\(\int \frac {-b+a x^2}{(b-a x^2+x^4) \sqrt [4]{b x^2+a x^4}} \, dx\) [1534]

Optimal result

Integrand size = 38, antiderivative size = 106 \[ \int \frac {-b+a x^2}{\left (b-a x^2+x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\frac {1}{2} \text {RootSum}\left [2 a^2+b-3 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-2 a \log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{3 a \text {$\#$1}-2 \text {$\#$1}^5}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(631\) vs. \(2(106)=212\).

Time = 1.23 (sec) , antiderivative size = 631, normalized size of antiderivative = 5.95, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {2081, 6847, 6860, 385, 218, 214, 211} \[ \int \frac {-b+a x^2}{\left (b-a x^2+x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\frac {\sqrt {x} \left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \left (\frac {a^2-2 b}{\sqrt {a^2-4 b}}+a\right ) \sqrt [4]{a x^2+b} \arctan \left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2+b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2+2 b} \sqrt [4]{a x^4+b x^2}}-\frac {\sqrt {x} \left (\frac {a^2-2 b}{\sqrt {a^2-4 b}}+a\right ) \sqrt [4]{a x^2+b} \text {arctanh}\left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2+b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2-4 b}+a^2+2 b} \sqrt [4]{a x^4+b x^2}} \]

[In]

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

[Out]

-(((a - (a^2 - 2*b)/Sqrt[a^2 - 4*b])*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[((a^2 - a*Sqrt[a^2 - 4*b] + 2*b)^(1/4)*S
qrt[x])/((a - Sqrt[a^2 - 4*b])^(1/4)*(b + a*x^2)^(1/4))])/((a - Sqrt[a^2 - 4*b])^(3/4)*(a^2 - a*Sqrt[a^2 - 4*b
] + 2*b)^(1/4)*(b*x^2 + a*x^4)^(1/4))) - ((a + (a^2 - 2*b)/Sqrt[a^2 - 4*b])*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)*(a^2 + a*Sqrt[a^2 - 4*b] + 2*b)^(1/4)*(b*x^2 + a*x^4)^(1/4)) - ((a - (a^2 - 2*b)/Sqrt[a^2 - 4*
b])*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)*(a^2 - a*Sqrt[a^2 - 4*b] + 2*b)^(1/4)*(b*x^2 + a*x^4)^(
1/4)) - ((a + (a^2 - 2*b)/Sqrt[a^2 - 4*b])*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)*(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 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]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {-b+a x^2}{\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 {-b+a x^4}{\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 (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \left (\frac {a+\frac {a^2-2 b}{\sqrt {a^2-4 b}}}{\left (-a-\sqrt {a^2-4 b}+2 x^4\right ) \sqrt [4]{b+a x^4}}+\frac {a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}}{\left (-a+\sqrt {a^2-4 b}+2 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 (2 \left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}}+\frac {\left (2 \left (a+\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}} \\ & = \frac {\left (2 \left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}}+\frac {\left (2 \left (a+\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}} \\ & = -\frac {\left (\left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}}-\frac {\left (\left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}}-\frac {\left (\left (a+\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}}-\frac {\left (\left (a+\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{b x^2+a x^4}} \\ & = -\frac {\left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{a^2-a \sqrt {a^2-4 b}+2 b} \sqrt [4]{b x^2+a x^4}}-\frac {\left (a+\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{a^2+a \sqrt {a^2-4 b}+2 b} \sqrt [4]{b x^2+a x^4}}-\frac {\left (a-\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [4]{a^2-a \sqrt {a^2-4 b}+2 b} \sqrt [4]{b x^2+a x^4}}-\frac {\left (a+\frac {a^2-2 b}{\sqrt {a^2-4 b}}\right ) \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 [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 = 0.88 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.30 \[ \int \frac {-b+a x^2}{\left (b-a x^2+x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\frac {\sqrt {x} \sqrt [4]{b+a x^2} \text {RootSum}\left [2 a^2+b-3 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-4 a \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{3 a \text {$\#$1}-2 \text {$\#$1}^5}\&\right ]}{4 \sqrt [4]{x^2 \left (b+a x^2\right )}} \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-3 a \,\textit {\_Z}^{4}+2 a^{2}+b \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 a \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (2 \textit {\_R}^{4}-3 a \right )}\right )}{2}\)	\(68\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

integrate((a*x^2-b)/(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 = 6.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.29 \[ \int \frac {-b+a x^2}{\left (b-a x^2+x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {a x^{2} - b}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (- a x^{2} + b + x^{4}\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.36 \[ \int \frac {-b+a x^2}{\left (b-a x^2+x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {a x^{2} - b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - a x^{2} + b\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 8.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.36 \[ \int \frac {-b+a x^2}{\left (b-a x^2+x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {a x^{2} - b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - a x^{2} + b\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 5.96 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.36 \[ \int \frac {-b+a x^2}{\left (b-a x^2+x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int -\frac {b-a\,x^2}{{\left (a\,x^4+b\,x^2\right )}^{1/4}\,\left (x^4-a\,x^2+b\right )} \,d x \]

[In]

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

[Out]

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