\(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^{11}} \, dx\) [973]

Optimal result

Integrand size = 20, antiderivative size = 162 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {3 b \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{256 a^3 x^4}+\frac {b \left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{32 a^2 x^8}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{10 a x^{10}}+\frac {3 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{512 a^{7/2}} \] Output:

-3/256*b*(-4*a*c+b^2)*(b*x^2+2*a)*(c*x^4+b*x^2+a)^(1/2)/a^3/x^4+1/32*b*(b* 
x^2+2*a)*(c*x^4+b*x^2+a)^(3/2)/a^2/x^8-1/10*(c*x^4+b*x^2+a)^(5/2)/a/x^10+3 
/512*b*(-4*a*c+b^2)^2*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2 
))/a^(7/2)
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\frac {-\frac {\sqrt {a} \sqrt {a+b x^2+c x^4} \left (128 a^4+15 b^4 x^8-10 a b^2 x^6 \left (b+10 c x^2\right )+16 a^3 \left (11 b x^2+16 c x^4\right )+8 a^2 x^4 \left (b^2+7 b c x^2+16 c^2 x^4\right )\right )}{x^{10}}-15 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{1280 a^{7/2}} \] Input:

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^11,x]
 

Output:

(-((Sqrt[a]*Sqrt[a + b*x^2 + c*x^4]*(128*a^4 + 15*b^4*x^8 - 10*a*b^2*x^6*( 
b + 10*c*x^2) + 16*a^3*(11*b*x^2 + 16*c*x^4) + 8*a^2*x^4*(b^2 + 7*b*c*x^2 
+ 16*c^2*x^4)))/x^10) - 15*b*(b^2 - 4*a*c)^2*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a 
 + b*x^2 + c*x^4])/Sqrt[a]])/(1280*a^(7/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1434, 1157, 1152, 1152, 1154, 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.

\(\displaystyle \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx\)

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1157

\(\displaystyle \frac {1}{2} \left (-\frac {b \int \frac {\left (c x^4+b x^2+a\right )^{3/2}}{x^{10}}dx^2}{2 a}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 1152

\(\displaystyle \frac {1}{2} \left (-\frac {b \left (-\frac {3 \left (b^2-4 a c\right ) \int \frac {\sqrt {c x^4+b x^2+a}}{x^6}dx^2}{16 a}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}\right )}{2 a}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 1152

\(\displaystyle \frac {1}{2} \left (-\frac {b \left (-\frac {3 \left (b^2-4 a c\right ) \left (-\frac {\left (b^2-4 a c\right ) \int \frac {1}{x^2 \sqrt {c x^4+b x^2+a}}dx^2}{8 a}-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{4 a x^4}\right )}{16 a}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}\right )}{2 a}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (-\frac {b \left (-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 a-x^4}d\frac {b x^2+2 a}{\sqrt {c x^4+b x^2+a}}}{4 a}-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{4 a x^4}\right )}{16 a}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}\right )}{2 a}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{5 a x^{10}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {b \left (-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{8 a^{3/2}}-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{4 a x^4}\right )}{16 a}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{8 a x^8}\right )}{2 a}-\frac {\left (a+b x^2+c x^4\right )^{5/2}}{5 a x^{10}}\right )\)

Input:

Int[(a + b*x^2 + c*x^4)^(3/2)/x^11,x]
 

Output:

(-1/5*(a + b*x^2 + c*x^4)^(5/2)/(a*x^10) - (b*(-1/8*((2*a + b*x^2)*(a + b* 
x^2 + c*x^4)^(3/2))/(a*x^8) - (3*(b^2 - 4*a*c)*(-1/4*((2*a + b*x^2)*Sqrt[a 
 + b*x^2 + c*x^4])/(a*x^4) + ((b^2 - 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[ 
a]*Sqrt[a + b*x^2 + c*x^4])])/(8*a^(3/2))))/(16*a)))/(2*a))/2
 

