Optimal. Leaf size=168 \[ \frac {x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}}+\frac {9 x \left (a-b x^2\right )}{32 a^3 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}}+\frac {19 \sqrt {a+b x^2} \sqrt {a-b x^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{32 \sqrt {2} a^3 \sqrt {b} \sqrt {a^2-b^2 x^4}} \]
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Rubi [A] time = 0.09, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1152, 414, 527, 12, 377, 205} \begin {gather*} \frac {9 x \left (a-b x^2\right )}{32 a^3 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}}+\frac {x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}}+\frac {19 \sqrt {a+b x^2} \sqrt {a-b x^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{32 \sqrt {2} a^3 \sqrt {b} \sqrt {a^2-b^2 x^4}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 377
Rule 414
Rule 527
Rule 1152
Rubi steps
\begin {align*} \int \frac {1}{\left (a+b x^2\right )^{5/2} \sqrt {a^2-b^2 x^4}} \, dx &=\frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {a-b x^2} \left (a+b x^2\right )^3} \, dx}{\sqrt {a^2-b^2 x^4}}\\ &=\frac {x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}}-\frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {-7 a b+2 b^2 x^2}{\sqrt {a-b x^2} \left (a+b x^2\right )^2} \, dx}{8 a^2 b \sqrt {a^2-b^2 x^4}}\\ &=\frac {x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}}+\frac {9 x \left (a-b x^2\right )}{32 a^3 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}}+\frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {19 a^2 b^2}{\sqrt {a-b x^2} \left (a+b x^2\right )} \, dx}{32 a^4 b^2 \sqrt {a^2-b^2 x^4}}\\ &=\frac {x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}}+\frac {9 x \left (a-b x^2\right )}{32 a^3 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}}+\frac {\left (19 \sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {a-b x^2} \left (a+b x^2\right )} \, dx}{32 a^2 \sqrt {a^2-b^2 x^4}}\\ &=\frac {x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}}+\frac {9 x \left (a-b x^2\right )}{32 a^3 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}}+\frac {\left (19 \sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 a b x^2} \, dx,x,\frac {x}{\sqrt {a-b x^2}}\right )}{32 a^2 \sqrt {a^2-b^2 x^4}}\\ &=\frac {x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}}+\frac {9 x \left (a-b x^2\right )}{32 a^3 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}}+\frac {19 \sqrt {a-b x^2} \sqrt {a+b x^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{32 \sqrt {2} a^3 \sqrt {b} \sqrt {a^2-b^2 x^4}}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 123, normalized size = 0.73 \begin {gather*} \frac {\sqrt {a^2-b^2 x^4} \left (2 \sqrt {b} x \sqrt {a-b x^2} \left (13 a+9 b x^2\right )+19 \sqrt {2} \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )\right )}{64 a^3 \sqrt {b} \sqrt {a-b x^2} \left (a+b x^2\right )^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 2.96, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a+b x^2\right )^{5/2} \sqrt {a^2-b^2 x^4}} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.74, size = 365, normalized size = 2.17 \begin {gather*} \left [-\frac {19 \, \sqrt {2} {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-b} \log \left (-\frac {3 \, b^{2} x^{4} + 2 \, a b x^{2} - 2 \, \sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {-b} x - a^{2}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, \sqrt {-b^{2} x^{4} + a^{2}} {\left (9 \, b^{2} x^{3} + 13 \, a b x\right )} \sqrt {b x^{2} + a}}{128 \, {\left (a^{3} b^{4} x^{6} + 3 \, a^{4} b^{3} x^{4} + 3 \, a^{5} b^{2} x^{2} + a^{6} b\right )}}, -\frac {19 \, \sqrt {2} {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {b}}{2 \, {\left (b^{2} x^{3} + a b x\right )}}\right ) - 2 \, \sqrt {-b^{2} x^{4} + a^{2}} {\left (9 \, b^{2} x^{3} + 13 \, a b x\right )} \sqrt {b x^{2} + a}}{64 \, {\left (a^{3} b^{4} x^{6} + 3 \, a^{4} b^{3} x^{4} + 3 \, a^{5} b^{2} x^{2} + a^{6} b\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {-b^{2} x^{4} + a^{2}} {\left (b x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 711, normalized size = 4.23 \begin {gather*} -\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (19 \sqrt {2}\, \sqrt {a}\, b^{\frac {5}{2}} x^{4} \ln \left (\frac {2 \left (a -\sqrt {-a b}\, x +\sqrt {2}\, \sqrt {-b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x -\sqrt {-a b}}\right )-19 \sqrt {2}\, \sqrt {a}\, b^{\frac {5}{2}} x^{4} \ln \left (\frac {2 \left (a +\sqrt {-a b}\, x +\sqrt {2}\, \sqrt {-b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x +\sqrt {-a b}}\right )-16 \sqrt {-a b}\, b^{2} x^{4} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right )+16 \sqrt {-a b}\, b^{2} x^{4} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )+38 \sqrt {2}\, a^{\frac {3}{2}} b^{\frac {3}{2}} x^{2} \ln \left (\frac {2 \left (a -\sqrt {-a b}\, x +\sqrt {2}\, \sqrt {-b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x -\sqrt {-a b}}\right )-38 \sqrt {2}\, a^{\frac {3}{2}} b^{\frac {3}{2}} x^{2} \ln \left (\frac {2 \left (a +\sqrt {-a b}\, x +\sqrt {2}\, \sqrt {-b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x +\sqrt {-a b}}\right )-32 \sqrt {-a b}\, a b \,x^{2} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right )+32 \sqrt {-a b}\, a b \,x^{2} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )-36 \sqrt {-a b}\, \sqrt {-b \,x^{2}+a}\, b^{\frac {3}{2}} x^{3}+19 \sqrt {2}\, a^{\frac {5}{2}} \sqrt {b}\, \ln \left (\frac {2 \left (a -\sqrt {-a b}\, x +\sqrt {2}\, \sqrt {-b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x -\sqrt {-a b}}\right )-19 \sqrt {2}\, a^{\frac {5}{2}} \sqrt {b}\, \ln \left (\frac {2 \left (a +\sqrt {-a b}\, x +\sqrt {2}\, \sqrt {-b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x +\sqrt {-a b}}\right )-16 \sqrt {-a b}\, a^{2} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right )+16 \sqrt {-a b}\, a^{2} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right )-52 \sqrt {-a b}\, \sqrt {-b \,x^{2}+a}\, a \sqrt {b}\, x \right ) b^{\frac {9}{2}}}{16 \sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{2}+a}\, \sqrt {-a b}\, \left (\sqrt {-a b}+\sqrt {a b}\right )^{3} \left (-\sqrt {-a b}+\sqrt {a b}\right )^{3} \left (b x +\sqrt {-a b}\right )^{2} \left (b x -\sqrt {-a b}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {-b^{2} x^{4} + a^{2}} {\left (b x^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \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}{\sqrt {a^2-b^2\,x^4}\,{\left (b\,x^2+a\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {- \left (- a + b x^{2}\right ) \left (a + b x^{2}\right )} \left (a + b x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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