Integrand size = 22, antiderivative size = 90 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=-\frac {A \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 A b-4 a B) \sqrt {a+b x^2}}{8 a^2 x^2}-\frac {b (3 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \]
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Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {457, 79, 44, 65, 214} \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=-\frac {b (3 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {\sqrt {a+b x^2} (3 A b-4 a B)}{8 a^2 x^2}-\frac {A \sqrt {a+b x^2}}{4 a x^4} \]
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Rule 44
Rule 65
Rule 79
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = -\frac {A \sqrt {a+b x^2}}{4 a x^4}+\frac {\left (-\frac {3 A b}{2}+2 a B\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )}{4 a} \\ & = -\frac {A \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 A b-4 a B) \sqrt {a+b x^2}}{8 a^2 x^2}+\frac {(b (3 A b-4 a B)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{16 a^2} \\ & = -\frac {A \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 A b-4 a B) \sqrt {a+b x^2}}{8 a^2 x^2}+\frac {(3 A b-4 a B) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{8 a^2} \\ & = -\frac {A \sqrt {a+b x^2}}{4 a x^4}+\frac {(3 A b-4 a B) \sqrt {a+b x^2}}{8 a^2 x^2}-\frac {b (3 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-2 a A+3 A b x^2-4 a B x^2\right )}{8 a^2 x^4}+\frac {b (-3 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \]
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Time = 2.83 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.81
method | result | size |
pseudoelliptic | \(\frac {-\frac {3 x^{4} \left (A b -\frac {4 B a}{3}\right ) b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{8}+\frac {3 \left (\frac {2 \left (-2 x^{2} B -A \right ) a^{\frac {3}{2}}}{3}+A \sqrt {a}\, b \,x^{2}\right ) \sqrt {b \,x^{2}+a}}{8}}{a^{\frac {5}{2}} x^{4}}\) | \(73\) |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (-3 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 a^{2} x^{4}}-\frac {\left (3 A b -4 B a \right ) b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8 a^{\frac {5}{2}}}\) | \(76\) |
default | \(B \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )+A \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\) | \(124\) |
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Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.90 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=\left [-\frac {{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a^{3} x^{4}}, -\frac {{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a^{3} x^{4}}\right ] \]
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Time = 14.92 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.67 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=- \frac {A}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A \sqrt {b}}{8 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 A b^{\frac {3}{2}}}{8 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {5}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 a x} + \frac {B b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {3}{2}}} \]
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Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=\frac {B b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {3 \, A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} - \frac {\sqrt {b x^{2} + a} B}{2 \, a x^{2}} + \frac {3 \, \sqrt {b x^{2} + a} A b}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} A}{4 \, a x^{4}} \]
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Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=-\frac {\frac {{\left (4 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x^{2} + a} B a^{2} b^{2} - 3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3} + 5 \, \sqrt {b x^{2} + a} A a b^{3}}{a^{2} b^{2} x^{4}}}{8 \, b} \]
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Time = 5.89 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx=\frac {3\,A\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {5\,A\,\sqrt {b\,x^2+a}}{8\,a\,x^4}-\frac {3\,A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {B\,\sqrt {b\,x^2+a}}{2\,a\,x^2}+\frac {B\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}} \]
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