\(\int \frac {(-b+a x^2) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx\) [2525]

Optimal result

Integrand size = 39, antiderivative size = 211 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=-a^{5/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )+a^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )-\frac {1}{2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {a b \log (x)-a b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )-a^2 \log (x) \text {$\#$1}^4+b \log (x) \text {$\#$1}^4+a^2 \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ] \]

[Out]

Unintegrable

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {2081, 1283, 1542, 525, 524} \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {4 b x \sqrt [4]{a x^4-b x^2} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a-\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (-a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {4 b x \sqrt [4]{a x^4-b x^2} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a+\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}} \]

[In]

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

[Out]

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

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 1283

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With
[{k = Denominator[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + e*(x^(2*k)/f^2))^q*(a + b*(x^(2*k)/f^k) + c*
(x^(4*k)/f^4))^p, x], x, (f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && Fra
ctionQ[m] && IntegerQ[p]

Rule 1542

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

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (4 a^{5/4} \left (-\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )+\text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-a b \log (x)+2 a b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )+a^2 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4-2 a^2 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4+2 b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ]\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}} \]

[In]

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

[Out]

((-(b*x^2) + a*x^4)^(1/4)*(4*a^(5/4)*(-ArcTan[(a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)] + ArcTanh[(a^(1/4)*Sqrt[x]
)/(-b + a*x^2)^(1/4)]) + RootSum[b - a*#1^4 + #1^8 & , (-(a*b*Log[x]) + 2*a*b*Log[(-b + a*x^2)^(1/4) - Sqrt[x]
*#1] + a^2*Log[x]*#1^4 - b*Log[x]*#1^4 - 2*a^2*Log[(-b + a*x^2)^(1/4) - Sqrt[x]*#1]*#1^4 + 2*b*Log[(-b + a*x^2
)^(1/4) - Sqrt[x]*#1]*#1^4)/(a*#1^3 - 2*#1^7) & ]))/(4*Sqrt[x]*(-b + a*x^2)^(1/4))

Maple [N/A] (verified)

Time = 0.54 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(a^{\frac {5}{4}} \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\frac {a^{\frac {5}{4}} \ln \left (\frac {x \,a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{-x \,a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right )}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-a \,\textit {\_Z}^{4}+b \right )}{\sum }\left (-\frac {\left (\left (a^{2}-b \right ) \textit {\_R}^{4}-a b \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (-2 \textit {\_R}^{4}+a \right )}\right )\right )}{2}\)	\(153\)

[In]

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

[Out]

a^(5/4)*arctan(1/a^(1/4)/x*(x^2*(a*x^2-b))^(1/4))+1/2*a^(5/4)*ln((x*a^(1/4)+(x^2*(a*x^2-b))^(1/4))/(-x*a^(1/4)
+(x^2*(a*x^2-b))^(1/4)))+1/2*sum(-((a^2-b)*_R^4-a*b)*ln((-_R*x+(x^2*(a*x^2-b))^(1/4))/x)/_R^3/(-2*_R^4+a),_R=R
ootOf(_Z^8-_Z^4*a+b))

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 13.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.15 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (a x^{2} - b\right )}{- a x^{2} + b + x^{4}}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int { \frac {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}}{x^{4} - a x^{2} + b} \,d x } \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 8.05 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) \]

[In]

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

[Out]

1/2*sqrt(2)*(-a)^(1/4)*a*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x^2)^(1/4))/(-a)^(1/4)) + 1/2*sqrt(
2)*(-a)^(1/4)*a*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a - b/x^2)^(1/4))/(-a)^(1/4)) + 1/4*sqrt(2)*(-a)^
(1/4)*a*log(sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2)) - 1/4*sqrt(2)*(-a)^(1/4)*a*log(
-sqrt(2)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(-a) + sqrt(a - b/x^2))

Mupad [N/A]

Not integrable

Time = 8.55 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int -\frac {\left (b-a\,x^2\right )\,{\left (a\,x^4-b\,x^2\right )}^{1/4}}{x^4-a\,x^2+b} \,d x \]

[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)