\(\int \frac {x^3 (a+b x^2+c x^4)}{\sqrt {d-e x} \sqrt {d+e x}} \, dx\) [18]

Optimal result

Integrand size = 35, antiderivative size = 159 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {d^2 \left (c d^4+b d^2 e^2+a e^4\right ) \sqrt {d-e x} \sqrt {d+e x}}{e^8}+\frac {\left (3 c d^4+2 b d^2 e^2+a e^4\right ) (d-e x)^{3/2} (d+e x)^{3/2}}{3 e^8}-\frac {\left (3 c d^2+b e^2\right ) (d-e x)^{5/2} (d+e x)^{5/2}}{5 e^8}+\frac {c (d-e x)^{7/2} (d+e x)^{7/2}}{7 e^8} \] Output:

-d^2*(a*e^4+b*d^2*e^2+c*d^4)*(-e*x+d)^(1/2)*(e*x+d)^(1/2)/e^8+1/3*(a*e^4+2 
*b*d^2*e^2+3*c*d^4)*(-e*x+d)^(3/2)*(e*x+d)^(3/2)/e^8-1/5*(b*e^2+3*c*d^2)*( 
-e*x+d)^(5/2)*(e*x+d)^(5/2)/e^8+1/7*c*(-e*x+d)^(7/2)*(e*x+d)^(7/2)/e^8
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {\sqrt {d-e x} \sqrt {d+e x} \left (35 a e^4 \left (2 d^2+e^2 x^2\right )+7 b \left (8 d^4 e^2+4 d^2 e^4 x^2+3 e^6 x^4\right )+3 c \left (16 d^6+8 d^4 e^2 x^2+6 d^2 e^4 x^4+5 e^6 x^6\right )\right )}{105 e^8} \] Input:

Integrate[(x^3*(a + b*x^2 + c*x^4))/(Sqrt[d - e*x]*Sqrt[d + e*x]),x]
 

Output:

-1/105*(Sqrt[d - e*x]*Sqrt[d + e*x]*(35*a*e^4*(2*d^2 + e^2*x^2) + 7*b*(8*d 
^4*e^2 + 4*d^2*e^4*x^2 + 3*e^6*x^4) + 3*c*(16*d^6 + 8*d^4*e^2*x^2 + 6*d^2* 
e^4*x^4 + 5*e^6*x^6)))/e^8
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1905, 1578, 1195, 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.

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

\(\Big \downarrow \) 1905

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

\(\Big \downarrow \) 1578

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

\(\Big \downarrow \) 1195

\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \int \left (-\frac {c \left (d^2-e^2 x^2\right )^{5/2}}{e^6}+\frac {\left (3 c d^2+b e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{e^6}+\frac {\left (-3 c d^4-2 b e^2 d^2-a e^4\right ) \sqrt {d^2-e^2 x^2}}{e^6}+\frac {c d^6+b e^2 d^4+a e^4 d^2}{e^6 \sqrt {d^2-e^2 x^2}}\right )dx^2}{2 \sqrt {d-e x} \sqrt {d+e x}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {d^2-e^2 x^2} \left (\frac {2 \left (d^2-e^2 x^2\right )^{3/2} \left (a e^4+2 b d^2 e^2+3 c d^4\right )}{3 e^8}-\frac {2 d^2 \sqrt {d^2-e^2 x^2} \left (a e^4+b d^2 e^2+c d^4\right )}{e^8}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2} \left (b e^2+3 c d^2\right )}{5 e^8}+\frac {2 c \left (d^2-e^2 x^2\right )^{7/2}}{7 e^8}\right )}{2 \sqrt {d-e x} \sqrt {d+e x}}\)

Input:

Int[(x^3*(a + b*x^2 + c*x^4))/(Sqrt[d - e*x]*Sqrt[d + e*x]),x]
 

Output:

(Sqrt[d^2 - e^2*x^2]*((-2*d^2*(c*d^4 + b*d^2*e^2 + a*e^4)*Sqrt[d^2 - e^2*x 
^2])/e^8 + (2*(3*c*d^4 + 2*b*d^2*e^2 + a*e^4)*(d^2 - e^2*x^2)^(3/2))/(3*e^ 
8) - (2*(3*c*d^2 + b*e^2)*(d^2 - e^2*x^2)^(5/2))/(5*e^8) + (2*c*(d^2 - e^2 
*x^2)^(7/2))/(7*e^8)))/(2*Sqrt[d - e*x]*Sqrt[d + e*x])
 

