\(\int \frac {x^9}{(a+b x^2+c x^4)^{3/2}} \, dx\) [1008]

Optimal result

Integrand size = 20, antiderivative size = 190 \[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {b x^4 \sqrt {a+b x^2+c x^4}}{c \left (b^2-4 a c\right )}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c \left (5 b^2-12 a c\right ) x^2\right ) \sqrt {a+b x^2+c x^4}}{8 c^3 \left (b^2-4 a c\right )}+\frac {3 \left (5 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 c^{7/2}} \] Output:

x^6*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^(1/2)-b*x^4*(c*x^4+b*x^2+a)^( 
1/2)/c/(-4*a*c+b^2)-1/8*(b*(-52*a*c+15*b^2)-2*c*(-12*a*c+5*b^2)*x^2)*(c*x^ 
4+b*x^2+a)^(1/2)/c^3/(-4*a*c+b^2)+3/16*(-4*a*c+5*b^2)*arctanh(1/2*(2*c*x^2 
+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.87 \[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {4 a^2 c \left (-13 b+6 c x^2\right )+b^2 x^2 \left (15 b^2+5 b c x^2-2 c^2 x^4\right )+a \left (15 b^3-62 b^2 c x^2-20 b c^2 x^4+8 c^3 x^6\right )}{8 c^3 \left (-b^2+4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {3 \left (-5 b^2+4 a c\right ) \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{16 c^{7/2}} \] Input:

Integrate[x^9/(a + b*x^2 + c*x^4)^(3/2),x]
 

Output:

(4*a^2*c*(-13*b + 6*c*x^2) + b^2*x^2*(15*b^2 + 5*b*c*x^2 - 2*c^2*x^4) + a* 
(15*b^3 - 62*b^2*c*x^2 - 20*b*c^2*x^4 + 8*c^3*x^6))/(8*c^3*(-b^2 + 4*a*c)* 
Sqrt[a + b*x^2 + c*x^4]) + (3*(-5*b^2 + 4*a*c)*Log[b + 2*c*x^2 - 2*Sqrt[c] 
*Sqrt[a + b*x^2 + c*x^4]])/(16*c^(7/2))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1434, 1164, 27, 1236, 27, 1225, 1092, 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.

\(\displaystyle \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {x^8}{\left (c x^4+b x^2+a\right )^{3/2}}dx^2\)

\(\Big \downarrow \) 1164

\(\displaystyle \frac {1}{2} \left (\frac {2 x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {2 \int \frac {3 x^4 \left (b x^2+2 a\right )}{\sqrt {c x^4+b x^2+a}}dx^2}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {2 x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {6 \int \frac {x^4 \left (b x^2+2 a\right )}{\sqrt {c x^4+b x^2+a}}dx^2}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 1236

\(\displaystyle \frac {1}{2} \left (\frac {2 x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {6 \left (\frac {\int -\frac {x^2 \left (\left (5 b^2-12 a c\right ) x^2+4 a b\right )}{2 \sqrt {c x^4+b x^2+a}}dx^2}{3 c}+\frac {b x^4 \sqrt {a+b x^2+c x^4}}{3 c}\right )}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {2 x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {6 \left (\frac {b x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\int \frac {x^2 \left (\left (5 b^2-12 a c\right ) x^2+4 a b\right )}{\sqrt {c x^4+b x^2+a}}dx^2}{6 c}\right )}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {1}{2} \left (\frac {2 x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {6 \left (\frac {b x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\frac {3 \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{8 c^2}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c x^2 \left (5 b^2-12 a c\right )\right ) \sqrt {a+b x^2+c x^4}}{4 c^2}}{6 c}\right )}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (\frac {2 x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {6 \left (\frac {b x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\frac {3 \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{4 c^2}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c x^2 \left (5 b^2-12 a c\right )\right ) \sqrt {a+b x^2+c x^4}}{4 c^2}}{6 c}\right )}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {2 x^6 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {6 \left (\frac {b x^4 \sqrt {a+b x^2+c x^4}}{3 c}-\frac {\frac {3 \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac {\left (b \left (15 b^2-52 a c\right )-2 c x^2 \left (5 b^2-12 a c\right )\right ) \sqrt {a+b x^2+c x^4}}{4 c^2}}{6 c}\right )}{b^2-4 a c}\right )\)

Input:

Int[x^9/(a + b*x^2 + c*x^4)^(3/2),x]
 

Output:

