Integrand size = 24, antiderivative size = 132 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=a^2 c^2 \sqrt {c+d x^2}+\frac {1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}-a^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 90, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=-a^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+a^2 c^2 \sqrt {c+d x^2}+\frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}+\frac {1}{3} a^2 c \left (c+d x^2\right )^{3/2}-\frac {b \left (c+d x^2\right )^{7/2} (b c-2 a d)}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2} \]
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Rule 52
Rule 65
Rule 90
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)^{5/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {b (b c-2 a d) (c+d x)^{5/2}}{d}+\frac {a^2 (c+d x)^{5/2}}{x}+\frac {b^2 (c+d x)^{7/2}}{d}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac {1}{2} a^2 \text {Subst}\left (\int \frac {(c+d x)^{5/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac {1}{2} \left (a^2 c\right ) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac {1}{2} \left (a^2 c^2\right ) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right ) \\ & = a^2 c^2 \sqrt {c+d x^2}+\frac {1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac {1}{2} \left (a^2 c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = a^2 c^2 \sqrt {c+d x^2}+\frac {1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}+\frac {\left (a^2 c^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d} \\ & = a^2 c^2 \sqrt {c+d x^2}+\frac {1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}-a^2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {\sqrt {c+d x^2} \left (90 a b d \left (c+d x^2\right )^3-5 b^2 \left (2 c-7 d x^2\right ) \left (c+d x^2\right )^3+21 a^2 d^2 \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{315 d^2}-a^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
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Time = 2.94 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95
method | result | size |
default | \(b^{2} \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{9 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{63 d^{2}}\right )+a^{2} \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )+\frac {2 a b \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{7 d}\) | \(125\) |
pseudoelliptic | \(\frac {-15 a^{2} c^{\frac {5}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+23 \left (\frac {3 x^{4} \left (\frac {5}{9} b^{2} x^{4}+\frac {10}{7} a b \,x^{2}+a^{2}\right ) d^{4}}{23}+\frac {11 x^{2} \left (\frac {95}{231} b^{2} x^{4}+\frac {90}{77} a b \,x^{2}+a^{2}\right ) c \,d^{3}}{23}+c^{2} \left (\frac {25}{161} b^{2} x^{4}+\frac {90}{161} a b \,x^{2}+a^{2}\right ) d^{2}+\frac {30 \left (\frac {b \,x^{2}}{18}+a \right ) b \,c^{3} d}{161}-\frac {10 b^{2} c^{4}}{483}\right ) \sqrt {d \,x^{2}+c}}{15 d^{2}}\) | \(148\) |
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Time = 0.27 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.73 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\left [\frac {315 \, a^{2} c^{\frac {5}{2}} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (35 \, b^{2} d^{4} x^{8} + 5 \, {\left (19 \, b^{2} c d^{3} + 18 \, a b d^{4}\right )} x^{6} - 10 \, b^{2} c^{4} + 90 \, a b c^{3} d + 483 \, a^{2} c^{2} d^{2} + 3 \, {\left (25 \, b^{2} c^{2} d^{2} + 90 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{4} + {\left (5 \, b^{2} c^{3} d + 270 \, a b c^{2} d^{2} + 231 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{630 \, d^{2}}, \frac {315 \, a^{2} \sqrt {-c} c^{2} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (35 \, b^{2} d^{4} x^{8} + 5 \, {\left (19 \, b^{2} c d^{3} + 18 \, a b d^{4}\right )} x^{6} - 10 \, b^{2} c^{4} + 90 \, a b c^{3} d + 483 \, a^{2} c^{2} d^{2} + 3 \, {\left (25 \, b^{2} c^{2} d^{2} + 90 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{4} + {\left (5 \, b^{2} c^{3} d + 270 \, a b c^{2} d^{2} + 231 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{315 \, d^{2}}\right ] \]
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Time = 22.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {\begin {cases} \frac {2 a^{2} c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a^{2} c^{2} \sqrt {c + d x^{2}} + \frac {2 a^{2} c \left (c + d x^{2}\right )^{\frac {3}{2}}}{3} + \frac {2 a^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}{5} + \frac {2 b^{2} \left (c + d x^{2}\right )^{\frac {9}{2}}}{9 d^{2}} + \frac {2 \left (c + d x^{2}\right )^{\frac {7}{2}} \cdot \left (2 a b d - b^{2} c\right )}{7 d^{2}} & \text {for}\: d \neq 0 \\a^{2} c^{\frac {5}{2}} \log {\left (x^{2} \right )} + 2 a b c^{\frac {5}{2}} x^{2} + \frac {b^{2} c^{\frac {5}{2}} x^{4}}{2} & \text {otherwise} \end {cases}}{2} \]
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Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x^{2}}{9 \, d} - a^{2} c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c + \sqrt {d x^{2} + c} a^{2} c^{2} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c}{63 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{7 \, d} \]
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Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {a^{2} c^{3} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {35 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} b^{2} d^{16} - 45 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c d^{16} + 90 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b d^{17} + 63 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{18} + 105 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{18} + 315 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{18}}{315 \, d^{18}} \]
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Time = 5.36 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.89 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx={\left (d\,x^2+c\right )}^{5/2}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{5\,d^2}-\frac {c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )}{5}\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{7\,d^2}-\frac {b^2\,c}{7\,d^2}\right )\,{\left (d\,x^2+c\right )}^{7/2}+c^2\,\sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{9/2}}{9\,d^2}+\frac {c\,{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )}{3}+a^2\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i} \]
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