Defintions of rubi rules used

rule 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 

rule 1905

Int[((f_.)*(x_))^(m_.)*((d1_) + (e1_.)*(x_)^(non2_.))^(q_.)*((d2_) + (e2_.) 
*(x_)^(non2_.))^(q_.)*((a_.) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_.), x 
_Symbol] :> Simp[(d1 + e1*x^(n/2))^FracPart[q]*((d2 + e2*x^(n/2))^FracPart[ 
q]/(d1*d2 + e1*e2*x^n)^FracPart[q])   Int[(f*x)^m*(d1*d2 + e1*e2*x^n)^q*(a 
+ b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, f, n, p, 
q}, x] && EqQ[n2, 2*n] && EqQ[non2, n/2] && EqQ[d2*e1 + d1*e2, 0]
 

rule 2009

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

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (15 c \,x^{6} e^{6}+21 b \,e^{6} x^{4}+18 c \,d^{2} e^{4} x^{4}+35 a \,e^{6} x^{2}+28 b \,d^{2} e^{4} x^{2}+24 c \,d^{4} e^{2} x^{2}+70 a \,d^{2} e^{4}+56 b \,d^{4} e^{2}+48 c \,d^{6}\right )}{105 e^{8}}\)	\(109\)
default	\(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (15 c \,x^{6} e^{6}+21 b \,e^{6} x^{4}+18 c \,d^{2} e^{4} x^{4}+35 a \,e^{6} x^{2}+28 b \,d^{2} e^{4} x^{2}+24 c \,d^{4} e^{2} x^{2}+70 a \,d^{2} e^{4}+56 b \,d^{4} e^{2}+48 c \,d^{6}\right )}{105 e^{8}}\)	\(109\)
risch	\(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (15 c \,x^{6} e^{6}+21 b \,e^{6} x^{4}+18 c \,d^{2} e^{4} x^{4}+35 a \,e^{6} x^{2}+28 b \,d^{2} e^{4} x^{2}+24 c \,d^{4} e^{2} x^{2}+70 a \,d^{2} e^{4}+56 b \,d^{4} e^{2}+48 c \,d^{6}\right )}{105 e^{8}}\)	\(109\)
orering	\(-\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (15 c \,x^{6} e^{6}+21 b \,e^{6} x^{4}+18 c \,d^{2} e^{4} x^{4}+35 a \,e^{6} x^{2}+28 b \,d^{2} e^{4} x^{2}+24 c \,d^{4} e^{2} x^{2}+70 a \,d^{2} e^{4}+56 b \,d^{4} e^{2}+48 c \,d^{6}\right )}{105 e^{8}}\)	\(109\)

Input:

int(x^3*(c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x,method=_RETURNVERBO 
SE)
 

Output:

-1/105*(e*x+d)^(1/2)*(-e*x+d)^(1/2)*(15*c*e^6*x^6+21*b*e^6*x^4+18*c*d^2*e^ 
4*x^4+35*a*e^6*x^2+28*b*d^2*e^4*x^2+24*c*d^4*e^2*x^2+70*a*d^2*e^4+56*b*d^4 
*e^2+48*c*d^6)/e^8
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.65 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {{\left (15 \, c e^{6} x^{6} + 48 \, c d^{6} + 56 \, b d^{4} e^{2} + 70 \, a d^{2} e^{4} + 3 \, {\left (6 \, c d^{2} e^{4} + 7 \, b e^{6}\right )} x^{4} + {\left (24 \, c d^{4} e^{2} + 28 \, b d^{2} e^{4} + 35 \, a e^{6}\right )} x^{2}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{105 \, e^{8}} \] Input:

integrate(x^3*(c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="f 
ricas")
 

Output:

-1/105*(15*c*e^6*x^6 + 48*c*d^6 + 56*b*d^4*e^2 + 70*a*d^2*e^4 + 3*(6*c*d^2 
*e^4 + 7*b*e^6)*x^4 + (24*c*d^4*e^2 + 28*b*d^2*e^4 + 35*a*e^6)*x^2)*sqrt(e 
*x + d)*sqrt(-e*x + d)/e^8
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 13.02 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.31 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=- \frac {i a d^{3} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{4}} - \frac {a d^{3} {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{4}} - \frac {i b d^{5} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {9}{4}, - \frac {7}{4} & -2, -2, - \frac {3}{2}, 1 \\- \frac {5}{2}, - \frac {9}{4}, -2, - \frac {7}{4}, - \frac {3}{2}, 0 & \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{6}} - \frac {b d^{5} {G_{6, 6}^{2, 6}\left (\begin {matrix} -3, - \frac {11}{4}, - \frac {5}{2}, - \frac {9}{4}, -2, 1 & \\- \frac {11}{4}, - \frac {9}{4} & -3, - \frac {5}{2}, - \frac {5}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{6}} - \frac {i c d^{7} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {13}{4}, - \frac {11}{4} & -3, -3, - \frac {5}{2}, 1 \\- \frac {7}{2}, - \frac {13}{4}, -3, - \frac {11}{4}, - \frac {5}{2}, 0 & \end {matrix} \middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{8}} - \frac {c d^{7} {G_{6, 6}^{2, 6}\left (\begin {matrix} -4, - \frac {15}{4}, - \frac {7}{2}, - \frac {13}{4}, -3, 1 & \\- \frac {15}{4}, - \frac {13}{4} & -4, - \frac {7}{2}, - \frac {7}{2}, 0 \end {matrix} \middle | {\frac {d^{2} e^{- 2 i \pi }}{e^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{8}} \] Input:

integrate(x**3*(c*x**4+b*x**2+a)/(-e*x+d)**(1/2)/(e*x+d)**(1/2),x)
 

Output:

-I*a*d**3*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)), ((-3/2, -5/4, -1, -3/ 
4, -1/2, 0), ()), d**2/(e**2*x**2))/(4*pi**(3/2)*e**4) - a*d**3*meijerg((( 
-2, -7/4, -3/2, -5/4, -1, 1), ()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), d* 
*2*exp_polar(-2*I*pi)/(e**2*x**2))/(4*pi**(3/2)*e**4) - I*b*d**5*meijerg(( 
(-9/4, -7/4), (-2, -2, -3/2, 1)), ((-5/2, -9/4, -2, -7/4, -3/2, 0), ()), d 
**2/(e**2*x**2))/(4*pi**(3/2)*e**6) - b*d**5*meijerg(((-3, -11/4, -5/2, -9 
/4, -2, 1), ()), ((-11/4, -9/4), (-3, -5/2, -5/2, 0)), d**2*exp_polar(-2*I 
*pi)/(e**2*x**2))/(4*pi**(3/2)*e**6) - I*c*d**7*meijerg(((-13/4, -11/4), ( 
-3, -3, -5/2, 1)), ((-7/2, -13/4, -3, -11/4, -5/2, 0), ()), d**2/(e**2*x** 
2))/(4*pi**(3/2)*e**8) - c*d**7*meijerg(((-4, -15/4, -7/2, -13/4, -3, 1), 
()), ((-15/4, -13/4), (-4, -7/2, -7/2, 0)), d**2*exp_polar(-2*I*pi)/(e**2* 
x**2))/(4*pi**(3/2)*e**8)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.36 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}} c x^{6}}{7 \, e^{2}} - \frac {6 \, \sqrt {-e^{2} x^{2} + d^{2}} c d^{2} x^{4}}{35 \, e^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} b x^{4}}{5 \, e^{2}} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} c d^{4} x^{2}}{35 \, e^{6}} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} b d^{2} x^{2}}{15 \, e^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} a x^{2}}{3 \, e^{2}} - \frac {16 \, \sqrt {-e^{2} x^{2} + d^{2}} c d^{6}}{35 \, e^{8}} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} b d^{4}}{15 \, e^{6}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} a d^{2}}{3 \, e^{4}} \] Input:

integrate(x^3*(c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="m 
axima")
 

Output:

-1/7*sqrt(-e^2*x^2 + d^2)*c*x^6/e^2 - 6/35*sqrt(-e^2*x^2 + d^2)*c*d^2*x^4/ 
e^4 - 1/5*sqrt(-e^2*x^2 + d^2)*b*x^4/e^2 - 8/35*sqrt(-e^2*x^2 + d^2)*c*d^4 
*x^2/e^6 - 4/15*sqrt(-e^2*x^2 + d^2)*b*d^2*x^2/e^4 - 1/3*sqrt(-e^2*x^2 + d 
^2)*a*x^2/e^2 - 16/35*sqrt(-e^2*x^2 + d^2)*c*d^6/e^8 - 8/15*sqrt(-e^2*x^2 
+ d^2)*b*d^4/e^6 - 2/3*sqrt(-e^2*x^2 + d^2)*a*d^2/e^4
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {{\left (105 \, c d^{6} + 105 \, b d^{4} e^{2} + 105 \, a d^{2} e^{4} - {\left (210 \, c d^{5} + 140 \, b d^{3} e^{2} + 70 \, a d e^{4} - {\left (357 \, c d^{4} + 154 \, b d^{2} e^{2} + 35 \, a e^{4} - 3 \, {\left (124 \, c d^{3} + 28 \, b d e^{2} - {\left (81 \, c d^{2} + 7 \, b e^{2} + 5 \, {\left ({\left (e x + d\right )} c - 6 \, c d\right )} {\left (e x + d\right )}\right )} {\left (e x + d\right )}\right )} {\left (e x + d\right )}\right )} {\left (e x + d\right )}\right )} {\left (e x + d\right )}\right )} \sqrt {e x + d} \sqrt {-e x + d}}{105 \, e^{8}} \] Input:

integrate(x^3*(c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="g 
iac")
 

