Integrand size = 22, antiderivative size = 88 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=\frac {(A b-4 a B) \sqrt {a+b x^2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}+\frac {b (A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}} \]
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Time = 0.05 (sec) , antiderivative size = 88, 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, 43, 65, 214} \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=\frac {b (A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {\sqrt {a+b x^2} (A b-4 a B)}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4} \]
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Rule 43
Rule 65
Rule 79
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}+\frac {\left (-\frac {A b}{2}+2 a B\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,x^2\right )}{4 a} \\ & = \frac {(A b-4 a B) \sqrt {a+b x^2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}-\frac {(b (A b-4 a B)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{16 a} \\ & = \frac {(A b-4 a B) \sqrt {a+b x^2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}-\frac {(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} \\ & = \frac {(A b-4 a B) \sqrt {a+b x^2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}+\frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=\frac {\sqrt {a+b x^2} \left (-2 a A-A b x^2-4 a B x^2\right )}{8 a x^4}-\frac {b (-A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}} \]
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Time = 2.88 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82
method | result | size |
pseudoelliptic | \(-\frac {-b \,x^{4} \left (A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\sqrt {b \,x^{2}+a}\, \left (\left (4 x^{2} B +2 A \right ) a^{\frac {3}{2}}+A \sqrt {a}\, b \,x^{2}\right )}{8 a^{\frac {3}{2}} x^{4}}\) | \(72\) |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 x^{4} a}+\frac {\left (A b -4 B a \right ) b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8 a^{\frac {3}{2}}}\) | \(74\) |
default | \(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )\) | \(154\) |
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Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.93 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=\left [-\frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a^{2} x^{4}}, \frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a^{2} x^{4}}\right ] \]
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Time = 33.41 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.64 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=- \frac {A a}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A b^{\frac {3}{2}}}{8 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {3}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {B b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 \sqrt {a}} \]
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Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=-\frac {B b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, \sqrt {a}} + \frac {A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {3}{2}}} + \frac {\sqrt {b x^{2} + a} B b}{2 \, a} - \frac {\sqrt {b x^{2} + a} A b^{2}}{8 \, a^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{2 \, a x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{4 \, a x^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=\frac {\frac {{\left (4 \, B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x^{2} + a} B a^{2} b^{2} + {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3} + \sqrt {b x^{2} + a} A a b^{3}}{a b^{2} x^{4}}}{8 \, b} \]
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Time = 6.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx=\frac {A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{3/2}}-\frac {B\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {A\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a\,x^4}-\frac {B\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,\sqrt {a}} \]
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