\(\int \frac {-b^2+a^2 x^8}{x^2 (b+a x^4)^{3/4} (b^2+a^2 x^8)} \, dx\) [1209]

Optimal result

Integrand size = 41, antiderivative size = 88 \[ \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\frac {\sqrt [4]{b+a x^4}}{b x}+\frac {a^2 \text {RootSum}\left [2 a^2-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{a \text {$\#$1}^3-\text {$\#$1}^7}\&\right ]}{4 b} \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(226\) vs. \(2(88)=176\).

Time = 1.05 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.57, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {1600, 6857, 283, 338, 304, 209, 212, 1543, 525, 524} \[ \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\frac {a x^3 \sqrt [4]{a x^4+b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},-\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{\frac {a x^4}{b}+1}}+\frac {a x^3 \sqrt [4]{a x^4+b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{\frac {a x^4}{b}+1}}+\frac {\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 b}-\frac {\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 b}+\frac {\sqrt [4]{a x^4+b}}{b x} \]

[In]

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

[Out]

(b + a*x^4)^(1/4)/(b*x) + (a*x^3*(b + a*x^4)^(1/4)*AppellF1[3/4, 1, -5/4, 7/4, -((Sqrt[-a^2]*x^4)/b), -((a*x^4
)/b)])/(6*b^2*(1 + (a*x^4)/b)^(1/4)) + (a*x^3*(b + a*x^4)^(1/4)*AppellF1[3/4, 1, -5/4, 7/4, (Sqrt[-a^2]*x^4)/b
, -((a*x^4)/b)])/(6*b^2*(1 + (a*x^4)/b)^(1/4)) + (a^(1/4)*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(2*b) - (a^(1
/4)*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(2*b)

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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 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]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6857

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2 \left (b^2+a^2 x^8\right )} \, dx \\ & = \int \left (-\frac {\sqrt [4]{b+a x^4}}{b x^2}+\frac {a x^2 \left (b+a x^4\right )^{5/4}}{b \left (b^2+a^2 x^8\right )}\right ) \, dx \\ & = -\frac {\int \frac {\sqrt [4]{b+a x^4}}{x^2} \, dx}{b}+\frac {a \int \frac {x^2 \left (b+a x^4\right )^{5/4}}{b^2+a^2 x^8} \, dx}{b} \\ & = \frac {\sqrt [4]{b+a x^4}}{b x}-\frac {a \int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx}{b}+\frac {a \int \left (-\frac {a^2 x^2 \left (b+a x^4\right )^{5/4}}{2 \sqrt {-a^2} b \left (\sqrt {-a^2} b-a^2 x^4\right )}-\frac {a^2 x^2 \left (b+a x^4\right )^{5/4}}{2 \sqrt {-a^2} b \left (\sqrt {-a^2} b+a^2 x^4\right )}\right ) \, dx}{b} \\ & = \frac {\sqrt [4]{b+a x^4}}{b x}+\frac {\left (a \sqrt {-a^2}\right ) \int \frac {x^2 \left (b+a x^4\right )^{5/4}}{\sqrt {-a^2} b-a^2 x^4} \, dx}{2 b^2}+\frac {\left (a \sqrt {-a^2}\right ) \int \frac {x^2 \left (b+a x^4\right )^{5/4}}{\sqrt {-a^2} b+a^2 x^4} \, dx}{2 b^2}-\frac {a \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{b} \\ & = \frac {\sqrt [4]{b+a x^4}}{b x}-\frac {\sqrt {a} \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 b}+\frac {\sqrt {a} \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 b}+\frac {\left (a \sqrt {-a^2} \sqrt [4]{b+a x^4}\right ) \int \frac {x^2 \left (1+\frac {a x^4}{b}\right )^{5/4}}{\sqrt {-a^2} b-a^2 x^4} \, dx}{2 b \sqrt [4]{1+\frac {a x^4}{b}}}+\frac {\left (a \sqrt {-a^2} \sqrt [4]{b+a x^4}\right ) \int \frac {x^2 \left (1+\frac {a x^4}{b}\right )^{5/4}}{\sqrt {-a^2} b+a^2 x^4} \, dx}{2 b \sqrt [4]{1+\frac {a x^4}{b}}} \\ & = \frac {\sqrt [4]{b+a x^4}}{b x}+\frac {a x^3 \sqrt [4]{b+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},-\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{1+\frac {a x^4}{b}}}+\frac {a x^3 \sqrt [4]{b+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{1+\frac {a x^4}{b}}}+\frac {\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 b}-\frac {\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\frac {\sqrt [4]{b+a x^4}}{b x}+\frac {a^2 \text {RootSum}\left [2 a^2-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x)-\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{-a \text {$\#$1}^3+\text {$\#$1}^7}\&\right ]}{4 b} \]

[In]

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

[Out]

(b + a*x^4)^(1/4)/(b*x) + (a^2*RootSum[2*a^2 - 2*a*#1^4 + #1^8 & , (Log[x] - Log[(b + a*x^4)^(1/4) - x*#1])/(-
(a*#1^3) + #1^7) & ])/(4*b)

Maple [N/A] (verified)

Time = 1.86 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(\frac {-a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 \textit {\_Z}^{4} a +2 a^{2}\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (\textit {\_R}^{4}-a \right )}\right ) x +4 \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{4 b x}\)	\(78\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

integrate((a^2*x^8-b^2)/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 = 30.64 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.35 \[ \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int \frac {\left (a x^{4} - b\right ) \sqrt [4]{a x^{4} + b}}{x^{2} \left (a^{2} x^{8} + b^{2}\right )}\, dx \]

[In]

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

[Out]

Integral((a*x**4 - b)*(a*x**4 + b)**(1/4)/(x**2*(a**2*x**8 + b**2)), x)

Maxima [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.47 \[ \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int { \frac {a^{2} x^{8} - b^{2}}{{\left (a^{2} x^{8} + b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}} x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.47 \[ \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int { \frac {a^{2} x^{8} - b^{2}}{{\left (a^{2} x^{8} + b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}} x^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 6.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.47 \[ \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx=\int -\frac {b^2-a^2\,x^8}{x^2\,\left (a^2\,x^8+b^2\right )\,{\left (a\,x^4+b\right )}^{3/4}} \,d x \]

[In]

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

[Out]

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