3.28.30 \(\int \frac {1}{(-b+a x^4)^2 \sqrt [4]{-b x^2+a x^4}} \, dx\)

Optimal. Leaf size=251 \[ \frac {\left (-a x^2-b\right ) \left (a x^4-b x^2\right )^{3/4}}{4 b^2 x (b-a) \left (b-a x^4\right )}-\frac {\text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4 a+a^2-a b\& ,\frac {-8 \text {$\#$1}^4 a \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )+7 \text {$\#$1}^4 b \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )+8 \text {$\#$1}^4 a \log (x)-7 \text {$\#$1}^4 b \log (x)+8 a^2 \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )-6 a b \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )-8 a^2 \log (x)+6 a b \log (x)}{\text {$\#$1} a-\text {$\#$1}^5}\& \right ]}{32 b^2 (a-b)} \]

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Rubi [F]  time = 0.11, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[1/((-b + a*x^4)^2*(-(b*x^2) + a*x^4)^(1/4)),x]

[Out]

(2*Sqrt[x]*(-b + a*x^2)^(1/4)*Defer[Subst][Defer[Int][1/((-b + a*x^4)^(1/4)*(-b + a*x^8)^2), x], x, Sqrt[x]])/
(-(b*x^2) + a*x^4)^(1/4)

Rubi steps

\begin {align*} \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{-b+a x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt [4]{-b+a x^2} \left (-b+a x^4\right )^2} \, dx}{\sqrt [4]{-b x^2+a x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^4} \left (-b+a x^8\right )^2} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^4}}\\ \end {align*}

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Mathematica [B]  time = 1.80, size = 644, normalized size = 2.57 \begin {gather*} \frac {\left (a x^2-b\right ) \left (\frac {7 b \sqrt [4]{\sqrt {b}-\sqrt {a}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{-\sqrt {a}-\sqrt {b}}}\right )+\sqrt {a} \sqrt {b} \sqrt [4]{\sqrt {b}-\sqrt {a}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{-\sqrt {a}-\sqrt {b}}}\right )-8 a \sqrt [4]{\sqrt {b}-\sqrt {a}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{-\sqrt {a}-\sqrt {b}}}\right )+7 b \sqrt [4]{-\sqrt {a}-\sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{\sqrt {b}-\sqrt {a}}}\right )-\sqrt {a} \sqrt {b} \sqrt [4]{-\sqrt {a}-\sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{\sqrt {b}-\sqrt {a}}}\right )-8 a \sqrt [4]{-\sqrt {a}-\sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{\sqrt {b}-\sqrt {a}}}\right )+\sqrt [4]{\sqrt {b}-\sqrt {a}} \left (-\sqrt {a} \sqrt {b}+8 a-7 b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{-\sqrt {a}-\sqrt {b}}}\right )+\sqrt [4]{-\sqrt {a}-\sqrt {b}} \left (\sqrt {a} \sqrt {b}+8 a-7 b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {b}{x^2}-a}}{\sqrt [8]{a} \sqrt [4]{\sqrt {b}-\sqrt {a}}}\right )}{\sqrt [8]{a} \sqrt [4]{-\sqrt {a}-\sqrt {b}} \sqrt [4]{\sqrt {b}-\sqrt {a}} \left (\frac {b}{x^2}-a\right )^{3/4}}+\frac {4 x^2 \left (a x^2+b\right )}{a x^4-b}\right )}{16 b^2 x (b-a) \sqrt [4]{a x^4-b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((-b + a*x^4)^2*(-(b*x^2) + a*x^4)^(1/4)),x]

[Out]

