Integrand size = 22, antiderivative size = 143 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=\frac {5}{8} b (3 A b+4 a B) \sqrt {a+b x^2}+\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}-\frac {5}{8} \sqrt {a} b (3 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {457, 79, 43, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=-\frac {5}{8} \sqrt {a} b (4 a B+3 A b) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 a B+3 A b)}{8 a x^2}+\frac {5 b \left (a+b x^2\right )^{3/2} (4 a B+3 A b)}{24 a}+\frac {5}{8} b \sqrt {a+b x^2} (4 a B+3 A b)-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4} \]
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Rule 43
Rule 52
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)^{5/2} (A+B x)}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {\left (\frac {3 A b}{2}+2 a B\right ) \text {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^2} \, dx,x,x^2\right )}{4 a} \\ & = -\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {(5 b (3 A b+4 a B)) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^2\right )}{16 a} \\ & = \frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {1}{16} (5 b (3 A b+4 a B)) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right ) \\ & = \frac {5}{8} b (3 A b+4 a B) \sqrt {a+b x^2}+\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {1}{16} (5 a b (3 A b+4 a B)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {5}{8} b (3 A b+4 a B) \sqrt {a+b x^2}+\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac {1}{8} (5 a (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 ) \\ & = \frac {5}{8} b (3 A b+4 a B) \sqrt {a+b x^2}+\frac {5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac {(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{4 a x^4}-\frac {5}{8} \sqrt {a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=\frac {\sqrt {a+b x^2} \left (-6 a^2 A-27 a A b x^2-12 a^2 B x^2+24 A b^2 x^4+56 a b B x^4+8 b^2 B x^6\right )}{24 x^4}-\frac {5}{8} \sqrt {a} b (3 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
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Time = 2.88 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(\frac {-\frac {15 x^{4} b a \left (A b +\frac {4 B a}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{8}+\sqrt {b \,x^{2}+a}\, \left (-\frac {9 x^{2} \left (-\frac {56 x^{2} B}{27}+A \right ) b \,a^{\frac {3}{2}}}{8}+\left (-\frac {x^{2} B}{2}-\frac {A}{4}\right ) a^{\frac {5}{2}}+b^{2} x^{4} \sqrt {a}\, \left (\frac {x^{2} B}{3}+A \right )\right )}{\sqrt {a}\, x^{4}}\) | \(98\) |
risch | \(-\frac {a \sqrt {b \,x^{2}+a}\, \left (9 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 x^{4}}+\frac {b \left (8 B \,b^{2} \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+8 A b \sqrt {b \,x^{2}+a}+24 B a \sqrt {b \,x^{2}+a}-5 \sqrt {a}\, \left (3 A b +4 B a \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{8}\) | \(142\) |
default | \(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )\) | \(210\) |
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Time = 0.28 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=\left [\frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, B b^{2} x^{6} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, x^{4}}, \frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{2} x^{6} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{24 \, x^{4}}\right ] \]
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Time = 43.71 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.07 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=- \frac {15 A \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8} - \frac {A a^{3}}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A a^{2} \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{x} + \frac {7 A a b^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {5}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 B a^{\frac {3}{2}} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {B a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {2 B a^{2} \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {2 B a b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + B b^{2} \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=-\frac {5}{2} \, B a^{\frac {3}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) - \frac {15}{8} \, A \sqrt {a} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {5}{6} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{2 \, a} + \frac {5}{2} \, \sqrt {b x^{2} + a} B a b + \frac {15}{8} \, \sqrt {b x^{2} + a} A b^{2} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{8 \, a^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{8 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{2 \, a x^{2}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{4 \, a x^{4}} \]
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Time = 0.30 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=\frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2} + 48 \, \sqrt {b x^{2} + a} B a b^{2} + 24 \, \sqrt {b x^{2} + a} A b^{3} + \frac {15 \, {\left (4 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, {\left (4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} b^{2} - 4 \, \sqrt {b x^{2} + a} B a^{3} b^{2} + 9 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b^{3} - 7 \, \sqrt {b x^{2} + a} A a^{2} b^{3}\right )}}{b^{2} x^{4}}}{24 \, b} \]
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Time = 6.87 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^5} \, dx=A\,b^2\,\sqrt {b\,x^2+a}+\frac {B\,b\,{\left (b\,x^2+a\right )}^{3/2}}{3}+2\,B\,a\,b\,\sqrt {b\,x^2+a}+\frac {A\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{8}-\frac {9\,A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4}+\frac {7\,A\,a^2\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {B\,a^2\,\sqrt {b\,x^2+a}}{2\,x^2}+\frac {B\,a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{2} \]
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