\(\int \frac {1}{(b-a x^2+x^4) \sqrt [4]{-b x^2+a x^4}} \, dx\) [1481]

Optimal result

Integrand size = 30, antiderivative size = 104 \[ \int \frac {1}{\left (b-a x^2+x^4\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {\text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {a \log (x)-a \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-2 \text {$\#$1}^5}\&\right ]}{2 b} \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(609\) vs. \(2(104)=208\).

Time = 0.28 (sec) , antiderivative size = 609, normalized size of antiderivative = 5.86, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2081, 1283, 1442, 385, 218, 214, 211} \[ \int \frac {1}{\left (b-a x^2+x^4\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {2 \sqrt {x} \sqrt [4]{a x^2-b} \arctan \left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2-b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4-b x^2}}-\frac {2 \sqrt {x} \sqrt [4]{a x^2-b} \arctan \left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2-b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4-b x^2}}+\frac {2 \sqrt {x} \sqrt [4]{a x^2-b} \text {arctanh}\left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2-b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4-b x^2}}-\frac {2 \sqrt {x} \sqrt [4]{a x^2-b} \text {arctanh}\left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2-b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4-b x^2}} \]

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1283

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With
[{k = Denominator[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + e*(x^(2*k)/f^2))^q*(a + b*(x^(2*k)/f^k) + c*
(x^(4*k)/f^4))^p, x], x, (f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && Fra
ctionQ[m] && IntegerQ[p]

Rule 1442

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[2*(c/r), Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[2*(c/r), Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

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^2+x^4\right )} \, 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^4+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-b x^2+a x^4}} \\ & = \frac {\left (4 \sqrt {x} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (-a-\sqrt {a^2-4 b}+2 x^4\right ) \sqrt [4]{-b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2-4 b} \sqrt [4]{-b x^2+a x^4}}-\frac {\left (4 \sqrt {x} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (-a+\sqrt {a^2-4 b}+2 x^4\right ) \sqrt [4]{-b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2-4 b} \sqrt [4]{-b x^2+a x^4}} \\ & = \frac {\left (4 \sqrt {x} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{-a-\sqrt {a^2-4 b}-\left (a \left (-a-\sqrt {a^2-4 b}\right )+2 b\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {a^2-4 b} \sqrt [4]{-b x^2+a x^4}}-\frac {\left (4 \sqrt {x} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{-a+\sqrt {a^2-4 b}-\left (a \left (-a+\sqrt {a^2-4 b}\right )+2 b\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {a^2-4 b} \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 {a-\sqrt {a^2-4 b}}-\sqrt {a^2-a \sqrt {a^2-4 b}-2 b} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {a-\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \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 {a-\sqrt {a^2-4 b}}+\sqrt {a^2-a \sqrt {a^2-4 b}-2 b} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {a-\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \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 {a+\sqrt {a^2-4 b}}-\sqrt {a^2+a \sqrt {a^2-4 b}-2 b} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {a+\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \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 {a+\sqrt {a^2-4 b}}+\sqrt {a^2+a \sqrt {a^2-4 b}-2 b} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\sqrt {a+\sqrt {a^2-4 b}} \sqrt {a^2-4 b} \sqrt [4]{-b x^2+a x^4}} \\ & = \frac {2 \sqrt {x} \sqrt [4]{-b+a x^2} \arctan \left (\frac {\sqrt [4]{a^2-a \sqrt {a^2-4 b}-2 b} \sqrt {x}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{-b+a x^2}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a^2-a \sqrt {a^2-4 b}-2 b} \sqrt [4]{-b x^2+a x^4}}-\frac {2 \sqrt {x} \sqrt [4]{-b+a x^2} \arctan \left (\frac {\sqrt [4]{a^2+a \sqrt {a^2-4 b}-2 b} \sqrt {x}}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt [4]{-b+a x^2}}\right )}{\left (a+\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a^2+a \sqrt {a^2-4 b}-2 b} \sqrt [4]{-b x^2+a x^4}}+\frac {2 \sqrt {x} \sqrt [4]{-b+a x^2} \text {arctanh}\left (\frac {\sqrt [4]{a^2-a \sqrt {a^2-4 b}-2 b} \sqrt {x}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{-b+a x^2}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a^2-a \sqrt {a^2-4 b}-2 b} \sqrt [4]{-b x^2+a x^4}}-\frac {2 \sqrt {x} \sqrt [4]{-b+a x^2} \text {arctanh}\left (\frac {\sqrt [4]{a^2+a \sqrt {a^2-4 b}-2 b} \sqrt {x}}{\sqrt [4]{a+\sqrt {a^2-4 b}} \sqrt [4]{-b+a x^2}}\right )}{\left (a+\sqrt {a^2-4 b}\right )^{3/4} \sqrt {a^2-4 b} \sqrt [4]{a^2+a \sqrt {a^2-4 b}-2 b} \sqrt [4]{-b x^2+a x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (b-a x^2+x^4\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {\sqrt {x} \sqrt [4]{-b+a x^2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {a \log (x)-2 a \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-2 \text {$\#$1}^5}\&\right ]}{4 b \sqrt [4]{-b x^2+a x^4}} \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-a \,\textit {\_Z}^{4}+b \right )}{\sum }\frac {\left (\textit {\_R}^{4}-a \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (2 \textit {\_R}^{4}-a \right )}}{2 b}\)	\(68\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 1.98 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.25 \[ \int \frac {1}{\left (b-a x^2+x^4\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\int \frac {1}{\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (- a x^{2} + b + x^{4}\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (b-a x^2+x^4\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\int { \frac {1}{{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - a x^{2} + b\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 3.39 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (b-a x^2+x^4\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\int { \frac {1}{{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - a x^{2} + b\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (b-a x^2+x^4\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\int \frac {1}{{\left (a\,x^4-b\,x^2\right )}^{1/4}\,\left (x^4-a\,x^2+b\right )} \,d x \]

[In]

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

[Out]

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