Integrand size = 24, antiderivative size = 181 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=\frac {\left (8 b^2 c^2+3 a d (8 b c+a d)\right ) \sqrt {c+d x^2}}{8 c}+\frac {\left (8 b^2 c^2+3 a d (8 b c+a d)\right ) \left (c+d x^2\right )^{3/2}}{24 c^2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}-\frac {\left (8 b^2 c^2+3 a d (8 b c+a d)\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}} \]
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Time = 0.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 91, 79, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {\left (3 a d (a d+8 b c)+8 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}}+\frac {1}{24} \left (c+d x^2\right )^{3/2} \left (\frac {3 a d (a d+8 b c)}{c^2}+8 b^2\right )+\frac {\sqrt {c+d x^2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{8 c}-\frac {a \left (c+d x^2\right )^{5/2} (a d+8 b c)}{8 c^2 x^2} \]
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Rule 52
Rule 65
Rule 79
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^3} \, dx,x,x^2\right ) \\ & = -\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac {\text {Subst}\left (\int \frac {\left (\frac {1}{2} a (8 b c+a d)+2 b^2 c x\right ) (c+d x)^{3/2}}{x^2} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {1}{16} \left (c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{8} c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {1}{8} c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {\left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{8 d} \\ & = \frac {1}{8} c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}-\frac {\left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.64 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=\frac {1}{24} \left (\frac {\sqrt {c+d x^2} \left (-24 a b x^2 \left (c-2 d x^2\right )+8 b^2 x^4 \left (4 c+d x^2\right )-3 a^2 \left (2 c+5 d x^2\right )\right )}{x^4}-\frac {3 \left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}\right ) \]
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Time = 2.91 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.62
method | result | size |
pseudoelliptic | \(-\frac {3 \left (x^{4} \left (a^{2} d^{2}+8 a b c d +\frac {8}{3} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+\frac {5 \sqrt {d \,x^{2}+c}\, \left (\frac {2 \left (-\frac {16}{3} b^{2} x^{4}+4 a b \,x^{2}+a^{2}\right ) c^{\frac {3}{2}}}{5}+d \,x^{2} \sqrt {c}\, \left (-\frac {8}{15} b^{2} x^{4}-\frac {16}{5} a b \,x^{2}+a^{2}\right )\right )}{3}\right )}{8 \sqrt {c}\, x^{4}}\) | \(113\) |
risch | \(-\frac {\sqrt {d \,x^{2}+c}\, a \left (5 a d \,x^{2}+8 c b \,x^{2}+2 a c \right )}{8 x^{4}}+b^{2} d^{2} \left (\frac {x^{2} \sqrt {d \,x^{2}+c}}{3 d}-\frac {2 c \sqrt {d \,x^{2}+c}}{3 d^{2}}\right )+2 a b d \sqrt {d \,x^{2}+c}+2 b^{2} c \sqrt {d \,x^{2}+c}-\frac {\left (3 a^{2} d^{2}+24 a b c d +8 b^{2} c^{2}\right ) \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{8 \sqrt {c}}\) | \(156\) |
default | \(b^{2} \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{4 c \,x^{4}}+\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{2 c \,x^{2}}+\frac {3 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )}{2 c}\right )}{4 c}\right )+2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{2 c \,x^{2}}+\frac {3 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )}{2 c}\right )\) | \(242\) |
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Time = 0.27 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=\left [\frac {3 \, {\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {c} x^{4} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (8 \, b^{2} c d x^{6} + 16 \, {\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \, {\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c x^{4}}, \frac {3 \, {\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} c d x^{6} + 16 \, {\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \, {\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, c x^{4}}\right ] \]
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Time = 59.38 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.92 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=- \frac {a^{2} c^{2}}{4 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a^{2} c \sqrt {d}}{8 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} - \frac {a^{2} d^{\frac {3}{2}}}{8 x \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a^{2} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{8 \sqrt {c}} - 3 a b \sqrt {c} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} - \frac {a b c \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{x} + \frac {2 a b c \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} - b^{2} c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {b^{2} c^{2}}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {b^{2} c \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + b^{2} d \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=-b^{2} c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - 3 \, a b \sqrt {c} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, \sqrt {c}} + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} + \sqrt {d x^{2} + c} b^{2} c + 3 \, \sqrt {d x^{2} + c} a b d + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d}{c} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{8 \, c^{2}} + \frac {3 \, \sqrt {d x^{2} + c} a^{2} d^{2}}{8 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b}{c x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d}{8 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{4 \, c x^{4}} \]
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Time = 0.31 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=\frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d + 24 \, \sqrt {d x^{2} + c} b^{2} c d + 48 \, \sqrt {d x^{2} + c} a b d^{2} + \frac {3 \, {\left (8 \, b^{2} c^{2} d + 24 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {3 \, {\left (8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} - 8 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} - 3 \, \sqrt {d x^{2} + c} a^{2} c d^{3}\right )}}{d^{2} x^{4}}}{24 \, d} \]
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Time = 6.16 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx=\frac {\sqrt {d\,x^2+c}\,\left (\frac {3\,a^2\,c\,d^2}{8}+b\,a\,c^2\,d\right )-\left (\frac {5\,a^2\,d^2}{8}+b\,c\,a\,d\right )\,{\left (d\,x^2+c\right )}^{3/2}}{{\left (d\,x^2+c\right )}^2-2\,c\,\left (d\,x^2+c\right )+c^2}+\sqrt {d\,x^2+c}\,\left (c\,b^2+2\,a\,d\,b\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{3/2}}{3}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,\left (3\,a^2\,d^2+24\,a\,b\,c\,d+8\,b^2\,c^2\right )\,1{}\mathrm {i}}{4\,\sqrt {c}\,\left (\frac {3\,a^2\,d^2}{4}+6\,a\,b\,c\,d+2\,b^2\,c^2\right )}\right )\,\left (3\,a^2\,d^2+24\,a\,b\,c\,d+8\,b^2\,c^2\right )\,1{}\mathrm {i}}{8\,\sqrt {c}} \]
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