\(\int \frac {x^2 (c+d x^4)}{(a+b x^4)^{5/2}} \, dx\) [53]

Optimal result

Integrand size = 22, antiderivative size = 302 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {(b c-a d) x^3}{6 a b \left (a+b x^4\right )^{3/2}}+\frac {(b c+a d) x^3}{4 a^2 b \sqrt {a+b x^4}}-\frac {(b c+a d) x \sqrt {a+b x^4}}{4 a^2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {(b c+a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{7/4} b^{7/4} \sqrt {a+b x^4}}-\frac {(b c+a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{8 a^{7/4} b^{7/4} \sqrt {a+b x^4}} \] Output:

1/6*(-a*d+b*c)*x^3/a/b/(b*x^4+a)^(3/2)+1/4*(a*d+b*c)*x^3/a^2/b/(b*x^4+a)^( 
1/2)-1/4*(a*d+b*c)*x*(b*x^4+a)^(1/2)/a^2/b^(3/2)/(a^(1/2)+b^(1/2)*x^2)+1/4 
*(a*d+b*c)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2) 
*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/a^(7/4)/b^(7/4)/( 
b*x^4+a)^(1/2)-1/8*(a*d+b*c)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^( 
1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2) 
)/a^(7/4)/b^(7/4)/(b*x^4+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.26 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {x^3 \left (-a^2 d+(b c+a d) \left (a+b x^4\right ) \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{3 a^2 b \left (a+b x^4\right )^{3/2}} \] Input:

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

Output:

(x^3*(-(a^2*d) + (b*c + a*d)*(a + b*x^4)*Sqrt[1 + (b*x^4)/a]*Hypergeometri 
c2F1[3/4, 5/2, 7/4, -((b*x^4)/a)]))/(3*a^2*b*(a + b*x^4)^(3/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {957, 819, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 957

\(\displaystyle \frac {(a d+b c) \int \frac {x^2}{\left (b x^4+a\right )^{3/2}}dx}{2 a b}+\frac {x^3 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

\(\Big \downarrow \) 819

\(\displaystyle \frac {(a d+b c) \left (\frac {x^3}{2 a \sqrt {a+b x^4}}-\frac {\int \frac {x^2}{\sqrt {b x^4+a}}dx}{2 a}\right )}{2 a b}+\frac {x^3 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {(a d+b c) \left (\frac {x^3}{2 a \sqrt {a+b x^4}}-\frac {\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}}{2 a}\right )}{2 a b}+\frac {x^3 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a d+b c) \left (\frac {x^3}{2 a \sqrt {a+b x^4}}-\frac {\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}}{2 a}\right )}{2 a b}+\frac {x^3 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(a d+b c) \left (\frac {x^3}{2 a \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}}{2 a}\right )}{2 a b}+\frac {x^3 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {(a d+b c) \left (\frac {x^3}{2 a \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{\sqrt {b}}}{2 a}\right )}{2 a b}+\frac {x^3 (b c-a d)}{6 a b \left (a+b x^4\right )^{3/2}}\)

Input:

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

Output:

((b*c - a*d)*x^3)/(6*a*b*(a + b*x^4)^(3/2)) + ((b*c + a*d)*(x^3/(2*a*Sqrt[ 
a + b*x^4]) - (-((-((x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2)) + (a^(1/4 
)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elli 
pticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + b*x^4]))/Sqrt 
[b]) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b 
]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*b^(3/4)*Sqrt[a 
 + b*x^4]))/(2*a)))/(2*a*b)
 

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 761

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

rule 819

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( 
c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

rule 1510

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.90 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.58

method	result	size
elliptic	\(-\frac {x^{3} \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{6 a \,b^{3} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x^{3} \left (a d +c b \right )}{4 b \,a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \left (a d +c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{4 b^{\frac {3}{2}} a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(175\)
default	\(c \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{6 a \,b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x^{3}}{4 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{4 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+d \left (-\frac {x^{3} \sqrt {b \,x^{4}+a}}{6 b^{3} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x^{3}}{4 b a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{4 b^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(304\)

Input:

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

Output:

-1/6/a*x^3/b^3*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^2+1/4/b/a^2*x^3*(a*d+b* 
c)/((x^4+a/b)*b)^(1/2)-1/4*I/b^(3/2)/a^(3/2)*(a*d+b*c)/(I/a^(1/2)*b^(1/2)) 
^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b* 
x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1 
/2)*b^(1/2))^(1/2),I))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.74 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {3 \, {\left ({\left (b^{3} c + a b^{2} d\right )} x^{8} + 2 \, {\left (a b^{2} c + a^{2} b d\right )} x^{4} + a^{2} b c + a^{3} d\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 3 \, {\left ({\left (b^{3} c + a b^{2} d\right )} x^{8} + 2 \, {\left (a b^{2} c + a^{2} b d\right )} x^{4} + a^{2} b c + a^{3} d\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (3 \, {\left (b^{3} c + a b^{2} d\right )} x^{7} + {\left (5 \, a b^{2} c + a^{2} b d\right )} x^{3}\right )} \sqrt {b x^{4} + a}}{12 \, {\left (a^{2} b^{4} x^{8} + 2 \, a^{3} b^{3} x^{4} + a^{4} b^{2}\right )}} \] Input:

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

Output:

1/12*(3*((b^3*c + a*b^2*d)*x^8 + 2*(a*b^2*c + a^2*b*d)*x^4 + a^2*b*c + a^3 
*d)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^(1/4)), -1) - 3*((b^3* 
c + a*b^2*d)*x^8 + 2*(a*b^2*c + a^2*b*d)*x^4 + a^2*b*c + a^3*d)*sqrt(a)*(- 
b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) + (3*(b^3*c + a*b^2*d)*x 
^7 + (5*a*b^2*c + a^2*b*d)*x^3)*sqrt(b*x^4 + a))/(a^2*b^4*x^8 + 2*a^3*b^3* 
x^4 + a^4*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 19.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.26 \[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {c x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {d x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {5}{2} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {11}{4}\right )} \] Input:

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

Output:

c*x**3*gamma(3/4)*hyper((3/4, 5/2), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(5/2)*gamma(7/4)) + d*x**7*gamma(7/4)*hyper((7/4, 5/2), (11/4,), b*x**4* 
exp_polar(I*pi)/a)/(4*a**(5/2)*gamma(11/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^2 \left (c+d x^4\right )}{\left (a+b x^4\right )^{5/2}} \, dx=\frac {-\sqrt {b \,x^{4}+a}\, d \,x^{3}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{3} d +3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b c +6 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a^{2} b d \,x^{4}+6 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a \,b^{2} c \,x^{4}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) a \,b^{2} d \,x^{8}+3 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}}d x \right ) b^{3} c \,x^{8}}{3 b \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}\right )} \] Input:

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

Output:

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