Integrand size = 24, antiderivative size = 131 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=\frac {4 a b-\frac {2 b^2 c}{d}-\frac {5 a^2 d}{c}}{6 c \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}-\frac {a (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}} \]
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Time = 0.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 91, 79, 53, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {5 a^2 d^2-4 a b c d+2 b^2 c^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}-\frac {a (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}} \]
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Rule 53
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}{x^2 (c+d x)^{5/2}} \, dx,x,x^2\right ) \\ & = -\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (4 b c-5 a d)+b^2 c x}{x (c+d x)^{5/2}} \, dx,x,x^2\right )}{2 c} \\ & = -\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {(a (4 b c-5 a d)) \text {Subst}\left (\int \frac {1}{x (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 c^2} \\ & = -\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}+\frac {(a (4 b c-5 a d)) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c^3} \\ & = -\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}+\frac {(a (4 b c-5 a d)) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 c^3 d} \\ & = -\frac {2 b^2 c^2-4 a b c d+5 a^2 d^2}{6 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {a^2}{2 c x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (4 b c-5 a d)}{2 c^3 \sqrt {c+d x^2}}-\frac {a (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=\frac {-2 b^2 c^3 x^2+4 a b c d x^2 \left (4 c+3 d x^2\right )-a^2 d \left (3 c^2+20 c d x^2+15 d^2 x^4\right )}{6 c^3 d x^2 \left (c+d x^2\right )^{3/2}}+\frac {a (-4 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 c^{7/2}} \]
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Time = 2.98 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.91
method | result | size |
pseudoelliptic | \(-\frac {-5 x^{2} \left (a d -\frac {4 b c}{5}\right ) \left (d \,x^{2}+c \right )^{\frac {3}{2}} d a \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )+\frac {20 x^{2} d^{2} \left (-\frac {3 b \,x^{2}}{5}+a \right ) a \,c^{\frac {3}{2}}}{3}+a d \left (-\frac {16 b \,x^{2}}{3}+a \right ) c^{\frac {5}{2}}+5 \sqrt {c}\, a^{2} d^{3} x^{4}+\frac {2 c^{\frac {7}{2}} b^{2} x^{2}}{3}}{2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{\frac {7}{2}} d \,x^{2}}\) | \(119\) |
default | \(-\frac {b^{2}}{3 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+a^{2} \left (-\frac {1}{2 c \,x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {5 d \left (\frac {1}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}}{c}\right )}{2 c}\right )+2 a b \left (\frac {1}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}}{c}\right )\) | \(173\) |
risch | \(-\frac {a^{2} \sqrt {d \,x^{2}+c}}{2 c^{3} x^{2}}+\frac {5 a^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) d}{2 c^{\frac {7}{2}}}-\frac {2 a \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) b}{c^{\frac {5}{2}}}-\frac {13 d \sqrt {d \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}+2 \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}\, a^{2}}{12 c^{3} \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}+\frac {7 \sqrt {d \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}+2 \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}\, a b}{6 c^{2} \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}+\frac {13 d \sqrt {d \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}-2 \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}\, a^{2}}{12 c^{3} \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}-\frac {7 \sqrt {d \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}-2 \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}\, a b}{6 c^{2} \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}+\frac {\sqrt {d \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}-2 \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}\, a^{2}}{12 c^{3} \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}}-\frac {\sqrt {d \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}-2 \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}\, a b}{6 c^{2} d \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}}+\frac {\sqrt {d \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}-2 \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}\, b^{2}}{12 c \,d^{2} \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}}+\frac {\sqrt {d \left (x +\frac {\sqrt {-c d}}{d}\right )^{2}-2 \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}\, b^{2}}{12 c d \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}+\frac {\sqrt {d \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}+2 \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}\, a^{2}}{12 c^{3} \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}}-\frac {\sqrt {d \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}+2 \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}\, a b}{6 c^{2} d \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}}+\frac {\sqrt {d \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}+2 \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}\, b^{2}}{12 c \,d^{2} \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}}-\frac {\sqrt {d \left (x -\frac {\sqrt {-c d}}{d}\right )^{2}+2 \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}\, b^{2}}{12 c d \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}\) | \(886\) |
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Time = 0.26 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.25 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (4 \, a b c d^{3} - 5 \, a^{2} d^{4}\right )} x^{6} + 2 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + {\left (4 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (3 \, a^{2} c^{3} d - 3 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} - 8 \, a b c^{3} d + 10 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, {\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}}, \frac {3 \, {\left ({\left (4 \, a b c d^{3} - 5 \, a^{2} d^{4}\right )} x^{6} + 2 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + {\left (4 \, a b c^{3} d - 5 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (3 \, a^{2} c^{3} d - 3 \, {\left (4 \, a b c^{2} d^{2} - 5 \, a^{2} c d^{3}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} - 8 \, a b c^{3} d + 10 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, {\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}}\right ] \]
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\[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{x^{3} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {2 \, a b \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} + \frac {5 \, a^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, c^{\frac {7}{2}}} + \frac {2 \, a b}{\sqrt {d x^{2} + c} c^{2}} + \frac {2 \, a b}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {b^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {5 \, a^{2} d}{2 \, \sqrt {d x^{2} + c} c^{3}} - \frac {5 \, a^{2} d}{6 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} - \frac {a^{2}}{2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{2}} \]
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Time = 0.31 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left (4 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, \sqrt {-c} c^{3}} - \frac {\sqrt {d x^{2} + c} a^{2}}{2 \, c^{3} x^{2}} - \frac {b^{2} c^{3} - 6 \, {\left (d x^{2} + c\right )} a b c d - 2 \, a b c^{2} d + 6 \, {\left (d x^{2} + c\right )} a^{2} d^{2} + a^{2} c d^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3} d} \]
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Time = 5.86 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^2}{x^3 \left (c+d x^2\right )^{5/2}} \, dx=\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (5\,a\,d-4\,b\,c\right )}{2\,c^{7/2}}-\frac {\frac {\left (d\,x^2+c\right )\,\left (-5\,a^2\,d^2+4\,a\,b\,c\,d+b^2\,c^2\right )}{3\,c^2}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{3\,c}+\frac {d\,{\left (d\,x^2+c\right )}^2\,\left (5\,a^2\,d-4\,a\,b\,c\right )}{2\,c^3}}{d\,{\left (d\,x^2+c\right )}^{5/2}-c\,d\,{\left (d\,x^2+c\right )}^{3/2}} \]
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