\(\int \frac {(a+b x^4)^{5/4}}{(c+d x^4)^2} \, dx\) [122]

Optimal result

Integrand size = 21, antiderivative size = 298 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx=-\frac {(b c-a d) x \sqrt [4]{a+b x^4}}{4 c d \left (c+d x^4\right )}+\frac {\sqrt {a} b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{4 c d \left (a+b x^4\right )^{3/4}}+\frac {(2 b c+3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 d}+\frac {(2 b c+3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 d} \] Output:

-1/4*(-a*d+b*c)*x*(b*x^4+a)^(1/4)/c/d/(d*x^4+c)+1/4*a^(1/2)*b^(3/2)*(1+a/b 
/x^4)^(3/4)*x^3*InverseJacobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/c 
/d/(b*x^4+a)^(3/4)+1/8*(3*a*d+2*b*c)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1/2)*E 
llipticPi(b^(1/4)*x/(b*x^4+a)^(1/4),-(-a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)/b 
^(1/4)/c^2/d+1/8*(3*a*d+2*b*c)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1/2)*Ellipti 
cPi(b^(1/4)*x/(b*x^4+a)^(1/4),(-a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)/b^(1/4)/ 
c^2/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 10.37 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \left (2 b (b c+a d) x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+\frac {5 c \left (-5 a c \left (4 a^2 d-b^2 c x^4+a b d x^4\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+(-b c+a d) x^4 \left (a+b x^4\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}{\left (c+d x^4\right ) \left (-5 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+x^4 \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}\right )}{20 c^2 d \left (a+b x^4\right )^{3/4}} \] Input:

Integrate[(a + b*x^4)^(5/4)/(c + d*x^4)^2,x]
 

Output:

(x*(2*b*(b*c + a*d)*x^4*(1 + (b*x^4)/a)^(3/4)*AppellF1[5/4, 3/4, 1, 9/4, - 
((b*x^4)/a), -((d*x^4)/c)] + (5*c*(-5*a*c*(4*a^2*d - b^2*c*x^4 + a*b*d*x^4 
)*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + (-(b*c) + a*d)* 
x^4*(a + b*x^4)*(4*a*d*AppellF1[5/4, 3/4, 2, 9/4, -((b*x^4)/a), -((d*x^4)/ 
c)] + 3*b*c*AppellF1[5/4, 7/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))/((c 
+ d*x^4)*(-5*a*c*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + 
x^4*(4*a*d*AppellF1[5/4, 3/4, 2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + 3*b*c* 
AppellF1[5/4, 7/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))))/(20*c^2*d*(a + 
 b*x^4)^(3/4))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.88, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {930, 404, 768, 858, 807, 229, 923, 925, 27, 1542}

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.

\(\displaystyle \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx\)

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 404

\(\displaystyle \frac {(3 a d+2 b c) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx-a b \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {(3 a d+2 b c) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx-\frac {a b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{\left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {(3 a d+2 b c) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx+\frac {a b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{\left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(3 a d+2 b c) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx+\frac {a b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{2 \left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {(3 a d+2 b c) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx+\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 923

\(\displaystyle \frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (3 a d+2 b c) \int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (c-\frac {(b c-a d) x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 925

\(\displaystyle \frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (3 a d+2 b c) \left (\frac {\int \frac {\sqrt {c}}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 c}+\frac {\int \frac {\sqrt {c}}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 c}\right )+\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (3 a d+2 b c) \left (\frac {\int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )+\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

\(\Big \downarrow \) 1542

\(\displaystyle \frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (3 a d+2 b c) \left (\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c}+\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c}\right )+\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4}}}{4 c d}-\frac {x \sqrt [4]{a+b x^4} (b c-a d)}{4 c d \left (c+d x^4\right )}\)

Input:

Int[(a + b*x^4)^(5/4)/(c + d*x^4)^2,x]
 

Output:

-1/4*((b*c - a*d)*x*(a + b*x^4)^(1/4))/(c*d*(c + d*x^4)) + ((Sqrt[a]*b^(3/ 
2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2] 
)/(a + b*x^4)^(3/4) + (2*b*c + 3*a*d)*Sqrt[a/(a + b*x^4)]*Sqrt[a + b*x^4]* 
(EllipticPi[-(Sqrt[b*c - a*d]/(Sqrt[b]*Sqrt[c])), ArcSin[(b^(1/4)*x)/(a + 
b*x^4)^(1/4)], -1]/(2*b^(1/4)*c) + EllipticPi[Sqrt[b*c - a*d]/(Sqrt[b]*Sqr 
t[c]), ArcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1]/(2*b^(1/4)*c)))/(4*c*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 404

Int[((e_) + (f_.)*(x_)^4)/(((a_) + (b_.)*(x_)^4)^(3/4)*((c_) + (d_.)*(x_)^4 
)), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^4)^(3/4), x] 
, x] - Simp[(d*e - c*f)/(b*c - a*d)   Int[(a + b*x^4)^(1/4)/(c + d*x^4), x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 923

