Integrand size = 26, antiderivative size = 101 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}+\frac {4 c (5 b B-6 A c) \left (b+2 c x^2\right )}{15 b^4 \sqrt {b x^2+c x^4}} \]
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Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2059, 806, 672, 627} \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {4 c \left (b+2 c x^2\right ) (5 b B-6 A c)}{15 b^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}} \]
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Rule 627
Rule 672
Rule 806
Rule 2059
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}+\frac {\left (\frac {1}{2} (b B-2 A c)-2 (-b B+A c)\right ) \text {Subst}\left (\int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{5 b} \\ & = -\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {(2 c (5 b B-6 A c)) \text {Subst}\left (\int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{15 b^2} \\ & = -\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}+\frac {4 c (5 b B-6 A c) \left (b+2 c x^2\right )}{15 b^4 \sqrt {b x^2+c x^4}} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-5 b B x^2 \left (b^2-4 b c x^2-8 c^2 x^4\right )-3 A \left (b^3-2 b^2 c x^2+8 b c^2 x^4+16 c^3 x^6\right )}{15 b^4 x^4 \sqrt {x^2 \left (b+c x^2\right )}} \]
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Time = 2.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79
method | result | size |
pseudoelliptic | \(\frac {\left (-5 x^{2} B -3 A \right ) b^{3}+6 x^{2} \left (\frac {10 x^{2} B}{3}+A \right ) c \,b^{2}-24 x^{4} c^{2} \left (-\frac {5 x^{2} B}{3}+A \right ) b -48 A \,c^{3} x^{6}}{15 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, x^{4} b^{4}}\) | \(80\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (48 A \,c^{3} x^{6}-40 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-20 x^{4} B \,b^{2} c -6 A \,b^{2} c \,x^{2}+5 b^{3} B \,x^{2}+3 b^{3} A \right )}{15 x^{2} b^{4} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\) | \(94\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (48 A \,c^{3} x^{6}-40 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-20 x^{4} B \,b^{2} c -6 A \,b^{2} c \,x^{2}+5 b^{3} B \,x^{2}+3 b^{3} A \right )}{15 x^{2} b^{4} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\) | \(94\) |
trager | \(-\frac {\left (48 A \,c^{3} x^{6}-40 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-20 x^{4} B \,b^{2} c -6 A \,b^{2} c \,x^{2}+5 b^{3} B \,x^{2}+3 b^{3} A \right ) \sqrt {x^{4} c +b \,x^{2}}}{15 \left (c \,x^{2}+b \right ) b^{4} x^{6}}\) | \(96\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (33 A \,c^{2} x^{4}-25 x^{4} B b c -9 A b c \,x^{2}+5 b^{2} B \,x^{2}+3 b^{2} A \right )}{15 b^{4} x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {x^{2} c^{2} \left (A c -B b \right )}{b^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(103\) |
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Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (8 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{6} + 4 \, {\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} x^{4} - 3 \, A b^{3} - {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \]
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\[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {A + B x^{2}}{x^{3} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{3} \, B {\left (\frac {8 \, c^{2} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {4 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2}} - \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{2}}\right )} - \frac {1}{5} \, A {\left (\frac {16 \, c^{3} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{4}} + \frac {8 \, c^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} - \frac {2 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} + \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{4}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (89) = 178\).
Time = 0.92 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.99 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (B b c^{2} - A c^{3}\right )} x}{\sqrt {c x^{2} + b} b^{4} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b c^{\frac {3}{2}} - 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A c^{\frac {5}{2}} - 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{2} c^{\frac {3}{2}} + 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b c^{\frac {5}{2}} + 160 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{3} c^{\frac {3}{2}} - 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{2} c^{\frac {5}{2}} - 110 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{4} c^{\frac {3}{2}} + 150 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{3} c^{\frac {5}{2}} + 25 \, B b^{5} c^{\frac {3}{2}} - 33 \, A b^{4} c^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5} b^{3} \mathrm {sgn}\left (x\right )} \]
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Time = 9.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}\,\left (5\,B\,b^3\,x^2+3\,A\,b^3-20\,B\,b^2\,c\,x^4-6\,A\,b^2\,c\,x^2-40\,B\,b\,c^2\,x^6+24\,A\,b\,c^2\,x^4+48\,A\,c^3\,x^6\right )}{15\,b^4\,x^6\,\left (c\,x^2+b\right )} \]
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