((2*x^6*(2*a + b*x^2))/((b^2 - 4*a*c)*Sqrt[a + b*x^2 + c*x^4]) - (6*((b*x^ 
4*Sqrt[a + b*x^2 + c*x^4])/(3*c) - (-1/4*((b*(15*b^2 - 52*a*c) - 2*c*(5*b^ 
2 - 12*a*c)*x^2)*Sqrt[a + b*x^2 + c*x^4])/c^2 + (3*(b^2 - 4*a*c)*(5*b^2 - 
4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(8*c^(5 
/2)))/(6*c)))/(b^2 - 4*a*c))/2
 

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 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 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x 
+ c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* 
c))   Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* 
c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p 
+ 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

rule 1236

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 
1)/(c*(m + 2*p + 2))), x] + Simp[1/(c*(m + 2*p + 2))   Int[(d + e*x)^(m - 1 
)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m 
*(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ 
{a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])
 

rule 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(-\frac {48 \left (\left (-\frac {5}{24} b^{3} x^{4}+\frac {31}{12} a \,b^{2} x^{2}+\frac {13}{6} a^{2} b \right ) c^{\frac {3}{2}}+\left (\frac {1}{12} b^{2} x^{6}+\frac {5}{6} a b \,x^{4}-a^{2} x^{2}\right ) c^{\frac {5}{2}}-\frac {a \,c^{\frac {7}{2}} x^{6}}{3}-\frac {5 b^{3} \sqrt {c}\, \left (b \,x^{2}+a \right )}{8}+\left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )\right ) \left (a c -\frac {b^{2}}{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (a c -\frac {5 b^{2}}{4}\right )\right )}{c^{\frac {7}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (64 a c -16 b^{2}\right )}\)	\(183\)
risch	\(-\frac {\left (-2 c \,x^{2}+7 b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 c^{3}}-\frac {3 c \left (4 a c -5 b^{2}\right ) \left (-\frac {x^{2}}{2 c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (2 c \,x^{2}+b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}\right )}{4 c}+\frac {\ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {b \left (4 a c +7 b^{2}\right ) \left (b \,x^{2}+2 a \right )}{\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (4 a c -b^{2}\right )}+\frac {a \left (4 a c -7 b^{2}\right ) \left (2 c \,x^{2}+b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}}{8 c^{3}}\)	\(266\)
default	\(\frac {x^{6}}{4 c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {5 b \,x^{4}}{8 c^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {15 b^{2} x^{2}}{16 c^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{3}}{32 c^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{4} x^{2}}{16 c^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{5}}{32 c^{4} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {7}{2}}}-\frac {13 b a}{8 c^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {13 b^{2} a \,x^{2}}{4 c^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {13 b^{3} a}{8 c^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {3 a \,x^{2}}{4 c^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {3 a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {5}{2}}}\)	\(354\)
elliptic	\(\frac {x^{6}}{4 c \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {5 b \,x^{4}}{8 c^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {15 b^{2} x^{2}}{16 c^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{3}}{32 c^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{4} x^{2}}{16 c^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{5}}{32 c^{4} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {15 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 c^{\frac {7}{2}}}-\frac {13 b a}{8 c^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {13 b^{2} a \,x^{2}}{4 c^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {13 b^{3} a}{8 c^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {3 a \,x^{2}}{4 c^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {3 a \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {5}{2}}}\)	\(354\)

Input:

int(x^9/(c*x^4+b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-48/c^(7/2)/(c*x^4+b*x^2+a)^(1/2)*((-5/24*b^3*x^4+31/12*a*b^2*x^2+13/6*a^2 
*b)*c^(3/2)+(1/12*b^2*x^6+5/6*a*b*x^4-a^2*x^2)*c^(5/2)-1/3*a*c^(7/2)*x^6-5 
/8*b^3*c^(1/2)*(b*x^2+a)+(-ln(2)+ln((2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/ 
2)+b)/c^(1/2)))*(a*c-1/4*b^2)*(c*x^4+b*x^2+a)^(1/2)*(a*c-5/4*b^2))/(64*a*c 
-16*b^2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 591, normalized size of antiderivative = 3.11 \[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{6} - 15 \, a b^{3} c + 52 \, a^{2} b c^{2} - 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{4} - {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{4} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{2}\right )}}, -\frac {3 \, {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, {\left (2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{6} - 15 \, a b^{3} c + 52 \, a^{2} b c^{2} - 5 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{4} - {\left (15 \, b^{4} c - 62 \, a b^{2} c^{2} + 24 \, a^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, {\left (a b^{2} c^{4} - 4 \, a^{2} c^{5} + {\left (b^{2} c^{5} - 4 \, a c^{6}\right )} x^{4} + {\left (b^{3} c^{4} - 4 \, a b c^{5}\right )} x^{2}\right )}}\right ] \] Input:

