\(\int \frac {-b+2 a x^4}{(b+a x^4) \sqrt [4]{b x^2+a x^4}} \, dx\) [1751]

Optimal result

Integrand size = 35, antiderivative size = 118 \[ \int \frac {-b+2 a x^4}{\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}}+\frac {3}{4} \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \] Output:

Unintegrable
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.31 \[ \int \frac {-b+2 a x^4}{\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {\sqrt {x} \sqrt [4]{b+a x^2} \left (8 \left (\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )+3 \sqrt [4]{a} \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right )+\log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]\right )}{4 \sqrt [4]{a} \sqrt [4]{x^2 \left (b+a x^2\right )}} \] Input:

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

Output:

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(338\) vs. \(2(118)=236\).

Time = 1.14 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.86, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2467, 25, 2035, 7276, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7276

\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \int \left (\frac {3 b}{\sqrt [4]{a x^2+b} \left (a x^4+b\right )}-\frac {2}{\sqrt [4]{a x^2+b}}\right )d\sqrt {x}}{\sqrt [4]{a x^4+b x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \left (-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a}}+\frac {3 \arctan \left (\frac {\sqrt {x} \sqrt [4]{a-\sqrt {-a} \sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{4 \sqrt [4]{a-\sqrt {-a} \sqrt {b}}}+\frac {3 \arctan \left (\frac {\sqrt {x} \sqrt [4]{\sqrt {-a} \sqrt {b}+a}}{\sqrt [4]{a x^2+b}}\right )}{4 \sqrt [4]{\sqrt {-a} \sqrt {b}+a}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {x} \sqrt [4]{a-\sqrt {-a} \sqrt {b}}}{\sqrt [4]{a x^2+b}}\right )}{4 \sqrt [4]{a-\sqrt {-a} \sqrt {b}}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {x} \sqrt [4]{\sqrt {-a} \sqrt {b}+a}}{\sqrt [4]{a x^2+b}}\right )}{4 \sqrt [4]{\sqrt {-a} \sqrt {b}+a}}\right )}{\sqrt [4]{a x^4+b x^2}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

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]
 

rule 7276

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

Maple [N/A] (verified)

Time = 0.42 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+a b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right ) a^{\frac {1}{4}}-8 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+4 \ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )}{4 a^{\frac {1}{4}}}\)	\(127\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [N/A]

Not integrable

Time = 5.88 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.25 \[ \int \frac {-b+2 a x^4}{\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {2 a x^{4} - b}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} + b\right )}\, dx \] Input:

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

Output:

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

Maxima [N/A]

Not integrable

Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.30 \[ \int \frac {-b+2 a x^4}{\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {2 \, a x^{4} - b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}} \,d x } \] Input:

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

Output:

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

Giac [N/A]

Not integrable

Time = 2.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.30 \[ \int \frac {-b+2 a x^4}{\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {2 \, a x^{4} - b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}} \,d x } \] Input:

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

Output:

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

Mupad [N/A]

Not integrable

Time = 8.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29 \[ \int \frac {-b+2 a x^4}{\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int -\frac {b-2\,a\,x^4}{\left (a\,x^4+b\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}} \,d x \] Input:

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

Output:

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

Reduce [N/A]

Not integrable

Time = 2.31 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \frac {-b+2 a x^4}{\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {2 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {5}{4}}+2 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a \,x^{2}+2 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} b -3 \sqrt {a \,x^{2}+b}\, \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {3}{4}}}{a^{2} x^{7}+a b \,x^{5}+a b \,x^{3}+b^{2} x}d x \right ) a b \,x^{2}-3 \sqrt {a \,x^{2}+b}\, \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {3}{4}}}{a^{2} x^{7}+a b \,x^{5}+a b \,x^{3}+b^{2} x}d x \right ) b^{2}}{\sqrt {a \,x^{2}+b}\, \left (a \,x^{2}+b \right )} \] Input:

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

Output:

(2*sqrt(x)*(a*x**2 + b)**(5/4) + 2*sqrt(x)*(a*x**2 + b)**(1/4)*a*x**2 + 2* 
sqrt(x)*(a*x**2 + b)**(1/4)*b - 3*sqrt(a*x**2 + b)*int((sqrt(x)*(a*x**2 + 
b)**(3/4))/(a**2*x**7 + a*b*x**5 + a*b*x**3 + b**2*x),x)*a*b*x**2 - 3*sqrt 
(a*x**2 + b)*int((sqrt(x)*(a*x**2 + b)**(3/4))/(a**2*x**7 + a*b*x**5 + a*b 
*x**3 + b**2*x),x)*b**2)/(sqrt(a*x**2 + b)*(a*x**2 + b))