\(\int \frac {(-b+a x^2) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx\) [2526]

Optimal result

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

Unintegrable
 

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (4 a^{5/4} \left (-\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )+\text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-a b \log (x)+2 a b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )+a^2 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4-2 a^2 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4+2 b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ]\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}} \] Input:

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

Output:

((-(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))
 

Rubi [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2467, 1592, 1852, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [4]{a x^4-b x^2} \int \frac {\sqrt {x} \left (a x^2-b\right )^{5/4}}{x^4-a x^2+b}dx}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 1592

\(\displaystyle \frac {2 \sqrt [4]{a x^4-b x^2} \int \frac {x \left (a x^2-b\right )^{5/4}}{x^4-a x^2+b}d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 1852

\(\displaystyle \frac {2 \sqrt [4]{a x^4-b x^2} \int \left (-\frac {2 x \left (a x^2-b\right )^{5/4}}{\sqrt {a^2-4 b} \left (-2 x^2+a+\sqrt {a^2-4 b}\right )}-\frac {2 x \left (a x^2-b\right )^{5/4}}{\sqrt {a^2-4 b} \left (2 x^2-a+\sqrt {a^2-4 b}\right )}\right )d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \sqrt [4]{a x^4-b x^2} \left (\frac {2 b x^{3/2} \sqrt [4]{a x^2-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a-\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (-a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {2 b x^{3/2} \sqrt [4]{a x^2-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a+\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}\right )}{\sqrt {x} \sqrt [4]{a x^2-b}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 1592

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c 
_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[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] && FractionQ[m] && IntegerQ[p]
 

rule 1852

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [N/A] (verified)

Time = 0.39 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(a^{\frac {5}{4}} \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\frac {a^{\frac {5}{4}} \ln \left (\frac {-x \,a^{\frac {1}{4}}-\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}-\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right )}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-a \,\textit {\_Z}^{4}+b \right )}{\sum }\left (-\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right ) \left (\left (a^{2}-b \right ) \textit {\_R}^{4}-a b \right )}{\textit {\_R}^{3} \left (-2 \textit {\_R}^{4}+a \right )}\right )\right )}{2}\)	\(157\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [N/A]

Not integrable

Time = 12.87 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.15 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (a x^{2} - b\right )}{- a x^{2} + b + x^{4}}\, dx \] Input:

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

Output:

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 = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int { \frac {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}}{x^{4} - a x^{2} + b} \,d x } \] Input:

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

Output:

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 = 16.42 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) \] Input:

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

Output:

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)*(s 
qrt(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 = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int -\frac {\left (b-a\,x^2\right )\,{\left (a\,x^4-b\,x^2\right )}^{1/4}}{x^4-a\,x^2+b} \,d x \] Input:

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

Output:

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

Reduce [N/A]

Not integrable

Time = 0.48 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=-\left (\int \frac {\sqrt {x}\, \left (a \,x^{2}-b \right )^{\frac {1}{4}} x^{2}}{-x^{4}+a \,x^{2}-b}d x \right ) a +\left (\int \frac {\sqrt {x}\, \left (a \,x^{2}-b \right )^{\frac {1}{4}}}{-x^{4}+a \,x^{2}-b}d x \right ) b \] Input:

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

Output:

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