\(\int \frac {x^{3/2}}{(a+b x^2+c x^4)^2} \, dx\) [935]

Optimal result

Integrand size = 20, antiderivative size = 442 \[ \int \frac {x^{3/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=-\frac {\sqrt {x} \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {c^{3/4} \left (3+\frac {4 b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} \left (b^2-4 a c\right ) \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (3-\frac {4 b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} \left (b^2-4 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (3+\frac {4 b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} \left (b^2-4 a c\right ) \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (3-\frac {4 b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} \left (b^2-4 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.25 \[ \int \frac {x^{3/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {-\frac {4 \sqrt {x} \left (b+2 c x^2\right )}{a+b x^2+c x^4}+\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log \left (\sqrt {x}-\text {$\#$1}\right )-6 c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{8 \left (b^2-4 a c\right )} \] Input:

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

Output:

((-4*Sqrt[x]*(b + 2*c*x^2))/(a + b*x^2 + c*x^4) + RootSum[a + b*#1^4 + c*# 
1^8 & , (b*Log[Sqrt[x] - #1] - 6*c*Log[Sqrt[x] - #1]*#1^4)/(b*#1^3 + 2*c*# 
1^7) & ])/(8*(b^2 - 4*a*c))
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 385, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1435, 1700, 1752, 756, 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.

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

\(\Big \downarrow \) 1435

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

\(\Big \downarrow \) 1700

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

\(\Big \downarrow \) 1752

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

2*(-1/4*(Sqrt[x]*(b + 2*c*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (-(c 
*(3 + (4*b)/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - S 
qrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4))) 
 - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1 
/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4)))) - c*(3 - (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^(1/4)*c^(1/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/ 
4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b + Sqrt[b^ 
2 - 4*a*c])^(3/4))))/(4*(b^2 - 4*a*c)))
 

Defintions of rubi rules used

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 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 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 1752

Int[((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[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   I 
nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 
, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 
 - 4*a*c] ||  !IGtQ[n/2, 0])
 

Maple [C] (verified)

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

Time = 0.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.27

method	result	size
derivativedivides	\(\frac {\frac {2 c \,x^{\frac {5}{2}}}{8 a c -2 b^{2}}+\frac {2 b \sqrt {x}}{16 a c -4 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (6 \textit {\_R}^{4} c -b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{32 a c -8 b^{2}}\)	\(118\)
default	\(\frac {\frac {2 c \,x^{\frac {5}{2}}}{8 a c -2 b^{2}}+\frac {2 b \sqrt {x}}{16 a c -4 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (6 \textit {\_R}^{4} c -b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{32 a c -8 b^{2}}\)	\(118\)

Input:

int(x^(3/2)/(c*x^4+b*x^2+a)^2,x,method=_RETURNVERBOSE)
 

Output:

2*(1/2*c/(4*a*c-b^2)*x^(5/2)+1/4*b/(4*a*c-b^2)*x^(1/2))/(c*x^4+b*x^2+a)+1/ 
8/(4*a*c-b^2)*sum((6*_R^4*c-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. 8211 vs. \(2 (350) = 700\).

Time = 0.95 (sec) , antiderivative size = 8211, normalized size of antiderivative = 18.58 \[ \int \frac {x^{3/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(3/2)/(c*x**4+b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(x^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*(b*c*x^(9/2) + (b^2 - 2*a*c)*x^(5/2))/((a*b^2*c - 4*a^2*c^2)*x^4 + a^2 
*b^2 - 4*a^3*c + (a*b^3 - 4*a^2*b*c)*x^2) + integrate(-1/4*(b*c*x^(7/2) + 
(b^2 + 6*a*c)*x^(3/2))/((a*b^2*c - 4*a^2*c^2)*x^4 + a^2*b^2 - 4*a^3*c + (a 
*b^3 - 4*a^2*b*c)*x^2), x)
 

Giac [F]

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

integrate(x^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")
 

Output:

integrate(x^(3/2)/(c*x^4 + b*x^2 + a)^2, x)
 

Mupad [B] (verification not implemented)

Time = 23.64 (sec) , antiderivative size = 28713, normalized size of antiderivative = 64.96 \[ \int \frac {x^{3/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

int(x^(3/2)/(a + b*x^2 + c*x^4)^2,x)
 

Output:

atan((((((((b^4*(-(4*a*c - b^2)^15)^(1/2) - b^19 - 12386304*a^9*b*c^9 + 96 
*a^2*b^15*c^2 - 2752*a^3*b^13*c^3 + 55296*a^4*b^11*c^4 - 585216*a^5*b^9*c^ 
5 + 3350528*a^6*b^7*c^6 - 10665984*a^7*b^5*c^7 + 17891328*a^8*b^3*c^8 + 32 
4*a^2*c^2*(-(4*a*c - b^2)^15)^(1/2) + 3*a*b^17*c + 27*a*b^2*c*(-(4*a*c - b 
^2)^15)^(1/2))/(8192*(a^3*b^24 + 16777216*a^15*c^12 - 48*a^4*b^22*c + 1056 
*a^5*b^20*c^2 - 14080*a^6*b^18*c^3 + 126720*a^7*b^16*c^4 - 811008*a^8*b^14 
*c^5 + 3784704*a^9*b^12*c^6 - 12976128*a^10*b^10*c^7 + 32440320*a^11*b^8*c 
^8 - 57671680*a^12*b^6*c^9 + 69206016*a^13*b^4*c^10 - 50331648*a^14*b^2*c^ 
11)))^(1/4)*(100663296*a^8*c^11 + 4096*a*b^14*c^4 - 73728*a^2*b^12*c^5 + 3 
93216*a^3*b^10*c^6 + 655360*a^4*b^8*c^7 - 15728640*a^5*b^6*c^8 + 69206016* 
a^6*b^4*c^9 - 134217728*a^7*b^2*c^10))/(2*(b^8 + 256*a^4*c^4 + 96*a^2*b^4* 
c^2 - 256*a^3*b^2*c^3 - 16*a*b^6*c)) - (x^(1/2)*(2048*b^17*c^4 - 30720*a*b 
^15*c^5 + 100663296*a^8*b*c^12 + 73728*a^2*b^13*c^6 + 1212416*a^3*b^11*c^7 
 - 9830400*a^4*b^9*c^8 + 26738688*a^5*b^7*c^9 - 10485760*a^6*b^5*c^10 - 75 
497472*a^7*b^3*c^11))/(8*(b^12 + 4096*a^6*c^6 + 240*a^2*b^8*c^2 - 1280*a^3 
*b^6*c^3 + 3840*a^4*b^4*c^4 - 6144*a^5*b^2*c^5 - 24*a*b^10*c)))*((b^4*(-(4 
*a*c - b^2)^15)^(1/2) - b^19 - 12386304*a^9*b*c^9 + 96*a^2*b^15*c^2 - 2752 
*a^3*b^13*c^3 + 55296*a^4*b^11*c^4 - 585216*a^5*b^9*c^5 + 3350528*a^6*b^7* 
c^6 - 10665984*a^7*b^5*c^7 + 17891328*a^8*b^3*c^8 + 324*a^2*c^2*(-(4*a*c - 
 b^2)^15)^(1/2) + 3*a*b^17*c + 27*a*b^2*c*(-(4*a*c - b^2)^15)^(1/2))/(8...
 

Reduce [F]

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

int(x^(3/2)/(c*x^4+b*x^2+a)^2,x)
 

Output:

int(x^(3/2)/(c*x^4+b*x^2+a)^2,x)