\(\int \frac {x^2}{(b+a x^4)^{3/4} (b^2+a^2 x^8)} \, dx\) [720]

Optimal result

Integrand size = 28, antiderivative size = 55 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=-\frac {\text {RootSum}\left [2 a^2-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ]}{8 b^2} \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(325\) vs. \(2(55)=110\).

Time = 0.58 (sec) , antiderivative size = 325, normalized size of antiderivative = 5.91, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1543, 508, 304, 211} \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=-\frac {\left (-a^2\right )^{3/8} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{\sqrt {-a^2}-a} x}{\sqrt [8]{-a^2} \sqrt [4]{a x^4+b}}\right )}{4 a^{3/4} \left (\sqrt {-a^2}-a\right )^{3/4} b^2}+\frac {\sqrt [4]{a} \arctan \left (\frac {\left (-a^2\right )^{3/8} \sqrt [4]{\sqrt {-a^2}-a} x}{a^{3/4} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{-a^2} \left (\sqrt {-a^2}-a\right )^{3/4} b^2}-\frac {\left (-a^2\right )^{3/8} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{\sqrt {-a^2}+a} x}{\sqrt [8]{-a^2} \sqrt [4]{a x^4+b}}\right )}{4 a^{3/4} \left (\sqrt {-a^2}+a\right )^{3/4} b^2}+\frac {\sqrt [4]{a} \arctan \left (\frac {\left (-a^2\right )^{3/8} \sqrt [4]{\sqrt {-a^2}+a} x}{a^{3/4} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{-a^2} \left (\sqrt {-a^2}+a\right )^{3/4} b^2} \]

[In]

Int[x^2/((b + a*x^4)^(3/4)*(b^2 + a^2*x^8)),x]

[Out]

-1/4*((-a^2)^(3/8)*ArcTan[(a^(1/4)*(-a + Sqrt[-a^2])^(1/4)*x)/((-a^2)^(1/8)*(b + a*x^4)^(1/4))])/(a^(3/4)*(-a
+ Sqrt[-a^2])^(3/4)*b^2) + (a^(1/4)*ArcTan[((-a^2)^(3/8)*(-a + Sqrt[-a^2])^(1/4)*x)/(a^(3/4)*(b + a*x^4)^(1/4)
)])/(4*(-a^2)^(1/8)*(-a + Sqrt[-a^2])^(3/4)*b^2) - ((-a^2)^(3/8)*ArcTan[(a^(1/4)*(a + Sqrt[-a^2])^(1/4)*x)/((-
a^2)^(1/8)*(b + a*x^4)^(1/4))])/(4*a^(3/4)*(a + Sqrt[-a^2])^(3/4)*b^2) + (a^(1/4)*ArcTan[((-a^2)^(3/8)*(a + Sq
rt[-a^2])^(1/4)*x)/(a^(3/4)*(b + a*x^4)^(1/4))])/(4*(-a^2)^(1/8)*(a + Sqrt[-a^2])^(3/4)*b^2)

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 304

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

Rule 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

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]

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=-\frac {\text {RootSum}\left [2 a^2-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ]}{8 b^2} \]

[In]

Integrate[x^2/((b + a*x^4)^(3/4)*(b^2 + a^2*x^8)),x]

[Out]

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

Maple [N/A] (verified)

Time = 1.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+2 a^{2}\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3}}}{8 b^{2}}\)	\(48\)

[In]

int(x^2/(a*x^4+b)^(3/4)/(a^2*x^8+b^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\text {Timed out} \]

[In]

integrate(x^2/(a*x^4+b)^(3/4)/(a^2*x^8+b^2),x, algorithm="fricas")

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 4.71 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.44 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int \frac {x^{2}}{\left (a x^{4} + b\right )^{\frac {3}{4}} \left (a^{2} x^{8} + b^{2}\right )}\, dx \]

[In]

integrate(x**2/(a*x**4+b)**(3/4)/(a**2*x**8+b**2),x)

[Out]

Integral(x**2/((a*x**4 + b)**(3/4)*(a**2*x**8 + b**2)), x)

Maxima [N/A]

Not integrable

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.51 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int { \frac {x^{2}}{{\left (a^{2} x^{8} + b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^2/(a*x^4+b)^(3/4)/(a^2*x^8+b^2),x, algorithm="maxima")

[Out]

integrate(x^2/((a^2*x^8 + b^2)*(a*x^4 + b)^(3/4)), x)

Giac [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.51 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int { \frac {x^{2}}{{\left (a^{2} x^{8} + b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^2/(a*x^4+b)^(3/4)/(a^2*x^8+b^2),x, algorithm="giac")

[Out]

integrate(x^2/((a^2*x^8 + b^2)*(a*x^4 + b)^(3/4)), x)

Mupad [N/A]

Not integrable

Time = 5.92 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.51 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int \frac {x^2}{\left (a^2\,x^8+b^2\right )\,{\left (a\,x^4+b\right )}^{3/4}} \,d x \]

[In]

int(x^2/((b^2 + a^2*x^8)*(b + a*x^4)^(3/4)),x)

[Out]

int(x^2/((b^2 + a^2*x^8)*(b + a*x^4)^(3/4)), x)