integrate(x^9/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[-1/32*(3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2 + (5*b^4*c - 24*a*b^2*c^2 + 
 16*a^2*c^3)*x^4 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^2)*sqrt(c)*log(-8 
*c^2*x^4 - 8*b*c*x^2 - b^2 + 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt( 
c) - 4*a*c) - 4*(2*(b^2*c^3 - 4*a*c^4)*x^6 - 15*a*b^3*c + 52*a^2*b*c^2 - 5 
*(b^3*c^2 - 4*a*b*c^3)*x^4 - (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*x^2)*s 
qrt(c*x^4 + b*x^2 + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^4 + 
 (b^3*c^4 - 4*a*b*c^5)*x^2), -1/16*(3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2 
 + (5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^4 + (5*b^5 - 24*a*b^3*c + 16*a^ 
2*b*c^2)*x^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sq 
rt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) - 2*(2*(b^2*c^3 - 4*a*c^4)*x^6 - 15*a*b^ 
3*c + 52*a^2*b*c^2 - 5*(b^3*c^2 - 4*a*b*c^3)*x^4 - (15*b^4*c - 62*a*b^2*c^ 
2 + 24*a^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^ 
2*c^5 - 4*a*c^6)*x^4 + (b^3*c^4 - 4*a*b*c^5)*x^2)]
 

Sympy [F]

\[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{9}}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**9/(c*x**4+b*x**2+a)**(3/2),x)
 

Output:

Integral(x**9/(a + b*x**2 + c*x**4)**(3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^9/(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.12 \[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2}}{b^{2} c^{3} - 4 \, a c^{4}} - \frac {5 \, {\left (b^{3} c - 4 \, a b c^{2}\right )}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x^{2} - \frac {15 \, b^{4} - 62 \, a b^{2} c + 24 \, a^{2} c^{2}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x^{2} - \frac {15 \, a b^{3} - 52 \, a^{2} b c}{b^{2} c^{3} - 4 \, a c^{4}}}{8 \, \sqrt {c x^{4} + b x^{2} + a}} - \frac {3 \, {\left (5 \, b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {7}{2}}} \] Input:

integrate(x^9/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

1/8*(((2*(b^2*c^2 - 4*a*c^3)*x^2/(b^2*c^3 - 4*a*c^4) - 5*(b^3*c - 4*a*b*c^ 
2)/(b^2*c^3 - 4*a*c^4))*x^2 - (15*b^4 - 62*a*b^2*c + 24*a^2*c^2)/(b^2*c^3 
- 4*a*c^4))*x^2 - (15*a*b^3 - 52*a^2*b*c)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^4 
+ b*x^2 + a) - 3/16*(5*b^2 - 4*a*c)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + 
b*x^2 + a))*sqrt(c) + b))/c^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^9}{{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int(x^9/(a + b*x^2 + c*x^4)^(3/2),x)
 

Output:

int(x^9/(a + b*x^2 + c*x^4)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 5905, normalized size of antiderivative = 31.08 \[ \int \frac {x^9}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(x^9/(c*x^4+b*x^2+a)^(3/2),x)
 

Output:

( - 1536*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4 
) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**5*c**5 - 1536*sqrt(a + b*x**2 + c 
*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c 
 - b**2))*a**4*b**2*c**4 - 19968*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)* 
sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**4*b*c**5* 
x**2 - 19968*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c* 
x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**4*c**6*x**4 + 4800*sqrt(a + b 
*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/s 
qrt(4*a*c - b**2))*a**3*b**4*c**3 + 18432*sqrt(a + b*x**2 + c*x**4)*log((2 
*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a** 
3*b**3*c**4*x**2 - 24576*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + 
 b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**3*b**2*c**5*x**4 
- 86016*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) 
 + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**3*b*c**6*x**6 - 43008*sqrt(a + b*x 
**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqr 
t(4*a*c - b**2))*a**3*c**7*x**8 - 480*sqrt(a + b*x**2 + c*x**4)*log((2*sqr 
t(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b* 
*6*c**2 + 10560*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + 
 c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b**5*c**3*x**2 + 67392*s 
qrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b +...