\(\int \frac {(a-b x^2)^{5/4}}{(c+d x^2)^2} \, dx\) [464]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F(-1)]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 280 \[ \int \frac {\left (a-b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\frac {(b c+a d) x \sqrt [4]{a-b x^2}}{2 c d \left (c+d x^2\right )}+\frac {\sqrt {a} \sqrt {b} (3 b c-a d) \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{2 c d^2 \left (a-b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} (3 b c-2 a d) \sqrt {\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c+a d}},\arcsin \left (\frac {\sqrt [4]{a-b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c d^2 x}+\frac {\sqrt [4]{a} (3 b c-2 a d) \sqrt {\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c+a d}},\arcsin \left (\frac {\sqrt [4]{a-b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c d^2 x} \] Output:

1/2*(a*d+b*c)*x*(-b*x^2+a)^(1/4)/c/d/(d*x^2+c)+1/2*a^(1/2)*b^(1/2)*(-a*d+3 
*b*c)*(1-b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arcsin(b^(1/2)*x/a^(1/2)),2^(1 
/2))/c/d^2/(-b*x^2+a)^(3/4)+1/4*a^(1/4)*(-2*a*d+3*b*c)*(b*x^2/a)^(1/2)*Ell 
ipticPi((-b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d+b*c)^(1/2),I)/c/d^2 
/x+1/4*a^(1/4)*(-2*a*d+3*b*c)*(b*x^2/a)^(1/2)*EllipticPi((-b*x^2+a)^(1/4)/ 
a^(1/4),a^(1/2)*d^(1/2)/(a*d+b*c)^(1/2),I)/c/d^2/x
 

Mathematica [C] (warning: unable to verify)

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

Time = 10.26 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a-b x^2\right )^{5/4}}{\left (c+d x^2\right )^2} \, dx=\frac {x \left (-b (-3 b c+a d) x^2 \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {c \left (36 a c \left (2 a^2 d-b^2 c x^2-a b d x^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )-6 (b c+a d) x^2 \left (a-b x^2\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )-3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{\left (c+d x^2\right ) \left (6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (-4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{12 c^2 d \left (a-b x^2\right )^{3/4}} \] Input:

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

Output:

(x*(-(b*(-3*b*c + a*d)*x^2*(1 - (b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2 
, (b*x^2)/a, -((d*x^2)/c)]) + (c*(36*a*c*(2*a^2*d - b^2*c*x^2 - a*b*d*x^2) 
*AppellF1[1/2, 3/4, 1, 3/2, (b*x^2)/a, -((d*x^2)/c)] - 6*(b*c + a*d)*x^2*( 
a - b*x^2)*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, (b*x^2)/a, -((d*x^2)/c)] - 3* 
b*c*AppellF1[3/2, 7/4, 1, 5/2, (b*x^2)/a, -((d*x^2)/c)])))/((c + d*x^2)*(6 
*a*c*AppellF1[1/2, 3/4, 1, 3/2, (b*x^2)/a, -((d*x^2)/c)] + x^2*(-4*a*d*App 
ellF1[3/2, 3/4, 2, 5/2, (b*x^2)/a, -((d*x^2)/c)] + 3*b*c*AppellF1[3/2, 7/4 
, 1, 5/2, (b*x^2)/a, -((d*x^2)/c)])))))/(12*c^2*d*(a - b*x^2)^(3/4))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {315, 27, 405, 231, 230, 312, 118, 925, 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.

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 405

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

\(\Big \downarrow \) 231

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

\(\Big \downarrow \) 230

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

\(\Big \downarrow \) 312

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

\(\Big \downarrow \) 118

\(\displaystyle \frac {x \sqrt [4]{a-b x^2} (a d+b c)}{2 c d \left (c+d x^2\right )}-\frac {-\frac {2 \sqrt {\frac {b x^2}{a}} (3 b c-2 a d) (a d+b c) \int \frac {1}{\sqrt {1-\frac {x^8}{a}} \left (-d x^8+b c+a d\right )}d\sqrt [4]{a-b x^2}}{d x}-\frac {2 \sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} (3 b c-a d) \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a-b x^2\right )^{3/4}}}{4 c d}\)

\(\Big \downarrow \) 925

\(\displaystyle \frac {x \sqrt [4]{a-b x^2} (a d+b c)}{2 c d \left (c+d x^2\right )}-\frac {-\frac {2 \sqrt {\frac {b x^2}{a}} (3 b c-2 a d) (a d+b c) \left (\frac {\int \frac {1}{\left (1-\frac {\sqrt {d} x^4}{\sqrt {b c+a d}}\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{a-b x^2}}{2 (a d+b c)}+\frac {\int \frac {1}{\left (\frac {\sqrt {d} x^4}{\sqrt {b c+a d}}+1\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{a-b x^2}}{2 (a d+b c)}\right )}{d x}-\frac {2 \sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} (3 b c-a d) \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a-b x^2\right )^{3/4}}}{4 c d}\)

