\(\int \frac {x^{5/2}}{(a+b x^2+c x^4)^2} \, dx\) [934]

Optimal result

Integrand size = 20, antiderivative size = 450 \[ \int \frac {x^{5/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=-\frac {x^{3/2} \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\sqrt [4]{c} \left (4 b+\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (4 b-\sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b+\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (4 b+\sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}-\frac {\sqrt [4]{c} \left (4 b-\sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b+\sqrt {b^2-4 a c}}} \] Output:

-1/2*x^(3/2)*(2*c*x^2+b)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)-1/4*c^(1/4)*(4*b+(-4 
*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^(1 
/4))*2^(1/4)/(-4*a*c+b^2)^(3/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4)+1/4*c^(1/4)* 
(4*b-(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a*c+b^2)^( 
1/2))^(1/4))*2^(1/4)/(-4*a*c+b^2)^(3/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4)+1/4* 
c^(1/4)*(4*b+(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a 
*c+b^2)^(1/2))^(1/4))*2^(1/4)/(-4*a*c+b^2)^(3/2)/(-b-(-4*a*c+b^2)^(1/2))^( 
1/4)-1/4*c^(1/4)*(4*b-(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x^(1/2)/ 
(-b+(-4*a*c+b^2)^(1/2))^(1/4))*2^(1/4)/(-4*a*c+b^2)^(3/2)/(-b+(-4*a*c+b^2) 
^(1/2))^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.23 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.24 \[ \int \frac {x^{5/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {-\frac {4 x^{3/2} \left (b+2 c x^2\right )}{a+b x^2+c x^4}+\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {3 b \log \left (\sqrt {x}-\text {$\#$1}\right )-2 c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]}{8 \left (b^2-4 a c\right )} \] Input:

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

Output:

((-4*x^(3/2)*(b + 2*c*x^2))/(a + b*x^2 + c*x^4) + RootSum[a + b*#1^4 + c*# 
1^8 & , (3*b*Log[Sqrt[x] - #1] - 2*c*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 2*c*# 
1^5) & ])/(8*(b^2 - 4*a*c))
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1435, 1700, 1834, 27, 827, 218, 221}

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.

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

\(\Big \downarrow \) 1435

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

\(\Big \downarrow \) 1700

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

\(\Big \downarrow \) 1834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

2*(-1/4*(x^(3/2)*(b + 2*c*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (-2* 
c*(1 + (4*b)/Sqrt[b^2 - 4*a*c])*(ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sq 
rt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) 
 - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^ 
(3/4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1/4))) - 2*c*(1 - (4*b)/Sqrt[b^2 - 
 4*a*c])*(ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)] 
/(2*2^(3/4)*c^(3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - ArcTanh[(2^(1/4)*c^( 
1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b + Sqr 
t[b^2 - 4*a*c])^(1/4))))/(4*(b^2 - 4*a*c)))
 

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 218

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

rule 221

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

rule 827

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

rule 1435

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> With[{k = Denominator[m]}, Simp[k/d   Subst[Int[x^(k*(m + 1) - 1)*(a + b 
*(x^(2*k)/d^2) + c*(x^(4*k)/d^4))^p, x], x, (d*x)^(1/k)], x]] /; FreeQ[{a, 
b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && FractionQ[m] && IntegerQ[p]
 

rule 1700

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x 
_Symbol] :> Simp[d^(n - 1)*(d*x)^(m - n + 1)*(b + 2*c*x^n)*((a + b*x^n + c* 
x^(2*n))^(p + 1)/(n*(p + 1)*(b^2 - 4*a*c))), x] - Simp[d^n/(n*(p + 1)*(b^2 
- 4*a*c))   Int[(d*x)^(m - n)*(b*(m - n + 1) + 2*c*(m + 2*n*(p + 1) + 1)*x^ 
n)*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ 
[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && ILtQ[p, -1] && GtQ[m, n - 
 1] && LeQ[m, 2*n - 1]
 

rule 1834

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + 
 (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + 
 (2*c*d - b*e)/(2*q))   Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 
 - (2*c*d - b*e)/(2*q))   Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ 
[{a, b, c, d, e, f, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n 
, 0]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.50 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.27

method	result	size
derivativedivides	\(\frac {\frac {2 c \,x^{\frac {7}{2}}}{8 a c -2 b^{2}}+\frac {2 b \,x^{\frac {3}{2}}}{16 a c -4 b^{2}}}{c \,x^{4}+b \,x^{2}+a}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-2 \textit {\_R}^{6} c +3 \textit {\_R}^{2} b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{8 \left (4 a c -b^{2}\right )}\)	\(121\)
default	\(\frac {\frac {2 c \,x^{\frac {7}{2}}}{8 a c -2 b^{2}}+\frac {2 b \,x^{\frac {3}{2}}}{16 a c -4 b^{2}}}{c \,x^{4}+b \,x^{2}+a}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-2 \textit {\_R}^{6} c +3 \textit {\_R}^{2} b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{8 \left (4 a c -b^{2}\right )}\)	\(121\)

