\(\int \frac {1}{(-b+a x^4)^2 \sqrt [4]{-b x^2+a x^4}} \, dx\) [2730]

Optimal result

Integrand size = 28, antiderivative size = 251 \[ \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {\left (-b-a x^2\right ) \left (-b x^2+a x^4\right )^{3/4}}{4 b^2 (-a+b) x \left (b-a x^4\right )}-\frac {\text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-8 a^2 \log (x)+6 a b \log (x)+8 a^2 \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )-6 a b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )+8 a \log (x) \text {$\#$1}^4-7 b \log (x) \text {$\#$1}^4-8 a \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+7 b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ]}{32 (a-b) b^2} \]

[Out]

Unintegrable

Rubi [F]

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

[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*} \text {integral}& = \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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {\sqrt {x} \left (-8 \sqrt [4]{-b+a x^2} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right )+\log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]+\frac {\frac {16 \sqrt {x} \left (b^2-a^2 x^4\right )}{-b+a x^4}+b \sqrt [4]{-b+a x^2} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-4 a \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ]}{2 (a-b)}\right )}{32 b^2 \sqrt [4]{-b x^2+a x^4}} \]

[In]

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

[Out]

(Sqrt[x]*(-8*(-b + a*x^2)^(1/4)*RootSum[a^2 - a*b - 2*a*#1^4 + #1^8 & , (-Log[Sqrt[x]] + Log[(-b + a*x^2)^(1/4
) - Sqrt[x]*#1])/#1 & ] + ((16*Sqrt[x]*(b^2 - a^2*x^4))/(-b + a*x^4) + b*(-b + a*x^2)^(1/4)*RootSum[a^2 - a*b
- 2*a*#1^4 + #1^8 & , (2*a*Log[x] - 4*a*Log[(-b + a*x^2)^(1/4) - Sqrt[x]*#1] - Log[x]*#1^4 + 2*Log[(-b + a*x^2
)^(1/4) - Sqrt[x]*#1]*#1^4)/(a*#1 - #1^5) & ])/(2*(a - b))))/(32*b^2*(-(b*x^2) + a*x^4)^(1/4))

Maple [N/A]

Time = 0.47 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}-a b \right )}{\sum }\frac {\left (8 a \,\textit {\_R}^{4}-7 \textit {\_R}^{4} b -8 a^{2}+6 a b \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (\textit {\_R}^{4}-a \right )}\right ) a \,x^{5}-8 a \left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {3}{4}} x^{2}-8 b \left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {3}{4}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}-a b \right )}{\sum }\frac {\left (8 a \,\textit {\_R}^{4}-7 \textit {\_R}^{4} b -8 a^{2}+6 a b \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (\textit {\_R}^{4}-a \right )}\right ) b x}{32 \left (a \,x^{4}-b \right ) \left (a -b \right ) b^{2} x}\)	\(238\)

[In]

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

[Out]

1/32*(-sum(1/_R*(8*_R^4*a-7*_R^4*b-8*a^2+6*a*b)*ln((-_R*x+(x^2*(a*x^2-b))^(1/4))/x)/(_R^4-a),_R=RootOf(_Z^8-2*
_Z^4*a+a^2-a*b))*a*x^5-8*a*(x^2*(a*x^2-b))^(3/4)*x^2-8*b*(x^2*(a*x^2-b))^(3/4)+sum(1/_R*(8*_R^4*a-7*_R^4*b-8*a
^2+6*a*b)*ln((-_R*x+(x^2*(a*x^2-b))^(1/4))/x)/(_R^4-a),_R=RootOf(_Z^8-2*_Z^4*a+a^2-a*b))*b*x)/(a*x^4-b)/(a-b)/
b^2/x

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 9.62 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.10 \[ \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx=\int \frac {1}{\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (a x^{4} - b\right )^{2}}\, dx \]

[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)

Maxima [N/A]

Not integrable

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.11 \[ \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx=\int { \frac {1}{{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}^{2}} \,d x } \]

[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)

Giac [N/A]

Not integrable

Time = 0.93 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.11 \[ \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx=\int { \frac {1}{{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}^{2}} \,d x } \]

[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)

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.11 \[ \int \frac {1}{\left (-b+a x^4\right )^2 \sqrt [4]{-b x^2+a x^4}} \, dx=\int \frac {1}{{\left (b-a\,x^4\right )}^2\,{\left (a\,x^4-b\,x^2\right )}^{1/4}} \,d x \]

[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)