((-b + a*x^2)*((4*x^2*(b + a*x^2))/(-b + a*x^4) + (-8*a*(-Sqrt[a] + Sqrt[b])^(1/4)*ArcTan[(-a + b/x^2)^(1/4)/(
a^(1/8)*(-Sqrt[a] - Sqrt[b])^(1/4))] + Sqrt[a]*(-Sqrt[a] + Sqrt[b])^(1/4)*Sqrt[b]*ArcTan[(-a + b/x^2)^(1/4)/(a
^(1/8)*(-Sqrt[a] - Sqrt[b])^(1/4))] + 7*(-Sqrt[a] + Sqrt[b])^(1/4)*b*ArcTan[(-a + b/x^2)^(1/4)/(a^(1/8)*(-Sqrt
[a] - Sqrt[b])^(1/4))] - 8*a*(-Sqrt[a] - Sqrt[b])^(1/4)*ArcTan[(-a + b/x^2)^(1/4)/(a^(1/8)*(-Sqrt[a] + Sqrt[b]
)^(1/4))] - Sqrt[a]*(-Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[b]*ArcTan[(-a + b/x^2)^(1/4)/(a^(1/8)*(-Sqrt[a] + Sqrt[b])
^(1/4))] + 7*(-Sqrt[a] - Sqrt[b])^(1/4)*b*ArcTan[(-a + b/x^2)^(1/4)/(a^(1/8)*(-Sqrt[a] + Sqrt[b])^(1/4))] + (-
Sqrt[a] + Sqrt[b])^(1/4)*(8*a - Sqrt[a]*Sqrt[b] - 7*b)*ArcTanh[(-a + b/x^2)^(1/4)/(a^(1/8)*(-Sqrt[a] - Sqrt[b]
)^(1/4))] + (-Sqrt[a] - Sqrt[b])^(1/4)*(8*a + Sqrt[a]*Sqrt[b] - 7*b)*ArcTanh[(-a + b/x^2)^(1/4)/(a^(1/8)*(-Sqr
t[a] + Sqrt[b])^(1/4))])/(a^(1/8)*(-Sqrt[a] - Sqrt[b])^(1/4)*(-Sqrt[a] + Sqrt[b])^(1/4)*(-a + b/x^2)^(3/4))))/
(16*b^2*(-a + b)*x*(-(b*x^2) + a*x^4)^(1/4))

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IntegrateAlgebraic [A]  time = 0.00, size = 233, normalized size = 0.93 \begin {gather*} \frac {\left (-b-a x^2\right ) \left (-b x^2+a x^4\right )^{3/4}}{4 b^2 (-a+b) x \left (b-a x^4\right )}-\frac {\text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]}{4 b^2}+\frac {\text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 a \log (x)+2 a \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-a \text {$\#$1}+\text {$\#$1}^5}\&\right ]}{32 (a-b) b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((-b + a*x^4)^2*(-(b*x^2) + a*x^4)^(1/4)),x]

[Out]

((-b - a*x^2)*(-(b*x^2) + a*x^4)^(3/4))/(4*b^2*(-a + b)*x*(b - a*x^4)) - RootSum[a^2 - a*b - 2*a*#1^4 + #1^8 &
, (-Log[x] + Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1])/#1 & ]/(4*b^2) + RootSum[a^2 - a*b - 2*a*#1^4 + #1^8 & , (
-2*a*Log[x] + 2*a*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1] + Log[x]*#1^4 - Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1]*#1
^4)/(-(a*#1) + #1^5) & ]/(32*(a - b)*b)

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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.

[In]

integrate(1/(a*x^4-b)^2/(a*x^4-b*x^2)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x^4-b)^2/(a*x^4-b*x^2)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((a*x^4 - b*x^2)^(1/4)*(a*x^4 - b)^2), x)

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a \,x^{4}-b \right )^{2} \left (a \,x^{4}-b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x^4-b)^2/(a*x^4-b*x^2)^(1/4),x)

[Out]

int(1/(a*x^4-b)^2/(a*x^4-b*x^2)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x^4-b)^2/(a*x^4-b*x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((a*x^4 - b*x^2)^(1/4)*(a*x^4 - b)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b-a\,x^4\right )}^2\,{\left (a\,x^4-b\,x^2\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b - a*x^4)^2*(a*x^4 - b*x^2)^(1/4)),x)

[Out]

int(1/((b - a*x^4)^2*(a*x^4 - b*x^2)^(1/4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (a x^{4} - b\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x**4-b)**2/(a*x**4-b*x**2)**(1/4),x)

[Out]

Integral(1/((x**2*(a*x**2 - b))**(1/4)*(a*x**4 - b)**2), x)

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