Optimal. Leaf size=130 \[ \frac {\tan ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}-\frac {\sqrt [4]{n} \sqrt {a x^2+b}}{2^{3/4}}}{x \sqrt [4]{a x^2+b}}\right )}{2^{3/4} n^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt [4]{n} \sqrt {a x^2+b}}{2^{3/4}}+\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}}{x \sqrt [4]{a x^2+b}}\right )}{2^{3/4} n^{3/4}} \]
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Rubi [C] time = 0.84, antiderivative size = 571, normalized size of antiderivative = 4.39, number of steps used = 10, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1692, 399, 490, 1218} \begin {gather*} \frac {\sqrt [4]{b} \left (\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {-n a^2-\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {-a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}}-\frac {\sqrt [4]{b} \left (\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {-n a^2-\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {-a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}}+\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {-n a^2+\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}}-\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {-n a^2+\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}} \end {gather*}
Antiderivative was successfully verified.
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Rule 399
Rule 490
Rule 1218
Rule 1692
Rubi steps
\begin {align*} \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (b n+a n x^2+2 x^4\right )} \, dx &=\int \left (\frac {a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}}{\left (a n-\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}}+\frac {a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}}{\left (a n+\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}}\right ) \, dx\\ &=\left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \int \frac {1}{\left (a n-\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}} \, dx+\left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \int \frac {1}{\left (a n+\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}} \, dx\\ &=\frac {\left (2 \left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-4 b+a \left (a n-\sqrt {n} \sqrt {-8 b+a^2 n}\right )+4 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x}+\frac {\left (2 \left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-4 b+a \left (a n+\sqrt {n} \sqrt {-8 b+a^2 n}\right )+4 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x}\\ &=-\frac {\left (\left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}-2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}+2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}-\frac {\left (\left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}-2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}+2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}\\ &=\frac {\sqrt [4]{b} \left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}} x}-\frac {\sqrt [4]{b} \left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}} x}+\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}} x}-\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}} x}\\ \end {align*}
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Mathematica [F] time = 0.30, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (b n+a n x^2+2 x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 0.42, size = 130, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}-\frac {\sqrt [4]{n} \sqrt {b+a x^2}}{2^{3/4}}}{x \sqrt [4]{b+a x^2}}\right )}{2^{3/4} n^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}+\frac {\sqrt [4]{n} \sqrt {b+a x^2}}{2^{3/4}}}{x \sqrt [4]{b+a x^2}}\right )}{2^{3/4} n^{3/4}} \end {gather*}
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 {a x^{2} + 2 \, b}{{\left (2 \, x^{4} + 1881 \, a x^{2} + 1881 \, b\right )} {\left (a x^{2} + b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}+2 b}{\left (a \,x^{2}+b \right )^{\frac {1}{4}} \left (2 x^{4}+1881 a \,x^{2}+1881 b \right )}\, dx\]
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 {a x^{2} + 2 \, b}{{\left (2 \, x^{4} + 1881 \, a x^{2} + 1881 \, b\right )} {\left (a x^{2} + b\right )}^{\frac {1}{4}}}\,{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 {a\,x^2+2\,b}{{\left (a\,x^2+b\right )}^{1/4}\,\left (2\,x^4+1881\,a\,x^2+1881\,b\right )} \,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 {a x^{2} + 2 b}{\sqrt [4]{a x^{2} + b} \left (1881 a x^{2} + 1881 b + 2 x^{4}\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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