\(\int \frac {\sqrt [4]{b x^2+a x^4} (b+a x^4+x^8)}{b+a x^4} \, dx\) [2624]

Optimal result

Integrand size = 35, antiderivative size = 230 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (b+a x^4+x^8\right )}{b+a x^4} \, dx=\frac {\sqrt [4]{b x^2+a x^4} \left (96 a^3 x-96 a b x-7 b^2 x+4 a b x^3+32 a^2 x^5\right )}{192 a^3}+\frac {\left (-32 a^3 b+32 a b^2-7 b^3\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{128 a^{15/4}}+\frac {\left (32 a^3 b-32 a b^2+7 b^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{128 a^{15/4}}-\frac {b^2 \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]}{4 a^2} \] Output:

Unintegrable
 

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (b+a x^4+x^8\right )}{b+a x^4} \, dx=\frac {x^{3/2} \left (b+a x^2\right )^{3/4} \left (2 a^{3/4} x^{3/2} \sqrt [4]{b+a x^2} \left (96 a^3-7 b^2+32 a^2 x^4+4 a b \left (-24+x^2\right )\right )-3 b \left (32 a^3-32 a b+7 b^2\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+3 b \left (32 a^3-32 a b+7 b^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )-96 a^{7/4} b^2 \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right ) \text {$\#$1}+\log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]\right )}{384 a^{15/4} \left (x^2 \left (b+a x^2\right )\right )^{3/4}} \] Input:

Integrate[((b*x^2 + a*x^4)^(1/4)*(b + a*x^4 + x^8))/(b + a*x^4),x]
 

Output:

(x^(3/2)*(b + a*x^2)^(3/4)*(2*a^(3/4)*x^(3/2)*(b + a*x^2)^(1/4)*(96*a^3 - 
7*b^2 + 32*a^2*x^4 + 4*a*b*(-24 + x^2)) - 3*b*(32*a^3 - 32*a*b + 7*b^2)*Ar 
cTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] + 3*b*(32*a^3 - 32*a*b + 7*b^2)* 
ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] - 96*a^(7/4)*b^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) & ]))/(384*a^(15/4)*(x^2*(b + a*x^2))^(3/4))
 

Rubi [A] (warning: unable to verify)

Time = 1.34 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2467, 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.

\(\displaystyle \int \frac {\sqrt [4]{a x^4+b x^2} \left (a x^4+b+x^8\right )}{a x^4+b} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [4]{a x^4+b x^2} \int \frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (x^8+a x^4+b\right )}{a x^4+b}dx}{\sqrt {x} \sqrt [4]{a x^2+b}}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \int \frac {x \sqrt [4]{a x^2+b} \left (x^8+a x^4+b\right )}{a x^4+b}d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2+b}}\)

\(\Big \downarrow \) 7276

\(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \int \left (\frac {\sqrt [4]{a x^2+b} x^5}{a}+\left (1-\frac {b}{a^2}\right ) \sqrt [4]{a x^2+b} x+\frac {b^2 \sqrt [4]{a x^2+b} x}{a^2 \left (a x^4+b\right )}\right )d\sqrt {x}}{\sqrt {x} \sqrt [4]{a x^2+b}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \sqrt [4]{a x^4+b x^2} \left (-\frac {7 b^3 \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{256 a^{15/4}}+\frac {7 b^3 \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{256 a^{15/4}}-\frac {7 b^2 x^{3/2} \sqrt [4]{a x^2+b}}{384 a^3}+\frac {b x^{3/2} \sqrt [4]{a x^2+b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {\sqrt {-a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{6 a^2 \sqrt [4]{\frac {a x^2}{b}+1}}+\frac {b x^{3/2} \sqrt [4]{a x^2+b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {\sqrt {-a} x^2}{\sqrt {b}},-\frac {a x^2}{b}\right )}{6 a^2 \sqrt [4]{\frac {a x^2}{b}+1}}+\frac {b x^{7/2} \sqrt [4]{a x^2+b}}{96 a^2}+\frac {1}{4} x^{3/2} \left (1-\frac {b}{a^2}\right ) \sqrt [4]{a x^2+b}-\frac {b \left (a^2-b\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{8 a^{11/4}}+\frac {b \left (a^2-b\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{8 a^{11/4}}+\frac {x^{11/2} \sqrt [4]{a x^2+b}}{12 a}\right )}{\sqrt {x} \sqrt [4]{a x^2+b}}\)