Int[((a_) + (b_.)*(x_)^4)^(1/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Simp[Sq 
rt[a + b*x^4]*Sqrt[a/(a + b*x^4)]   Subst[Int[1/(Sqrt[1 - b*x^4]*(c - (b*c 
- a*d)*x^4)), x], x, x/(a + b*x^4)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[b*c - a*d, 0]
 

rule 925

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Simp[ 
1/(2*c)   Int[1/(Sqrt[a + b*x^4]*(1 - Rt[-d/c, 2]*x^2)), x], x] + Simp[1/(2 
*c)   Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a, b, 
 c, d}, x] && NeQ[b*c - a*d, 0]
 

rule 930

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(a*d - c*b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 
1))), x] - Simp[1/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q 
- 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*( 
p + q) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]
 

rule 1542

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x 
], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
 

Maple [F]

\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{\left (d \,x^{4}+c \right )^{2}}d x\]

Input:

int((b*x^4+a)^(5/4)/(d*x^4+c)^2,x)
 

Output:

int((b*x^4+a)^(5/4)/(d*x^4+c)^2,x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((b*x^4+a)^(5/4)/(d*x^4+c)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {\left (a + b x^{4}\right )^{\frac {5}{4}}}{\left (c + d x^{4}\right )^{2}}\, dx \] Input:

integrate((b*x**4+a)**(5/4)/(d*x**4+c)**2,x)
 

Output:

Integral((a + b*x**4)**(5/4)/(c + d*x**4)**2, x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:

integrate((b*x^4+a)^(5/4)/(d*x^4+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^(5/4)/(d*x^4 + c)^2, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:

integrate((b*x^4+a)^(5/4)/(d*x^4+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^4 + a)^(5/4)/(d*x^4 + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{5/4}}{{\left (d\,x^4+c\right )}^2} \,d x \] Input:

int((a + b*x^4)^(5/4)/(c + d*x^4)^2,x)
                                                                                    
                                                                                    
 

Output:

int((a + b*x^4)^(5/4)/(c + d*x^4)^2, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^4\right )^{5/4}}{\left (c+d x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int((b*x^4+a)^(5/4)/(d*x^4+c)^2,x)
 

Output:

( - 2*(a + b*x**4)**(1/4)*a*b*x + 9*int((a + b*x**4)**(1/4)/(3*a**2*c**2*d 
 + 6*a**2*c*d**2*x**4 + 3*a**2*d**3*x**8 - 2*a*b*c**3 - a*b*c**2*d*x**4 + 
4*a*b*c*d**2*x**8 + 3*a*b*d**3*x**12 - 2*b**2*c**3*x**4 - 4*b**2*c**2*d*x* 
*8 - 2*b**2*c*d**2*x**12),x)*a**4*c*d**2 + 9*int((a + b*x**4)**(1/4)/(3*a* 
*2*c**2*d + 6*a**2*c*d**2*x**4 + 3*a**2*d**3*x**8 - 2*a*b*c**3 - a*b*c**2* 
d*x**4 + 4*a*b*c*d**2*x**8 + 3*a*b*d**3*x**12 - 2*b**2*c**3*x**4 - 4*b**2* 
c**2*d*x**8 - 2*b**2*c*d**2*x**12),x)*a**4*d**3*x**4 - 6*int((a + b*x**4)* 
*(1/4)/(3*a**2*c**2*d + 6*a**2*c*d**2*x**4 + 3*a**2*d**3*x**8 - 2*a*b*c**3 
 - a*b*c**2*d*x**4 + 4*a*b*c*d**2*x**8 + 3*a*b*d**3*x**12 - 2*b**2*c**3*x* 
*4 - 4*b**2*c**2*d*x**8 - 2*b**2*c*d**2*x**12),x)*a**3*b*c**2*d - 6*int((a 
 + b*x**4)**(1/4)/(3*a**2*c**2*d + 6*a**2*c*d**2*x**4 + 3*a**2*d**3*x**8 - 
 2*a*b*c**3 - a*b*c**2*d*x**4 + 4*a*b*c*d**2*x**8 + 3*a*b*d**3*x**12 - 2*b 
**2*c**3*x**4 - 4*b**2*c**2*d*x**8 - 2*b**2*c*d**2*x**12),x)*a**3*b*c*d**2 
*x**4 - 3*int(((a + b*x**4)**(1/4)*x**8)/(3*a**2*c**2*d + 6*a**2*c*d**2*x* 
*4 + 3*a**2*d**3*x**8 - 2*a*b*c**3 - a*b*c**2*d*x**4 + 4*a*b*c*d**2*x**8 + 
 3*a*b*d**3*x**12 - 2*b**2*c**3*x**4 - 4*b**2*c**2*d*x**8 - 2*b**2*c*d**2* 
x**12),x)*a**2*b**2*c*d**2 - 3*int(((a + b*x**4)**(1/4)*x**8)/(3*a**2*c**2 
*d + 6*a**2*c*d**2*x**4 + 3*a**2*d**3*x**8 - 2*a*b*c**3 - a*b*c**2*d*x**4 
+ 4*a*b*c*d**2*x**8 + 3*a*b*d**3*x**12 - 2*b**2*c**3*x**4 - 4*b**2*c**2*d* 
x**8 - 2*b**2*c*d**2*x**12),x)*a**2*b**2*d**3*x**4 - 4*int(((a + b*x**4...