Defintions of rubi rules used

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 1152

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/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a 
*c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)))   Int[(d + e*x)^(m + 2)*(a + b*x + 
 c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] 
 && GtQ[p, 0]
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1157

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e^2)) 
  Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && EqQ[m + 2*p + 3, 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 = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(\frac {\frac {3 \left (a c -\frac {b^{2}}{4}\right )^{2} b \,x^{10} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (\frac {x^{4} \left (16 c^{2} x^{4}+7 b c \,x^{2}+b^{2}\right ) a^{\frac {5}{2}}}{16}-\frac {5 b^{2} x^{6} \left (10 c \,x^{2}+b \right ) a^{\frac {3}{2}}}{64}+\left (2 c \,x^{4}+\frac {11}{8} b \,x^{2}\right ) a^{\frac {7}{2}}+\frac {15 \sqrt {a}\, b^{4} x^{8}}{128}+a^{\frac {9}{2}}\right )}{10}}{a^{\frac {7}{2}} x^{10}}\)	\(153\)
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (128 a^{2} c^{2} x^{8}-100 a \,b^{2} c \,x^{8}+15 b^{4} x^{8}+56 a^{2} b c \,x^{6}-10 a \,b^{3} x^{6}+256 a^{3} c \,x^{4}+8 a^{2} b^{2} x^{4}+176 a^{3} b \,x^{2}+128 a^{4}\right )}{1280 x^{10} a^{3}}+\frac {3 b \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{512 a^{\frac {7}{2}}}\)	\(165\)
default	\(-\frac {3 b^{3} c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{64 a^{\frac {5}{2}}}+\frac {5 b^{2} c \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 a^{2} x^{2}}-\frac {7 b c \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 a \,x^{4}}+\frac {3 b \,c^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}+\frac {b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 a^{2} x^{4}}-\frac {3 b^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{256 a^{3} x^{2}}+\frac {3 b^{5} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{512 a^{\frac {7}{2}}}-\frac {c^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{10 a \,x^{2}}-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 a \,x^{6}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{10 x^{10}}-\frac {11 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{80 x^{8}}-\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{5 x^{6}}\)	\(337\)
elliptic	\(-\frac {3 b^{3} c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{64 a^{\frac {5}{2}}}+\frac {5 b^{2} c \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 a^{2} x^{2}}-\frac {7 b c \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 a \,x^{4}}+\frac {3 b \,c^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}+\frac {b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 a^{2} x^{4}}-\frac {3 b^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}}{256 a^{3} x^{2}}+\frac {3 b^{5} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{512 a^{\frac {7}{2}}}-\frac {c^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{10 a \,x^{2}}-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{160 a \,x^{6}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{10 x^{10}}-\frac {11 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{80 x^{8}}-\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{5 x^{6}}\)	\(337\)

Input:

int((c*x^4+b*x^2+a)^(3/2)/x^11,x,method=_RETURNVERBOSE)
 

Output:

3/32/a^(7/2)*((a*c-1/4*b^2)^2*b*x^10*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+ 
a)^(1/2))/x^2)-16/15*(c*x^4+b*x^2+a)^(1/2)*(1/16*x^4*(16*c^2*x^4+7*b*c*x^2 
+b^2)*a^(5/2)-5/64*b^2*x^6*(10*c*x^2+b)*a^(3/2)+(2*c*x^4+11/8*b*x^2)*a^(7/ 
2)+15/128*a^(1/2)*b^4*x^8+a^(9/2)))/x^10
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.36 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {a} x^{10} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{8} + 176 \, a^{4} b x^{2} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{6} + 128 \, a^{5} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{5120 \, a^{4} x^{10}}, -\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (15 \, a b^{4} - 100 \, a^{2} b^{2} c + 128 \, a^{3} c^{2}\right )} x^{8} + 176 \, a^{4} b x^{2} - 2 \, {\left (5 \, a^{2} b^{3} - 28 \, a^{3} b c\right )} x^{6} + 128 \, a^{5} + 8 \, {\left (a^{3} b^{2} + 32 \, a^{4} c\right )} x^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{2560 \, a^{4} x^{10}}\right ] \] Input:

integrate((c*x^4+b*x^2+a)^(3/2)/x^11,x, algorithm="fricas")
 

