\(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^9} \, dx\) [972]

Optimal result

Integrand size = 20, antiderivative size = 133 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {3 \left (b^2-4 a c\right ) \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{128 a^2 x^4}-\frac {\left (2 a+b x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{16 a x^8}-\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{256 a^{5/2}} \] Output:

3/128*(-4*a*c+b^2)*(b*x^2+2*a)*(c*x^4+b*x^2+a)^(1/2)/a^2/x^4-1/16*(b*x^2+2 
*a)*(c*x^4+b*x^2+a)^(3/2)/a/x^8-3/256*(-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^(5/2)
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {-\frac {\sqrt {a} \left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4} \left (8 a^2+8 a b x^2-3 b^2 x^4+20 a c x^4\right )}{x^8}+3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{128 a^{5/2}} \] Input:

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

Output:

(-((Sqrt[a]*(2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4]*(8*a^2 + 8*a*b*x^2 - 3*b 
^2*x^4 + 20*a*c*x^4))/x^8) + 3*(b^2 - 4*a*c)^2*ArcTanh[(Sqrt[c]*x^2 - Sqrt 
[a + b*x^2 + c*x^4])/Sqrt[a]])/(128*a^(5/2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1434, 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.

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1152

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

\(\Big \downarrow \) 1152

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-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)*Ar 
cTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(8*a^(3/2))))/(1 
6*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 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 = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92

method	result	size
pseudoelliptic	\(-\frac {3 \left (\left (a c -\frac {b^{2}}{4}\right )^{2} x^{8} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )-\frac {\left (-\frac {2 b \,x^{4} \left (10 c \,x^{2}+b \right ) a^{\frac {3}{2}}}{3}-8 \left (\frac {5 c \,x^{2}}{3}+b \right ) x^{2} a^{\frac {5}{2}}+\sqrt {a}\, b^{3} x^{6}-\frac {16 a^{\frac {7}{2}}}{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}\right )}{16 a^{\frac {5}{2}} x^{8}}\)	\(123\)
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (20 a b c \,x^{6}-3 b^{3} x^{6}+40 a^{2} c \,x^{4}+2 b^{2} x^{4} a +24 a^{2} b \,x^{2}+16 a^{3}\right )}{128 x^{8} a^{2}}-\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{256 a^{\frac {5}{2}}}\)	\(130\)
default	\(\frac {3 b^{2} c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}-\frac {5 b c \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 a \,x^{2}}-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 a \,x^{4}}+\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 a^{2} x^{2}}-\frac {3 b^{4} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{256 a^{\frac {5}{2}}}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{6}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{8}}-\frac {5 c \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{4}}-\frac {3 c^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}\)	\(260\)
elliptic	\(\frac {3 b^{2} c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{32 a^{\frac {3}{2}}}-\frac {5 b c \sqrt {c \,x^{4}+b \,x^{2}+a}}{32 a \,x^{2}}-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{64 a \,x^{4}}+\frac {3 b^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{128 a^{2} x^{2}}-\frac {3 b^{4} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{256 a^{\frac {5}{2}}}-\frac {3 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{6}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{8}}-\frac {5 c \sqrt {c \,x^{4}+b \,x^{2}+a}}{16 x^{4}}-\frac {3 c^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}\)	\(260\)

Input:

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

Output:

-3/16/a^(5/2)*((a*c-1/4*b^2)^2*x^8*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a) 
^(1/2))/x^2)-1/8*(-2/3*b*x^4*(10*c*x^2+b)*a^(3/2)-8*(5/3*c*x^2+b)*x^2*a^(5 
/2)+a^(1/2)*b^3*x^6-16/3*a^(7/2))*(c*x^4+b*x^2+a)^(1/2))/x^8
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.40 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{8} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{6} - 24 \, a^{3} b x^{2} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{4} - 16 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{512 \, a^{3} x^{8}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{6} - 24 \, a^{3} b x^{2} - 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{4} - 16 \, a^{4}\right )} \sqrt {c x^{4} + b x^{2} + a}}{256 \, a^{3} x^{8}}\right ] \] Input:

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

Output:

[1/512*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(a)*x^8*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*((3*a*b^3 - 20*a^2*b*c)*x^6 - 24*a^3*b*x^2 - 2*(a^2*b^2 + 20*a^3* 
c)*x^4 - 16*a^4)*sqrt(c*x^4 + b*x^2 + a))/(a^3*x^8), 1/256*(3*(b^4 - 8*a*b 
^2*c + 16*a^2*c^2)*sqrt(-a)*x^8*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 
+ 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*((3*a*b^3 - 20*a^2*b*c)*x^6 
 - 24*a^3*b*x^2 - 2*(a^2*b^2 + 20*a^3*c)*x^4 - 16*a^4)*sqrt(c*x^4 + b*x^2 
+ a))/(a^3*x^8)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((c*x^4+b*x^2+a)^(3/2)/x^9,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. 606 vs. \(2 (115) = 230\).

Time = 0.17 (sec) , antiderivative size = 606, normalized size of antiderivative = 4.56 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^9} \, dx=\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{128 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} b^{4} - 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a b^{2} c - 80 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{7} a^{2} c^{2} - 256 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{6} a^{2} b c^{\frac {3}{2}} - 11 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a b^{4} - 168 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{2} b^{2} c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{5} a^{3} c^{2} - 128 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{4} a^{2} b^{3} \sqrt {c} - 11 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{2} b^{4} - 168 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{3} b^{2} c - 48 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a^{4} c^{2} - 256 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a^{4} b c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{3} b^{4} - 24 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{4} b^{2} c - 80 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{5} c^{2}}{128 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{4} a^{2}} \] Input:

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

Output:

3/128*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b 
*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/128*(3*(sqrt(c)*x^2 - sqrt(c*x^4 + 
 b*x^2 + a))^7*b^4 - 24*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a*b^2*c 
- 80*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^7*a^2*c^2 - 256*(sqrt(c)*x^2 
- sqrt(c*x^4 + b*x^2 + a))^6*a^2*b*c^(3/2) - 11*(sqrt(c)*x^2 - sqrt(c*x^4 
+ b*x^2 + a))^5*a*b^4 - 168*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^2* 
b^2*c - 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^5*a^3*c^2 - 128*(sqrt(c 
)*x^2 - sqrt(c*x^4 + b*x^2 + a))^4*a^2*b^3*sqrt(c) - 11*(sqrt(c)*x^2 - sqr 
t(c*x^4 + b*x^2 + a))^3*a^2*b^4 - 168*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + 
a))^3*a^3*b^2*c - 48*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a^4*c^2 - 2 
56*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a^4*b*c^(3/2) + 3*(sqrt(c)*x^ 
2 - sqrt(c*x^4 + b*x^2 + a))*a^3*b^4 - 24*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^ 
2 + a))*a^4*b^2*c - 80*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^5*c^2)/(( 
(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^4*a^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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