Input:

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

Output:

2*(1/2*c/(4*a*c-b^2)*x^(7/2)+1/4*b/(4*a*c-b^2)*x^(3/2))/(c*x^4+b*x^2+a)-1/ 
8/(4*a*c-b^2)*sum((-2*_R^6*c+3*_R^2*b)/(2*_R^7*c+_R^3*b)*ln(x^(1/2)-_R),_R 
=RootOf(_Z^8*c+_Z^4*b+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 10601 vs. \(2 (350) = 700\).

Time = 1.88 (sec) , antiderivative size = 10601, normalized size of antiderivative = 23.56 \[ \int \frac {x^{5/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-1/2*(2*c*x^(7/2) + b*x^(3/2))/((b^2*c - 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + 
(b^3 - 4*a*b*c)*x^2) + integrate(-1/4*(2*c*x^(5/2) - 3*b*sqrt(x))/((b^2*c 
- 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^2), x)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 19.23 (sec) , antiderivative size = 21913, normalized size of antiderivative = 48.70 \[ \int \frac {x^{5/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((b*x^(3/2))/(2*(4*a*c - b^2)) + (c*x^(7/2))/(4*a*c - b^2))/(a + b*x^2 + c 
*x^4) - atan(((((110592*a*b^16*c^4 - 134217728*a^9*c^12 - 2433024*a^2*b^14 
*c^5 + 21200896*a^3*b^12*c^6 - 87687168*a^4*b^10*c^7 + 133693440*a^5*b^8*c 
^8 + 211812352*a^6*b^6*c^9 - 1031798784*a^7*b^4*c^10 + 1107296256*a^8*b^2* 
c^11)/(128*(b^14 - 16384*a^7*c^7 + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3 + 8 
960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 28*a*b^12*c)) - 
(x^(1/2)*(-(81*b^17 + 81*b^2*(-(4*a*c - b^2)^15)^(1/2) - 983040*a^8*b*c^8 
+ 960*a^2*b^13*c^2 + 84480*a^3*b^11*c^3 - 719360*a^4*b^9*c^4 + 2727936*a^5 
*b^7*c^5 - 5259264*a^6*b^5*c^6 + 4587520*a^7*b^3*c^7 - 1184*a*b^15*c - 4*a 
*c*(-(4*a*c - b^2)^15)^(1/2))/(8192*(a*b^24 + 16777216*a^13*c^12 - 48*a^2* 
b^22*c + 1056*a^3*b^20*c^2 - 14080*a^4*b^18*c^3 + 126720*a^5*b^16*c^4 - 81 
1008*a^6*b^14*c^5 + 3784704*a^7*b^12*c^6 - 12976128*a^8*b^10*c^7 + 3244032 
0*a^9*b^8*c^8 - 57671680*a^10*b^6*c^9 + 69206016*a^11*b^4*c^10 - 50331648* 
a^12*b^2*c^11)))^(1/4)*(134217728*a^9*c^12 + 36864*a*b^16*c^4 - 909312*a^2 
*b^14*c^5 + 9469952*a^3*b^12*c^6 - 53870592*a^4*b^10*c^7 + 180879360*a^5*b 
^8*c^8 - 362807296*a^6*b^6*c^9 + 427819008*a^7*b^4*c^10 - 301989888*a^8*b^ 
2*c^11))/(16*(b^12 + 4096*a^6*c^6 + 240*a^2*b^8*c^2 - 1280*a^3*b^6*c^3 + 3 
840*a^4*b^4*c^4 - 6144*a^5*b^2*c^5 - 24*a*b^10*c)))*(-(81*b^17 + 81*b^2*(- 
(4*a*c - b^2)^15)^(1/2) - 983040*a^8*b*c^8 + 960*a^2*b^13*c^2 + 84480*a^3* 
b^11*c^3 - 719360*a^4*b^9*c^4 + 2727936*a^5*b^7*c^5 - 5259264*a^6*b^5*c...
 

Reduce [F]

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

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

Output:

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