Output:

[1/5120*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(a)*x^10*log(-((b^2 + 4*a 
*c)*x^4 + 8*a*b*x^2 + 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8* 
a^2)/x^4) - 4*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*x^8 + 176*a^4*b*x^ 
2 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^6 + 128*a^5 + 8*(a^3*b^2 + 32*a^4*c)*x^4) 
*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^10), -1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^ 
2*b*c^2)*sqrt(-a)*x^10*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sq 
rt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*((15*a*b^4 - 100*a^2*b^2*c + 128*a^3 
*c^2)*x^8 + 176*a^4*b*x^2 - 2*(5*a^2*b^3 - 28*a^3*b*c)*x^6 + 128*a^5 + 8*( 
a^3*b^2 + 32*a^4*c)*x^4)*sqrt(c*x^4 + b*x^2 + a))/(a^4*x^10)]
 

Sympy [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x^{11}}\, dx \] Input:

integrate((c*x**4+b*x**2+a)**(3/2)/x**11,x)
 

Output:

Integral((a + b*x**2 + c*x**4)**(3/2)/x**11, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*x^4+b*x^2+a)^(3/2)/x^11,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 832 vs. \(2 (140) = 280\).

Time = 0.20 (sec) , antiderivative size = 832, normalized size of antiderivative = 5.14 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx =\text {Too large to display} \] Input:

integrate((c*x^4+b*x^2+a)^(3/2)/x^11,x, algorithm="giac")
 

Output:

-3/256*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 
+ b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^3) + 1/1280*(15*(sqrt(c)*x^2 - sqrt(c* 
x^4 + b*x^2 + a))^9*b^5 - 120*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^9*a* 
b^3*c + 240*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^9*a^2*b*c^2 + 1280*(sq 
rt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^8*a^3*c^(5/2) - 70*(sqrt(c)*x^2 - sqr 
t(c*x^4 + b*x^2 + a))^7*a*b^5 + 560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a) 
)^7*a^2*b^3*c + 2720*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a^3*b*c^2 + 
 5120*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^6*a^3*b^2*c^(3/2) + 128*(sqr 
t(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^2*b^5 + 2560*(sqrt(c)*x^2 - sqrt(c 
*x^4 + b*x^2 + a))^5*a^3*b^3*c + 3840*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + 
a))^5*a^4*b*c^2 + 1280*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^3*b^4*s 
qrt(c) + 2560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^4*b^2*c^(3/2) + 
2560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^5*c^(5/2) + 70*(sqrt(c)*x 
^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^3*b^5 + 2000*(sqrt(c)*x^2 - sqrt(c*x^4 + 
 b*x^2 + a))^3*a^4*b^3*c + 2400*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3* 
a^5*b*c^2 + 2560*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^5*b^2*c^(3/2) 
 - 15*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^4*b^5 + 120*(sqrt(c)*x^2 - 
 sqrt(c*x^4 + b*x^2 + a))*a^5*b^3*c + 1040*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x 
^2 + a))*a^6*b*c^2 + 256*a^7*c^(5/2))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 
+ a))^2 - a)^5*a^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x^{11}} \,d x \] Input:

int((a + b*x^2 + c*x^4)^(3/2)/x^11,x)
 

Output:

int((a + b*x^2 + c*x^4)^(3/2)/x^11, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\int \frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{x^{11}}d x \] Input:

int((c*x^4+b*x^2+a)^(3/2)/x^11,x)
 

Output:

int((c*x^4+b*x^2+a)^(3/2)/x^11,x)