \left (2 a (b c+a d)+b^2 c x^2\right ) \sqrt {c+d x^2} \, dx}{c} \\ & = -\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {1}{4} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) \int \sqrt {c+d x^2} \, dx \\ & = -\frac {1}{8} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) x \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {1}{8} \left (\frac {b^2 c^2}{d}-8 a (b c+a d)\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx \\ & = -\frac {1}{8} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) x \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {1}{8} \left (\frac {b^2 c^2}{d}-8 a (b c+a d)\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right ) \\ & = -\frac {1}{8} \left (\frac {b^2 c}{d}-\frac {8 a (b c+a d)}{c}\right ) x \sqrt {c+d x^2}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {\left (\frac {b^2 c^2}{d}-8 a (b c+a d)\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 \sqrt {d}} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=\frac {\sqrt {c+d x^2} \left (-8 a^2 d+b^2 c x^2+8 a b d x^2+2 b^2 d x^4\right )}{8 d x}+\frac {\left (-b^2 c^2+8 a b c d+8 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{4 d^{3/2}} \]
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Time = 2.92 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {\sqrt {d \,x^{2}+c}\, \left (-2 b^{2} d \,x^{4}-8 x^{2} a b d -b^{2} c \,x^{2}+8 a^{2} d \right )}{8 d x}+\frac {\left (8 a^{2} d^{2}+8 a b c d -b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {3}{2}}}\) | \(96\) |
pseudoelliptic | \(-\frac {-x \left (a^{2} d^{2}+a b c d -\frac {1}{8} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+\sqrt {d \,x^{2}+c}\, \left (\left (-\frac {1}{4} b^{2} x^{4}-a b \,x^{2}+a^{2}\right ) d^{\frac {3}{2}}-\frac {\sqrt {d}\, b^{2} c \,x^{2}}{8}\right )}{d^{\frac {3}{2}} x}\) | \(97\) |
default | \(b^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{c x}+\frac {2 d \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{c}\right )+2 a b \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )\) | \(165\) |
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Time = 0.27 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=\left [-\frac {{\left (b^{2} c^{2} - 8 \, a b c d - 8 \, a^{2} d^{2}\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{4} - 8 \, a^{2} d^{2} + {\left (b^{2} c d + 8 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, d^{2} x}, \frac {{\left (b^{2} c^{2} - 8 \, a b c d - 8 \, a^{2} d^{2}\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, b^{2} d^{2} x^{4} - 8 \, a^{2} d^{2} + {\left (b^{2} c d + 8 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, d^{2} x}\right ] \]
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Time = 1.40 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.75 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=- \frac {a^{2} \sqrt {c}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + a^{2} \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {a^{2} d x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + 2 a b \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} - \frac {c^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d} + \frac {c x \sqrt {c + d x^{2}}}{8 d} + \frac {x^{3} \sqrt {c + d x^{2}}}{4} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=\sqrt {d x^{2} + c} a b x + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x}{4 \, d} - \frac {\sqrt {d x^{2} + c} b^{2} c x}{8 \, d} - \frac {b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {3}{2}}} + \frac {a b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} + a^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {\sqrt {d x^{2} + c} a^{2}}{x} \]
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Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=\frac {2 \, a^{2} c \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {1}{8} \, {\left (2 \, b^{2} x^{2} + \frac {b^{2} c d + 8 \, a b d^{2}}{d^{2}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (b^{2} c^{2} - 8 \, a b c d - 8 \, a^{2} d^{2}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{16 \, d^{\frac {3}{2}}} \]
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Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{x^2} \,d x \]
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