Optimal. Leaf size=95 \[ \frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b^{7/2}} \]
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Rubi [A]
time = 0.04, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {342, 294, 327,
223, 212} \begin {gather*} -\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{8 b^{7/2}}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {1}{b x^5 \sqrt {a+\frac {b}{x^2}}} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 223
Rule 294
Rule 327
Rule 342
Rubi steps
\begin {align*} \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^8} \, dx &=-\text {Subst}\left (\int \frac {x^6}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \text {Subst}\left (\int \frac {x^4}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{b}\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {(15 a) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{4 b^2}\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{8 b^3}\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b^3}\\ &=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^5}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{4 b^2 x^3}+\frac {15 a \sqrt {a+\frac {b}{x^2}}}{8 b^3 x}-\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{8 b^{7/2}}\\ \end {align*}
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Mathematica [A]
time = 0.10, size = 88, normalized size = 0.93 \begin {gather*} \frac {\sqrt {b} \left (-2 b^2+5 a b x^2+15 a^2 x^4\right )-15 a^2 x^4 \sqrt {b+a x^2} \tanh ^{-1}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {a+\frac {b}{x^2}} x^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.07, size = 94, normalized size = 0.99
method | result | size |
default | \(-\frac {\left (a \,x^{2}+b \right ) \left (-15 b^{\frac {3}{2}} a^{2} x^{4}+15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) \sqrt {a \,x^{2}+b}\, a^{2} b \,x^{4}-5 b^{\frac {5}{2}} a \,x^{2}+2 b^{\frac {7}{2}}\right )}{8 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} x^{7} b^{\frac {9}{2}}}\) | \(94\) |
risch | \(\frac {\left (a \,x^{2}+b \right ) \left (7 a \,x^{2}-2 b \right )}{8 b^{3} x^{5} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {\left (\frac {a^{2}}{b^{3} \sqrt {a \,x^{2}+b}}-\frac {15 a^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right )}{8 b^{\frac {7}{2}}}\right ) \sqrt {a \,x^{2}+b}}{\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\) | \(114\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 139, normalized size = 1.46 \begin {gather*} \frac {15 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} a^{2} x^{4} - 25 \, {\left (a + \frac {b}{x^{2}}\right )} a^{2} b x^{2} + 8 \, a^{2} b^{2}}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} b^{3} x^{5} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{4} x^{3} + \sqrt {a + \frac {b}{x^{2}}} b^{5} x\right )}} + \frac {15 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{16 \, b^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 228, normalized size = 2.40 \begin {gather*} \left [\frac {15 \, {\left (a^{3} x^{5} + a^{2} b x^{3}\right )} \sqrt {b} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \, {\left (15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} - 2 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, {\left (a b^{4} x^{5} + b^{5} x^{3}\right )}}, \frac {15 \, {\left (a^{3} x^{5} + a^{2} b x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} - 2 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, {\left (a b^{4} x^{5} + b^{5} x^{3}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 4.22, size = 102, normalized size = 1.07 \begin {gather*} \frac {15 a^{\frac {3}{2}}}{8 b^{3} x \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {5 \sqrt {a}}{8 b^{2} x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{8 b^{\frac {7}{2}}} - \frac {1}{4 \sqrt {a} b x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 3.71, size = 99, normalized size = 1.04 \begin {gather*} \frac {15 \, a^{2} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{3} \mathrm {sgn}\left (x\right )} + \frac {a^{2}}{\sqrt {a x^{2} + b} b^{3} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{2} - 9 \, \sqrt {a x^{2} + b} a^{2} b}{8 \, a^{2} b^{3} x^{4} \mathrm {sgn}\left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^8\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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