\(\int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 (b+a x^4)} \, dx\) [1509]

Optimal result

Integrand size = 28, antiderivative size = 105 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx=-\frac {2 \left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{5 b^2 x^3}+\frac {a \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]}{4 b} \]

[Out]

Unintegrable

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.91, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {2081, 1284, 1535, 277, 270, 6857, 1543, 525, 524} \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx=-\frac {a x \sqrt [4]{a x^4+b x^2} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {\sqrt {-a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{3 b^2 \sqrt [4]{\frac {a x^2}{b}+1}}-\frac {a x \sqrt [4]{a x^4+b x^2} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {\sqrt {-a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{3 b^2 \sqrt [4]{\frac {a x^2}{b}+1}}-\frac {2 a \sqrt [4]{a x^4+b x^2}}{5 b^2 x}-\frac {2 \sqrt [4]{a x^4+b x^2}}{5 b x^3} \]

[In]

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

[Out]

(-2*(b*x^2 + a*x^4)^(1/4))/(5*b*x^3) - (2*a*(b*x^2 + a*x^4)^(1/4))/(5*b^2*x) - (a*x*(b*x^2 + a*x^4)^(1/4)*Appe
llF1[3/4, 1, -1/4, 7/4, -((Sqrt[-a]*x^2)/Sqrt[b]), -((a*x^2)/b)])/(3*b^2*(1 + (a*x^2)/b)^(1/4)) - (a*x*(b*x^2
+ a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (Sqrt[-a]*x^2)/Sqrt[b], -((a*x^2)/b)])/(3*b^2*(1 + (a*x^2)/b)^(1/4)
)

Rule 270

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

Rule 277

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

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 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 1535

Int[(((f_.)*(x_))^(m_)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Dist[d/a, Int[
(f*x)^m*(d + e*x^n)^(q - 1), x], x] + Dist[1/(a*f^n), Int[(f*x)^(m + n)*(d + e*x^n)^(q - 1)*(Simp[a*e - c*d*x^
n, x]/(a + c*x^(2*n))), x], x] /; FreeQ[{a, c, d, e, f}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] &&  !IntegerQ[q] &&
GtQ[q, 0] && LtQ[m, 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 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 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}& = \frac {\sqrt [4]{b x^2+a x^4} \int \frac {\sqrt [4]{b+a x^2}}{x^{7/2} \left (b+a x^4\right )} \, dx}{\sqrt {x} \sqrt [4]{b+a x^2}} \\ & = \frac {\left (2 \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{b+a x^4}}{x^6 \left (b+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{b+a x^2}} \\ & = \frac {\left (2 \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {1}{x^6 \left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{b+a x^2}}+\frac {\left (2 \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {a b-a b x^4}{x^2 \left (b+a x^4\right )^{3/4} \left (b+a x^8\right )} \, dx,x,\sqrt {x}\right )}{b \sqrt {x} \sqrt [4]{b+a x^2}} \\ & = -\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}+\frac {\left (2 \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \left (\frac {a}{x^2 \left (b+a x^4\right )^{3/4}}-\frac {a x^2 \sqrt [4]{b+a x^4}}{b+a x^8}\right ) \, dx,x,\sqrt {x}\right )}{b \sqrt {x} \sqrt [4]{b+a x^2}}-\frac {\left (8 a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{5 b \sqrt {x} \sqrt [4]{b+a x^2}} \\ & = -\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}+\frac {8 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}+\frac {\left (2 a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{b \sqrt {x} \sqrt [4]{b+a x^2}}-\frac {\left (2 a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{b+a x^4}}{b+a x^8} \, dx,x,\sqrt {x}\right )}{b \sqrt {x} \sqrt [4]{b+a x^2}} \\ & = -\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}-\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {\left (2 a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \left (-\frac {a x^2 \sqrt [4]{b+a x^4}}{2 \sqrt {-a} \sqrt {b} \left (\sqrt {-a} \sqrt {b}-a x^4\right )}-\frac {a x^2 \sqrt [4]{b+a x^4}}{2 \sqrt {-a} \sqrt {b} \left (\sqrt {-a} \sqrt {b}+a x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{b \sqrt {x} \sqrt [4]{b+a x^2}} \\ & = -\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}-\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {\left (\sqrt {-a} a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{b+a x^4}}{\sqrt {-a} \sqrt {b}-a x^4} \, dx,x,\sqrt {x}\right )}{b^{3/2} \sqrt {x} \sqrt [4]{b+a x^2}}-\frac {\left (\sqrt {-a} a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{b+a x^4}}{\sqrt {-a} \sqrt {b}+a x^4} \, dx,x,\sqrt {x}\right )}{b^{3/2} \sqrt {x} \sqrt [4]{b+a x^2}} \\ & = -\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}-\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {\left (\sqrt {-a} a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1+\frac {a x^4}{b}}}{\sqrt {-a} \sqrt {b}-a x^4} \, dx,x,\sqrt {x}\right )}{b^{3/2} \sqrt {x} \sqrt [4]{1+\frac {a x^2}{b}}}-\frac {\left (\sqrt {-a} a \sqrt [4]{b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1+\frac {a x^4}{b}}}{\sqrt {-a} \sqrt {b}+a x^4} \, dx,x,\sqrt {x}\right )}{b^{3/2} \sqrt {x} \sqrt [4]{1+\frac {a x^2}{b}}} \\ & = -\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}-\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {a x \sqrt [4]{b x^2+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {\sqrt {-a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{3 b^2 \sqrt [4]{1+\frac {a x^2}{b}}}-\frac {a x \sqrt [4]{b x^2+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {\sqrt {-a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{3 b^2 \sqrt [4]{1+\frac {a x^2}{b}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx=\frac {\sqrt [4]{x^2 \left (b+a x^2\right )} \left (-8 \left (b+a x^2\right )^{5/4}+5 a b x^{5/2} \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right ) \text {$\#$1}+\log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]\right )}{20 b^2 x^3 \sqrt [4]{b+a x^2}} \]

[In]

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

[Out]

((x^2*(b + a*x^2))^(1/4)*(-8*(b + a*x^2)^(5/4) + 5*a*b*x^(5/2)*RootSum[a^2 + a*b - 2*a*#1^4 + #1^8 & , (-(Log[
Sqrt[x]]*#1) + Log[(b + a*x^2)^(1/4) - Sqrt[x]*#1]*#1)/(-a + #1^4) & ]))/(20*b^2*x^3*(b + a*x^2)^(1/4))

Maple [N/A]

Time = 0.00 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(\frac {5 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+a b \right )}{\sum }\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{4}-a}\right ) b \,x^{3}-8 \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} \left (a \,x^{2}+b \right )}{20 x^{3} b^{2}}\)	\(93\)

[In]

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

[Out]

1/20*(5*a*sum(_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))*b*x^3-8*(x^2*(
a*x^2+b))^(1/4)*(a*x^2+b))/x^3/b^2

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 8.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.23 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )}}{x^{4} \left (a x^{4} + b\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}}}{{\left (a x^{4} + b\right )} x^{4}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.90 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx=\int { \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}}}{{\left (a x^{4} + b\right )} x^{4}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt [4]{b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx=\int \frac {{\left (a\,x^4+b\,x^2\right )}^{1/4}}{x^4\,\left (a\,x^4+b\right )} \,d x \]

[In]

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

[Out]

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