\(\int \frac {x^2}{(-b+a x^4) \sqrt [4]{b x^2+a x^4}} \, dx\) [883]

Optimal result

Integrand size = 30, antiderivative size = 67 \[ \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\frac {1}{4} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(401\) vs. \(2(67)=134\).

Time = 0.29 (sec) , antiderivative size = 401, normalized size of antiderivative = 5.99, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2081, 1284, 1543, 385, 218, 212, 209} \[ \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {\sqrt {x} \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{5/8} \sqrt {b} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{5/8} \sqrt {b} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt [4]{a x^4+b x^2}}+\frac {\sqrt {x} \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{5/8} \sqrt {b} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt {x} \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{5/8} \sqrt {b} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt [4]{a x^4+b x^2}} \]

[In]

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

[Out]

(Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(5/8)*(
Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[b]*(b*x^2 + a*x^4)^(1/4)) - (Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/8)*(Sqrt[a]
+ Sqrt[b])^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(5/8)*(Sqrt[a] + Sqrt[b])^(1/4)*Sqrt[b]*(b*x^2 + a*x^4)^(1/
4)) + (Sqrt[x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^
(5/8)*(Sqrt[a] - Sqrt[b])^(1/4)*Sqrt[b]*(b*x^2 + a*x^4)^(1/4)) - (Sqrt[x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/8)*(
Sqrt[a] + Sqrt[b])^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(2*a^(5/8)*(Sqrt[a] + Sqrt[b])^(1/4)*Sqrt[b]*(b*x^2 + a*
x^4)^(1/4))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 1284

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

Rule 1543

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Int[ExpandInte
grand[(d + e*x^n)^q, (f*x)^m/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, f, q, n}, x] && EqQ[n2, 2*n] && IGt
Q[n, 0] &&  !IntegerQ[q] && IntegerQ[m]

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 {x^{3/2}}{\sqrt [4]{b+a x^2} \left (-b+a 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 {x^4}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \left (-\frac {1}{2 \sqrt {a} \left (\sqrt {b}-\sqrt {a} x^4\right ) \sqrt [4]{b+a x^4}}+\frac {1}{2 \sqrt {a} \left (\sqrt {b}+\sqrt {a} x^4\right ) \sqrt [4]{b+a x^4}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}} \\ & = -\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {b}-\sqrt {a} x^4\right ) \sqrt [4]{b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {a} \sqrt [4]{b x^2+a x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {b}+\sqrt {a} x^4\right ) \sqrt [4]{b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {a} \sqrt [4]{b x^2+a x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}-\left (a \sqrt {b}-\sqrt {a} b\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {a} \sqrt [4]{b x^2+a x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}-\left (a \sqrt {b}+\sqrt {a} b\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt {a} \sqrt [4]{b x^2+a x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 \sqrt {a} \sqrt {b} \sqrt [4]{b x^2+a x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt [4]{a} \sqrt {\sqrt {a}-\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 \sqrt {a} \sqrt {b} \sqrt [4]{b x^2+a x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt [4]{a} \sqrt {\sqrt {a}+\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 \sqrt {a} \sqrt {b} \sqrt [4]{b x^2+a x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt [4]{a} \sqrt {\sqrt {a}+\sqrt {b}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 \sqrt {a} \sqrt {b} \sqrt [4]{b x^2+a x^4}} \\ & = \frac {\sqrt {x} \sqrt [4]{b+a x^2} \arctan \left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 a^{5/8} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt {b} \sqrt [4]{b x^2+a x^4}}-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \arctan \left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 a^{5/8} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt {b} \sqrt [4]{b x^2+a x^4}}+\frac {\sqrt {x} \sqrt [4]{b+a x^2} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 a^{5/8} \sqrt [4]{\sqrt {a}-\sqrt {b}} \sqrt {b} \sqrt [4]{b x^2+a x^4}}-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{2 a^{5/8} \sqrt [4]{\sqrt {a}+\sqrt {b}} \sqrt {b} \sqrt [4]{b x^2+a x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.52 \[ \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right )+\log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ]}{4 \sqrt [4]{x^2 \left (b+a x^2\right )}} \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 1.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 \textit {\_Z}^{4} a +a^{2}-a b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (\textit {\_R}^{4}-a \right )}\right )}{4}\)	\(60\)

[In]

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

[Out]

1/4*sum(1/_R*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))

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 3.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.36 \[ \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {x^{2}}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} - b\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.45 \[ \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {x^{2}}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 1.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.45 \[ \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {x^{2}}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.46 \[ \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\int \frac {x^2}{\left (b-a\,x^4\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}} \,d x \]

[In]

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

[Out]

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