Input:

Int[((b*x^2 + a*x^4)^(1/4)*(b + a*x^4 + x^8))/(b + a*x^4),x]
 

Output:

(2*(b*x^2 + a*x^4)^(1/4)*((-7*b^2*x^(3/2)*(b + a*x^2)^(1/4))/(384*a^3) + ( 
(1 - b/a^2)*x^(3/2)*(b + a*x^2)^(1/4))/4 + (b*x^(7/2)*(b + a*x^2)^(1/4))/( 
96*a^2) + (x^(11/2)*(b + a*x^2)^(1/4))/(12*a) + (b*x^(3/2)*(b + a*x^2)^(1/ 
4)*AppellF1[3/4, 1, -1/4, 7/4, -((Sqrt[-a]*x^2)/Sqrt[b]), -((a*x^2)/b)])/( 
6*a^2*(1 + (a*x^2)/b)^(1/4)) + (b*x^(3/2)*(b + a*x^2)^(1/4)*AppellF1[3/4, 
1, -1/4, 7/4, (Sqrt[-a]*x^2)/Sqrt[b], -((a*x^2)/b)])/(6*a^2*(1 + (a*x^2)/b 
)^(1/4)) - ((a^2 - b)*b*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(8*a^ 
(11/4)) - (7*b^3*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(256*a^(15/4 
)) + ((a^2 - b)*b*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(8*a^(11/4 
)) + (7*b^3*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(256*a^(15/4)))) 
/(Sqrt[x]*(b + a*x^2)^(1/4))
 

Defintions of rubi rules used

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 = 0.94 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(\frac {-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+a b \right )}{\sum }\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{4}-a}\right ) a^{\frac {7}{4}} b^{2}+\frac {b \left (a^{3}-a b +\frac {7}{32} b^{2}\right ) \ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )}{2}+b \left (a^{3}-a b +\frac {7}{32} b^{2}\right ) \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\frac {2 \left (b \left (\frac {x^{2}}{8}-3\right ) a^{\frac {7}{4}}+a^{\frac {11}{4}} x^{4}-\frac {7 b^{2} a^{\frac {3}{4}}}{32}+3 a^{\frac {15}{4}}\right ) x \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{3}}{4 a^{\frac {15}{4}}}\)	\(213\)

Input:

int((a*x^4+b*x^2)^(1/4)*(x^8+a*x^4+b)/(a*x^4+b),x,method=_RETURNVERBOSE)
 

Output:

1/4/a^(15/4)*(-sum(_R*ln((-_R*x+(x^2*(a*x^2+b))^(1/4))/x)/(_R^4-a),_R=Root 
Of(_Z^8-2*_Z^4*a+a^2+a*b))*a^(7/4)*b^2+1/2*b*(a^3-a*b+7/32*b^2)*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)))+b*(a^3-a*b+ 
7/32*b^2)*arctan(1/a^(1/4)/x*(x^2*(a*x^2+b))^(1/4))+2/3*(b*(1/8*x^2-3)*a^( 
7/4)+a^(11/4)*x^4-7/32*b^2*a^(3/4)+3*a^(15/4))*x*(x^2*(a*x^2+b))^(1/4))
 

Fricas [F(-1)]

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

integrate((a*x^4+b*x^2)^(1/4)*(x^8+a*x^4+b)/(a*x^4+b),x, algorithm="fricas 
")
 

Output:

Timed out
 

Sympy [N/A]

Not integrable

Time = 40.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.13 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (b+a x^4+x^8\right )}{b+a x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} + b + x^{8}\right )}{a x^{4} + b}\, dx \] Input:

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

Output:

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

Maxima [N/A]

Not integrable

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.15 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (b+a x^4+x^8\right )}{b+a x^4} \, dx=\int { \frac {{\left (x^{8} + a x^{4} + b\right )} {\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}}}{a x^{4} + b} \,d x } \] Input:

integrate((a*x^4+b*x^2)^(1/4)*(x^8+a*x^4+b)/(a*x^4+b),x, algorithm="maxima 
")
 

Output:

integrate((x^8 + a*x^4 + b)*(a*x^4 + b*x^2)^(1/4)/(a*x^4 + b), x)
 

