Integrand size = 24, antiderivative size = 136 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=\frac {1}{2} a (4 b c+3 a d) \sqrt {c+d x^2}+\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}-\frac {1}{2} a \sqrt {c} (4 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 91, 81, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}-\frac {1}{2} a \sqrt {c} (3 a d+4 b c) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {a \left (c+d x^2\right )^{3/2} (3 a d+4 b c)}{6 c}+\frac {1}{2} a \sqrt {c+d x^2} (3 a d+4 b c)+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d} \]
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Rule 52
Rule 65
Rule 81
Rule 91
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2 (c+d x)^{3/2}}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {\text {Subst}\left (\int \frac {\left (\frac {1}{2} a (4 b c+3 a d)+b^2 c x\right ) (c+d x)^{3/2}}{x} \, dx,x,x^2\right )}{2 c} \\ & = \frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {(a (4 b c+3 a d)) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right )}{4 c} \\ & = \frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {1}{4} (a (4 b c+3 a d)) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} a (4 b c+3 a d) \sqrt {c+d x^2}+\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {1}{4} (a c (4 b c+3 a d)) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} a (4 b c+3 a d) \sqrt {c+d x^2}+\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}+\frac {(a c (4 b c+3 a d)) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 d} \\ & = \frac {1}{2} a (4 b c+3 a d) \sqrt {c+d x^2}+\frac {a (4 b c+3 a d) \left (c+d x^2\right )^{3/2}}{6 c}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{2 c x^2}-\frac {1}{2} a \sqrt {c} (4 b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {c+d x^2} \left (-15 a^2 d \left (c-2 d x^2\right )+6 b^2 x^2 \left (c+d x^2\right )^2+20 a b d x^2 \left (4 c+d x^2\right )\right )}{30 d x^2}-\frac {1}{2} a \sqrt {c} (4 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 2.92 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.86
method | result | size |
pseudoelliptic | \(\frac {-\frac {3 x^{2} \left (a d +\frac {4 b c}{3}\right ) d c a \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )}{2}+\left (-\frac {d \left (-\frac {4}{5} b^{2} x^{4}-\frac {16}{3} a b \,x^{2}+a^{2}\right ) c^{\frac {3}{2}}}{2}+x^{2} \left (\frac {b^{2} c^{\frac {5}{2}}}{5}+d^{2} \sqrt {c}\, \left (\frac {1}{5} b^{2} x^{4}+\frac {2}{3} a b \,x^{2}+a^{2}\right )\right )\right ) \sqrt {d \,x^{2}+c}}{\sqrt {c}\, d \,x^{2}}\) | \(117\) |
default | \(\frac {b^{2} \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5 d}+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{2 c \,x^{2}}+\frac {3 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )}{2 c}\right )+2 a b \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\) | \(155\) |
risch | \(-\frac {c \,a^{2} \sqrt {d \,x^{2}+c}}{2 x^{2}}+\frac {\sqrt {d \,x^{2}+c}\, b^{2} d \,x^{4}}{5}+\frac {2 b^{2} c \,x^{2} \sqrt {d \,x^{2}+c}}{5}+\frac {b^{2} c^{2} \sqrt {d \,x^{2}+c}}{5 d}+\sqrt {d \,x^{2}+c}\, a^{2} d +\frac {2 a b d \,x^{2} \sqrt {d \,x^{2}+c}}{3}+\frac {8 a b c \sqrt {d \,x^{2}+c}}{3}-\frac {3 \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a^{2} d}{2}-2 c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a b\) | \(183\) |
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Time = 0.27 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.96 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=\left [\frac {15 \, {\left (4 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (6 \, b^{2} d^{2} x^{6} + 4 \, {\left (3 \, b^{2} c d + 5 \, a b d^{2}\right )} x^{4} - 15 \, a^{2} c d + 2 \, {\left (3 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, d x^{2}}, \frac {15 \, {\left (4 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (6 \, b^{2} d^{2} x^{6} + 4 \, {\left (3 \, b^{2} c d + 5 \, a b d^{2}\right )} x^{4} - 15 \, a^{2} c d + 2 \, {\left (3 \, b^{2} c^{2} + 40 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, d x^{2}}\right ] \]
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Time = 21.31 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.48 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=- \frac {3 a^{2} \sqrt {c} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2} - \frac {a^{2} c \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} + \frac {a^{2} c \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a^{2} d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} - 2 a b c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {2 a b c^{2}}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b c \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + 2 a b d \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + b^{2} c \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + b^{2} d \left (\begin {cases} - \frac {2 c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {x^{4} \sqrt {c + d x^{2}}}{5} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \]
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Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=-2 \, a b c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {3}{2} \, a^{2} \sqrt {c} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {2}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b + 2 \, \sqrt {d x^{2} + c} a b c + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2}}{5 \, d} + \frac {3}{2} \, \sqrt {d x^{2} + c} a^{2} d + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{2 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{2 \, c x^{2}} \]
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Time = 0.30 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=\frac {6 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} + 20 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d + 60 \, \sqrt {d x^{2} + c} a b c d + 30 \, \sqrt {d x^{2} + c} a^{2} d^{2} - \frac {15 \, \sqrt {d x^{2} + c} a^{2} c d}{x^{2}} + \frac {15 \, {\left (4 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}}}{30 \, d} \]
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Time = 5.76 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^3} \, dx=\frac {b^2\,{\left (d\,x^2+c\right )}^{5/2}}{5\,d}-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{3\,d}-\frac {2\,b^2\,c}{3\,d}\right )\,{\left (d\,x^2+c\right )}^{3/2}-\sqrt {d\,x^2+c}\,\left (2\,c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )-\frac {{\left (a\,d-b\,c\right )}^2}{d}+\frac {b^2\,c^2}{d}\right )-\frac {a^2\,c\,\sqrt {d\,x^2+c}}{2\,x^2}+2\,a\,\mathrm {atan}\left (\frac {2\,a\,\sqrt {d\,x^2+c}\,\left (3\,a\,d+4\,b\,c\right )\,\sqrt {-\frac {c}{16}}}{\frac {3\,d\,a^2\,c}{2}+2\,b\,a\,c^2}\right )\,\left (3\,a\,d+4\,b\,c\right )\,\sqrt {-\frac {c}{16}} \]
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