\(\Big \downarrow \) 1542

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

Input:

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

Output:

((b*c + a*d)*x*(a - b*x^2)^(1/4))/(2*c*d*(c + d*x^2)) - ((-2*Sqrt[a]*Sqrt[ 
b]*(3*b*c - a*d)*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a 
]]/2, 2])/(d*(a - b*x^2)^(3/4)) - (2*(3*b*c - 2*a*d)*(b*c + a*d)*Sqrt[(b*x 
^2)/a]*((a^(1/4)*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[b*c + a*d]), ArcSin[( 
a - b*x^2)^(1/4)/a^(1/4)], -1])/(2*(b*c + a*d)) + (a^(1/4)*EllipticPi[(Sqr 
t[a]*Sqrt[d])/Sqrt[b*c + a*d], ArcSin[(a - b*x^2)^(1/4)/a^(1/4)], -1])/(2* 
(b*c + a*d))))/(d*x))/(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 118
Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^( 
3/4)), x_] :> Simp[-4   Subst[Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + 
 d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e, f}, x] & 
& GtQ[-f/(d*e - c*f), 0]
 

rule 230
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] 
))*EllipticF[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[b/a]
 

rule 231
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( 
a + b*x^2)^(3/4)   Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

rule 312
Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim 
p[Sqrt[(-b)*(x^2/a)]/(2*x)   Subst[Int[1/(Sqrt[(-b)*(x/a)]*(a + b*x)^(3/4)* 
(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
 

rule 315
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 405
Int[(((a_) + (b_.)*(x_)^2)^(p_)*((e_) + (f_.)*(x_)^2))/((c_) + (d_.)*(x_)^2 
), x_Symbol] :> Simp[f/d   Int[(a + b*x^2)^p, x], x] + Simp[(d*e - c*f)/d 
 Int[(a + b*x^2)^p/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, p}, x]
 

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 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]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(4*(a - b*x**2)**(1/4)*a*b*x + 4*int((a - b*x**2)**(1/4)/(2*a**2*c**2*d + 
4*a**2*c*d**2*x**2 + 2*a**2*d**3*x**4 + 3*a*b*c**3 + 4*a*b*c**2*d*x**2 - a 
*b*c*d**2*x**4 - 2*a*b*d**3*x**6 - 3*b**2*c**3*x**2 - 6*b**2*c**2*d*x**4 - 
 3*b**2*c*d**2*x**6),x)*a**4*c*d**2 + 4*int((a - b*x**2)**(1/4)/(2*a**2*c* 
*2*d + 4*a**2*c*d**2*x**2 + 2*a**2*d**3*x**4 + 3*a*b*c**3 + 4*a*b*c**2*d*x 
**2 - a*b*c*d**2*x**4 - 2*a*b*d**3*x**6 - 3*b**2*c**3*x**2 - 6*b**2*c**2*d 
*x**4 - 3*b**2*c*d**2*x**6),x)*a**4*d**3*x**2 + 4*int((a - b*x**2)**(1/4)/ 
(2*a**2*c**2*d + 4*a**2*c*d**2*x**2 + 2*a**2*d**3*x**4 + 3*a*b*c**3 + 4*a* 
b*c**2*d*x**2 - a*b*c*d**2*x**4 - 2*a*b*d**3*x**6 - 3*b**2*c**3*x**2 - 6*b 
**2*c**2*d*x**4 - 3*b**2*c*d**2*x**6),x)*a**3*b*c**2*d + 4*int((a - b*x**2 
)**(1/4)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**2 + 2*a**2*d**3*x**4 + 3*a*b*c* 
*3 + 4*a*b*c**2*d*x**2 - a*b*c*d**2*x**4 - 2*a*b*d**3*x**6 - 3*b**2*c**3*x 
**2 - 6*b**2*c**2*d*x**4 - 3*b**2*c*d**2*x**6),x)*a**3*b*c*d**2*x**2 - 3*i 
nt((a - b*x**2)**(1/4)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**2 + 2*a**2*d**3*x 
**4 + 3*a*b*c**3 + 4*a*b*c**2*d*x**2 - a*b*c*d**2*x**4 - 2*a*b*d**3*x**6 - 
 3*b**2*c**3*x**2 - 6*b**2*c**2*d*x**4 - 3*b**2*c*d**2*x**6),x)*a**2*b**2* 
c**3 - 3*int((a - b*x**2)**(1/4)/(2*a**2*c**2*d + 4*a**2*c*d**2*x**2 + 2*a 
**2*d**3*x**4 + 3*a*b*c**3 + 4*a*b*c**2*d*x**2 - a*b*c*d**2*x**4 - 2*a*b*d 
**3*x**6 - 3*b**2*c**3*x**2 - 6*b**2*c**2*d*x**4 - 3*b**2*c*d**2*x**6),x)* 
a**2*b**2*c**2*d*x**2 + 6*int(((a - b*x**2)**(1/4)*x**4)/(2*a**2*c**2*d...