Giac [F(-1)]

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

integrate((a*x^4+b*x^2)^(1/4)*(x^8+a*x^4+b)/(a*x^4+b),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [N/A]

Not integrable

Time = 8.49 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.15 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (b+a x^4+x^8\right )}{b+a x^4} \, dx=\int \frac {{\left (a\,x^4+b\,x^2\right )}^{1/4}\,\left (x^8+a\,x^4+b\right )}{a\,x^4+b} \,d x \] Input:

int(((a*x^4 + b*x^2)^(1/4)*(b + a*x^4 + x^8))/(b + a*x^4),x)
 

Output:

int(((a*x^4 + b*x^2)^(1/4)*(b + a*x^4 + x^8))/(b + a*x^4), x)
 

Reduce [N/A]

Not integrable

Time = 1.54 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt [4]{b x^2+a x^4} \left (b+a x^4+x^8\right )}{b+a x^4} \, dx=\frac {192 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a^{3} x +64 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a^{2} x^{5}+8 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a b \,x^{3}-192 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} a b x -14 \sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} b^{2} x +96 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{4}}{a^{2} x^{6}+a b \,x^{4}+a b \,x^{2}+b^{2}}d x \right ) a^{4} b -96 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{4}}{a^{2} x^{6}+a b \,x^{4}+a b \,x^{2}+b^{2}}d x \right ) a^{2} b^{2}+21 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{4}}{a^{2} x^{6}+a b \,x^{4}+a b \,x^{2}+b^{2}}d x \right ) a \,b^{3}+384 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}} x^{2}}{a^{2} x^{6}+a b \,x^{4}+a b \,x^{2}+b^{2}}d x \right ) a^{2} b^{2}+96 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}}}{a^{2} x^{6}+a b \,x^{4}+a b \,x^{2}+b^{2}}d x \right ) a^{3} b^{2}+288 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}}}{a^{2} x^{6}+a b \,x^{4}+a b \,x^{2}+b^{2}}d x \right ) a \,b^{3}+21 \left (\int \frac {\sqrt {x}\, \left (a \,x^{2}+b \right )^{\frac {1}{4}}}{a^{2} x^{6}+a b \,x^{4}+a b \,x^{2}+b^{2}}d x \right ) b^{4}}{384 a^{3}} \] Input:

int((a*x^4+b*x^2)^(1/4)*(x^8+a*x^4+b)/(a*x^4+b),x)
 

Output:

(192*sqrt(x)*(a*x**2 + b)**(1/4)*a**3*x + 64*sqrt(x)*(a*x**2 + b)**(1/4)*a 
**2*x**5 + 8*sqrt(x)*(a*x**2 + b)**(1/4)*a*b*x**3 - 192*sqrt(x)*(a*x**2 + 
b)**(1/4)*a*b*x - 14*sqrt(x)*(a*x**2 + b)**(1/4)*b**2*x + 96*int((sqrt(x)* 
(a*x**2 + b)**(1/4)*x**4)/(a**2*x**6 + a*b*x**4 + a*b*x**2 + b**2),x)*a**4 
*b - 96*int((sqrt(x)*(a*x**2 + b)**(1/4)*x**4)/(a**2*x**6 + a*b*x**4 + a*b 
*x**2 + b**2),x)*a**2*b**2 + 21*int((sqrt(x)*(a*x**2 + b)**(1/4)*x**4)/(a* 
*2*x**6 + a*b*x**4 + a*b*x**2 + b**2),x)*a*b**3 + 384*int((sqrt(x)*(a*x**2 
 + b)**(1/4)*x**2)/(a**2*x**6 + a*b*x**4 + a*b*x**2 + b**2),x)*a**2*b**2 + 
 96*int((sqrt(x)*(a*x**2 + b)**(1/4))/(a**2*x**6 + a*b*x**4 + a*b*x**2 + b 
**2),x)*a**3*b**2 + 288*int((sqrt(x)*(a*x**2 + b)**(1/4))/(a**2*x**6 + a*b 
*x**4 + a*b*x**2 + b**2),x)*a*b**3 + 21*int((sqrt(x)*(a*x**2 + b)**(1/4))/ 
(a**2*x**6 + a*b*x**4 + a*b*x**2 + b**2),x)*b**4)/(384*a**3)