Output:

-1/105*(105*c*d^6 + 105*b*d^4*e^2 + 105*a*d^2*e^4 - (210*c*d^5 + 140*b*d^3 
*e^2 + 70*a*d*e^4 - (357*c*d^4 + 154*b*d^2*e^2 + 35*a*e^4 - 3*(124*c*d^3 + 
 28*b*d*e^2 - (81*c*d^2 + 7*b*e^2 + 5*((e*x + d)*c - 6*c*d)*(e*x + d))*(e* 
x + d))*(e*x + d))*(e*x + d))*(e*x + d))*sqrt(e*x + d)*sqrt(-e*x + d)/e^8
 

Mupad [B] (verification not implemented)

Time = 5.11 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.35 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {\sqrt {d-e\,x}\,\left (\frac {48\,c\,d^7+56\,b\,d^5\,e^2+70\,a\,d^3\,e^4}{105\,e^8}+\frac {x^5\,\left (18\,c\,d^2\,e^5+21\,b\,e^7\right )}{105\,e^8}+\frac {c\,x^7}{7\,e}+\frac {x^3\,\left (24\,c\,d^4\,e^3+28\,b\,d^2\,e^5+35\,a\,e^7\right )}{105\,e^8}+\frac {x\,\left (48\,c\,d^6\,e+56\,b\,d^4\,e^3+70\,a\,d^2\,e^5\right )}{105\,e^8}+\frac {x^4\,\left (18\,c\,d^3\,e^4+21\,b\,d\,e^6\right )}{105\,e^8}+\frac {x^2\,\left (24\,c\,d^5\,e^2+28\,b\,d^3\,e^4+35\,a\,d\,e^6\right )}{105\,e^8}+\frac {c\,d\,x^6}{7\,e^2}\right )}{\sqrt {d+e\,x}} \] Input:

int((x^3*(a + b*x^2 + c*x^4))/((d + e*x)^(1/2)*(d - e*x)^(1/2)),x)
 

Output:

-((d - e*x)^(1/2)*((48*c*d^7 + 70*a*d^3*e^4 + 56*b*d^5*e^2)/(105*e^8) + (x 
^5*(21*b*e^7 + 18*c*d^2*e^5))/(105*e^8) + (c*x^7)/(7*e) + (x^3*(35*a*e^7 + 
 28*b*d^2*e^5 + 24*c*d^4*e^3))/(105*e^8) + (x*(70*a*d^2*e^5 + 56*b*d^4*e^3 
 + 48*c*d^6*e))/(105*e^8) + (x^4*(18*c*d^3*e^4 + 21*b*d*e^6))/(105*e^8) + 
(x^2*(28*b*d^3*e^4 + 24*c*d^5*e^2 + 35*a*d*e^6))/(105*e^8) + (c*d*x^6)/(7* 
e^2)))/(d + e*x)^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.67 \[ \int \frac {x^3 \left (a+b x^2+c x^4\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {\sqrt {e x +d}\, \sqrt {-e x +d}\, \left (-15 c \,e^{6} x^{6}-21 b \,e^{6} x^{4}-18 c \,d^{2} e^{4} x^{4}-35 a \,e^{6} x^{2}-28 b \,d^{2} e^{4} x^{2}-24 c \,d^{4} e^{2} x^{2}-70 a \,d^{2} e^{4}-56 b \,d^{4} e^{2}-48 c \,d^{6}\right )}{105 e^{8}} \] Input:

int(x^3*(c*x^4+b*x^2+a)/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)
 

Output:

(sqrt(d + e*x)*sqrt(d - e*x)*( - 70*a*d**2*e**4 - 35*a*e**6*x**2 - 56*b*d* 
*4*e**2 - 28*b*d**2*e**4*x**2 - 21*b*e**6*x**4 - 48*c*d**6 - 24*c*d**4*e** 
2*x**2 - 18*c*d**2*e**4*x**4 - 15*c*e**6*x**6))/(105*e**8)