\(\int \frac {x^8 (c+d x^4)}{(a+b x^4)^{9/4}} \, dx\) [150]

Optimal result

Integrand size = 22, antiderivative size = 151 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {a (b c-a d) x}{5 b^3 \left (a+b x^4\right )^{5/4}}-\frac {(6 b c-11 a d) x}{5 b^3 \sqrt [4]{a+b x^4}}+\frac {d x \left (a+b x^4\right )^{3/4}}{4 b^3}+\frac {(4 b c-9 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{13/4}}+\frac {(4 b c-9 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{13/4}} \] Output:

1/5*a*(-a*d+b*c)*x/b^3/(b*x^4+a)^(5/4)-1/5*(-11*a*d+6*b*c)*x/b^3/(b*x^4+a) 
^(1/4)+1/4*d*x*(b*x^4+a)^(3/4)/b^3+1/8*(-9*a*d+4*b*c)*arctan(b^(1/4)*x/(b* 
x^4+a)^(1/4))/b^(13/4)+1/8*(-9*a*d+4*b*c)*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4 
))/b^(13/4)
 

Mathematica [A] (verified)

Time = 2.04 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.83 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} x \left (45 a^2 d+b^2 x^4 \left (-24 c+5 d x^4\right )+a b \left (-20 c+54 d x^4\right )\right )}{\left (a+b x^4\right )^{5/4}}+5 (4 b c-9 a d) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+5 (4 b c-9 a d) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{40 b^{13/4}} \] Input:

Integrate[(x^8*(c + d*x^4))/(a + b*x^4)^(9/4),x]
 

Output:

((2*b^(1/4)*x*(45*a^2*d + b^2*x^4*(-24*c + 5*d*x^4) + a*b*(-20*c + 54*d*x^ 
4)))/(a + b*x^4)^(5/4) + 5*(4*b*c - 9*a*d)*ArcTan[(b^(1/4)*x)/(a + b*x^4)^ 
(1/4)] + 5*(4*b*c - 9*a*d)*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(40*b^( 
13/4))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {957, 817, 843, 770, 756, 216, 219}

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.

\(\displaystyle \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^9 (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(4 b c-9 a d) \int \frac {x^8}{\left (b x^4+a\right )^{5/4}}dx}{5 a b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {x^9 (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {5 \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 a b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^9 (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 a b}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {x^9 (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 a b}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {x^9 (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 a b}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {x^9 (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 a b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {x^9 (b c-a d)}{5 a b \left (a+b x^4\right )^{5/4}}-\frac {(4 b c-9 a d) \left (\frac {5 \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{b}-\frac {x^5}{b \sqrt [4]{a+b x^4}}\right )}{5 a b}\)

Input:

Int[(x^8*(c + d*x^4))/(a + b*x^4)^(9/4),x]
 

Output:

((b*c - a*d)*x^9)/(5*a*b*(a + b*x^4)^(5/4)) - ((4*b*c - 9*a*d)*(-(x^5/(b*( 
a + b*x^4)^(1/4))) + (5*((x*(a + b*x^4)^(3/4))/(4*b) - (a*(ArcTan[(b^(1/4) 
*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4) 
]/(2*b^(1/4))))/(4*b)))/b))/(5*a*b)
 

Defintions of rubi rules used

rule 216

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

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

rule 817

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

rule 843

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

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)]))
 

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(-\frac {9 \left (\frac {16 \left (-\frac {27 d \,x^{4}}{10}+c \right ) x a \,b^{\frac {5}{4}}}{9}+\left (-\frac {4}{9} d \,x^{9}+\frac {32}{15} c \,x^{5}\right ) b^{\frac {9}{4}}-4 a^{2} d x \,b^{\frac {1}{4}}+\left (a d -\frac {4 c b}{9}\right ) \left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (\ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right )\right )\right )}{16 b^{\frac {13}{4}} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}\)	\(131\)

Input:

int(x^8*(d*x^4+c)/(b*x^4+a)^(9/4),x,method=_RETURNVERBOSE)
 

Output:

-9/16/b^(13/4)*(16/9*(-27/10*d*x^4+c)*x*a*b^(5/4)+(-4/9*d*x^9+32/15*c*x^5) 
*b^(9/4)-4*a^2*d*x*b^(1/4)+(a*d-4/9*c*b)*(b*x^4+a)^(5/4)*(ln((x*b^(1/4)+(b 
*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))-2*arctan((b*x^4+a)^(1/4)/x/b^ 
(1/4))))/(b*x^4+a)^(5/4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 884, normalized size of antiderivative = 5.85 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx =\text {Too large to display} \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(9/4),x, algorithm="fricas")
 

Output:

1/80*(5*(b^5*x^8 + 2*a*b^4*x^4 + a^2*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d 
 + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)/b^13)^(1/4)*lo 
g(-(b^10*x*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664 
*a^3*b*c*d^3 + 6561*a^4*d^4)/b^13)^(3/4) + (64*b^3*c^3 - 432*a*b^2*c^2*d + 
 972*a^2*b*c*d^2 - 729*a^3*d^3)*(b*x^4 + a)^(1/4))/x) - 5*(b^5*x^8 + 2*a*b 
^4*x^4 + a^2*b^3)*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 
- 11664*a^3*b*c*d^3 + 6561*a^4*d^4)/b^13)^(1/4)*log((b^10*x*((256*b^4*c^4 
- 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d 
^4)/b^13)^(3/4) - (64*b^3*c^3 - 432*a*b^2*c^2*d + 972*a^2*b*c*d^2 - 729*a^ 
3*d^3)*(b*x^4 + a)^(1/4))/x) - 5*(-I*b^5*x^8 - 2*I*a*b^4*x^4 - I*a^2*b^3)* 
((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^ 
3 + 6561*a^4*d^4)/b^13)^(1/4)*log((I*b^10*x*((256*b^4*c^4 - 2304*a*b^3*c^3 
*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)/b^13)^(3/4) 
- (64*b^3*c^3 - 432*a*b^2*c^2*d + 972*a^2*b*c*d^2 - 729*a^3*d^3)*(b*x^4 + 
a)^(1/4))/x) - 5*(I*b^5*x^8 + 2*I*a*b^4*x^4 + I*a^2*b^3)*((256*b^4*c^4 - 2 
304*a*b^3*c^3*d + 7776*a^2*b^2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4) 
/b^13)^(1/4)*log((-I*b^10*x*((256*b^4*c^4 - 2304*a*b^3*c^3*d + 7776*a^2*b^ 
2*c^2*d^2 - 11664*a^3*b*c*d^3 + 6561*a^4*d^4)/b^13)^(3/4) - (64*b^3*c^3 - 
432*a*b^2*c^2*d + 972*a^2*b*c*d^2 - 729*a^3*d^3)*(b*x^4 + a)^(1/4))/x) + 4 
*(5*b^2*d*x^9 - 6*(4*b^2*c - 9*a*b*d)*x^5 - 5*(4*a*b*c - 9*a^2*d)*x)*(b...
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 38.70 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.53 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\frac {c x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {9}{4}} \Gamma \left (\frac {13}{4}\right )} + \frac {d x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {9}{4}} \Gamma \left (\frac {17}{4}\right )} \] Input:

integrate(x**8*(d*x**4+c)/(b*x**4+a)**(9/4),x)
 

Output:

c*x**9*gamma(9/4)*hyper((9/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4* 
a**(9/4)*gamma(13/4)) + d*x**13*gamma(13/4)*hyper((9/4, 13/4), (17/4,), b* 
x**4*exp_polar(I*pi)/a)/(4*a**(9/4)*gamma(17/4))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (123) = 246\).

Time = 0.12 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.72 \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=-\frac {1}{20} \, {\left (\frac {4 \, {\left (b + \frac {5 \, {\left (b x^{4} + a\right )}}{x^{4}}\right )} x^{5}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2}} + \frac {5 \, {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{2}}\right )} c + \frac {1}{80} \, d {\left (\frac {4 \, {\left (4 \, a b^{2} + \frac {36 \, {\left (b x^{4} + a\right )} a b}{x^{4}} - \frac {45 \, {\left (b x^{4} + a\right )}^{2} a}{x^{8}}\right )}}{\frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{4}}{x^{5}} - \frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} b^{3}}{x^{9}}} + \frac {45 \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{3}}\right )} \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(9/4),x, algorithm="maxima")
 

Output:

-1/20*(4*(b + 5*(b*x^4 + a)/x^4)*x^5/((b*x^4 + a)^(5/4)*b^2) + 5*(2*arctan 
((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(b^(1/4) - (b*x^4 + a)^(1/4 
)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(1/4))/b^2)*c + 1/80*d*(4*(4*a*b^2 
 + 36*(b*x^4 + a)*a*b/x^4 - 45*(b*x^4 + a)^2*a/x^8)/((b*x^4 + a)^(5/4)*b^4 
/x^5 - (b*x^4 + a)^(9/4)*b^3/x^9) + 45*a*(2*arctan((b*x^4 + a)^(1/4)/(b^(1 
/4)*x))/b^(1/4) + log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + 
 a)^(1/4)/x))/b^(1/4))/b^3)
 

Giac [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{8}}{{\left (b x^{4} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(x^8*(d*x^4+c)/(b*x^4+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*x^8/(b*x^4 + a)^(9/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\int \frac {x^8\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{9/4}} \,d x \] Input:

int((x^8*(c + d*x^4))/(a + b*x^4)^(9/4),x)
 

Output:

int((x^8*(c + d*x^4))/(a + b*x^4)^(9/4), x)
 

Reduce [F]

\[ \int \frac {x^8 \left (c+d x^4\right )}{\left (a+b x^4\right )^{9/4}} \, dx=\left (\int \frac {x^{12}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{8}}d x \right ) d +\left (\int \frac {x^{8}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} x^{8}}d x \right ) c \] Input:

int(x^8*(d*x^4+c)/(b*x^4+a)^(9/4),x)
 

Output:

int(x**12/((a + b*x**4)**(1/4)*a**2 + 2*(a + b*x**4)**(1/4)*a*b*x**4 + (a 
+ b*x**4)**(1/4)*b**2*x**8),x)*d + int(x**8/((a + b*x**4)**(1/4)*a**2 + 2* 
(a + b*x**4)**(1/4)*a*b*x**4 + (a + b*x**4)**(1/4)*b**2*x**8),x)*c