Integrand size = 24, antiderivative size = 145 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {8 b^2 c^2-3 a d (8 b c-5 a d)}{8 c^3 \sqrt {c+d x^2}}-\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}}-\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}} \]
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Time = 0.12 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, 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} \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=-\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}-\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}}+\frac {8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt {c+d x^2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}} \]
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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^3 (c+d x)^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (8 b c-5 a d)+2 b^2 c x}{x^2 (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}}+\frac {1}{16} \left (8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}\right ) \text {Subst}\left (\int \frac {1}{x (c+d x)^{3/2}} \, dx,x,x^2\right ) \\ & = \frac {8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt {c+d x^2}}-\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}}+\frac {\left (8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{16 c} \\ & = \frac {8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt {c+d x^2}}-\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}}+\frac {\left (8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{8 c d} \\ & = \frac {8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt {c+d x^2}}-\frac {a^2}{4 c x^4 \sqrt {c+d x^2}}-\frac {a (8 b c-5 a d)}{8 c^2 x^2 \sqrt {c+d x^2}}-\frac {\left (8 b^2-\frac {3 a d (8 b c-5 a d)}{c^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{3/2}} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {c} \left (8 b^2 c^2 x^4-8 a b c x^2 \left (c+3 d x^2\right )+a^2 \left (-2 c^2+5 c d x^2+15 d^2 x^4\right )\right )}{x^4 \sqrt {c+d x^2}}+\left (-8 b^2 c^2+24 a b c d-15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}} \]
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Time = 2.96 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.85
method | result | size |
pseudoelliptic | \(-\frac {15 \left (-\frac {x^{2} \left (-\frac {24 b \,x^{2}}{5}+a \right ) d a \,c^{\frac {3}{2}}}{3}+\frac {2 \left (-4 b^{2} x^{4}+4 a b \,x^{2}+a^{2}\right ) c^{\frac {5}{2}}}{15}+x^{4} \left (-a^{2} d^{2} \sqrt {c}+\left (a^{2} d^{2}-\frac {8}{5} a b c d +\frac {8}{15} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right ) \sqrt {d \,x^{2}+c}\right )\right )}{8 \sqrt {d \,x^{2}+c}\, c^{\frac {7}{2}} x^{4}}\) | \(123\) |
risch | \(-\frac {\sqrt {d \,x^{2}+c}\, a \left (-7 a d \,x^{2}+8 c b \,x^{2}+2 a c \right )}{8 c^{3} x^{4}}+\frac {-\frac {a d \left (7 a d -8 b c \right )}{\sqrt {d \,x^{2}+c}}+c \left (15 a^{2} d^{2}-24 a b c d +8 b^{2} c^{2}\right ) \left (\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}\right )}{8 c^{3}}\) | \(134\) |
default | \(b^{2} \left (\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}\right )+a^{2} \left (-\frac {1}{4 c \,x^{4} \sqrt {d \,x^{2}+c}}-\frac {5 d \left (-\frac {1}{2 c \,x^{2} \sqrt {d \,x^{2}+c}}-\frac {3 d \left (\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )+2 a b \left (-\frac {1}{2 c \,x^{2} \sqrt {d \,x^{2}+c}}-\frac {3 d \left (\frac {1}{c \sqrt {d \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(212\) |
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Time = 0.28 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.51 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, a^{2} c^{3} - {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} + {\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, {\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}, \frac {{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, a^{2} c^{3} - {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} + {\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}\right ] \]
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\[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{x^{5} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=-\frac {b^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {3}{2}}} + \frac {3 \, a b d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} - \frac {15 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, c^{\frac {7}{2}}} + \frac {b^{2}}{\sqrt {d x^{2} + c} c} - \frac {3 \, a b d}{\sqrt {d x^{2} + c} c^{2}} + \frac {15 \, a^{2} d^{2}}{8 \, \sqrt {d x^{2} + c} c^{3}} - \frac {a b}{\sqrt {d x^{2} + c} c x^{2}} + \frac {5 \, a^{2} d}{8 \, \sqrt {d x^{2} + c} c^{2} x^{2}} - \frac {a^{2}}{4 \, \sqrt {d x^{2} + c} c x^{4}} \]
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Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left (8 \, b^{2} c^{2} - 24 \, a b c d + 15 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{8 \, \sqrt {-c} c^{3}} + \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{\sqrt {d x^{2} + c} c^{3}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d - 8 \, \sqrt {d x^{2} + c} a b c^{2} d - 7 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2} + 9 \, \sqrt {d x^{2} + c} a^{2} c d^{2}}{8 \, c^{3} d^{2} x^{4}} \]
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Time = 5.93 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{c}-\frac {\left (d\,x^2+c\right )\,\left (25\,a^2\,d^2-40\,a\,b\,c\,d+16\,b^2\,c^2\right )}{8\,c^2}+\frac {{\left (d\,x^2+c\right )}^2\,\left (15\,a^2\,d^2-24\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^3}}{{\left (d\,x^2+c\right )}^{5/2}-2\,c\,{\left (d\,x^2+c\right )}^{3/2}+c^2\,\sqrt {d\,x^2+c}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (15\,a^2\,d^2-24\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^{7/2}} \]
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