3.10.42 \(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^5} \, dx\) [942]

3.10.42.1 Optimal result
3.10.42.2 Mathematica [A] (verified)
3.10.42.3 Rubi [A] (verified)
3.10.42.4 Maple [A] (verified)
3.10.42.5 Fricas [A] (verification not implemented)
3.10.42.6 Sympy [F]
3.10.42.7 Maxima [F(-2)]
3.10.42.8 Giac [B] (verification not implemented)
3.10.42.9 Mupad [F(-1)]

3.10.42.1 Optimal result

Integrand size = 20, antiderivative size = 151 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx=-\frac {3 \left (b-2 c x^2\right ) \sqrt {a+b x^2+c x^4}}{8 x^2}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{4 x^4}-\frac {3 \left (b^2+4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 \sqrt {a}}+\frac {3}{4} b \sqrt {c} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right ) \]

output
-1/4*(c*x^4+b*x^2+a)^(3/2)/x^4-3/16*(4*a*c+b^2)*arctanh(1/2*(b*x^2+2*a)/a^ 
(1/2)/(c*x^4+b*x^2+a)^(1/2))/a^(1/2)+3/4*b*arctanh(1/2*(2*c*x^2+b)/c^(1/2) 
/(c*x^4+b*x^2+a)^(1/2))*c^(1/2)-3/8*(-2*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/x^2
 
3.10.42.2 Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\frac {1}{8} \left (\frac {\sqrt {a+b x^2+c x^4} \left (-2 a-5 b x^2+4 c x^4\right )}{x^4}+\frac {3 \left (b^2+4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a}}-6 b \sqrt {c} \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )\right ) \]

input
Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^5,x]
 
output
((Sqrt[a + b*x^2 + c*x^4]*(-2*a - 5*b*x^2 + 4*c*x^4))/x^4 + (3*(b^2 + 4*a* 
c)*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/Sqrt[a] - 6*b 
*Sqrt[c]*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/8
 
3.10.42.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1434, 1161, 1230, 25, 1269, 1092, 219, 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.

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1161

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

\(\Big \downarrow \) 1230

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

input
Int[(a + b*x^2 + c*x^4)^(3/2)/x^5,x]
 
output
(-1/2*(a + b*x^2 + c*x^4)^(3/2)/x^4 + (3*(-(((b - 2*c*x^2)*Sqrt[a + b*x^2 
+ c*x^4])/x^2) + (-(((b^2 + 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a 
 + b*x^2 + c*x^4])])/Sqrt[a]) + 4*b*Sqrt[c]*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[ 
c]*Sqrt[a + b*x^2 + c*x^4])])/2))/4)/2
 

3.10.42.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 1161
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[p/(e*(m + 1))   Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^(p - 
 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && GtQ[p, 0] && (IntegerQ[p] || 
 LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, 
 c, d, e, m, p, x]
 

rule 1230
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - 
 d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2*p 
+ 2))), x] + Simp[p/(e^2*(m + 1)*(m + 2*p + 2))   Int[(d + e*x)^(m + 1)*(a 
+ b*x + c*x^2)^(p - 1)*Simp[g*(b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m 
+ 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, 
 x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, - 
1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&  !ILtQ 
[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1269
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 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]
 
3.10.42.4 Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99

method result size
pseudoelliptic \(-\frac {3 \left (x^{4} \left (a c +\frac {b^{2}}{4}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )-\ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right ) b \,x^{4} \sqrt {a}\, \sqrt {c}+\frac {\left (a^{\frac {3}{2}}+\left (-2 c \,x^{4}+\frac {5}{2} b \,x^{2}\right ) \sqrt {a}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{3}+\ln \left (2\right ) b \,x^{4} \sqrt {a}\, \sqrt {c}\right )}{4 \sqrt {a}\, x^{4}}\) \(150\)
risch \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (5 b \,x^{2}+2 a \right )}{8 x^{4}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}-\frac {3 b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}\) \(163\)
default \(\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 x^{4}}-\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}\) \(174\)
elliptic \(\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{2}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 x^{4}}-\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 x^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 \sqrt {a}}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}\) \(174\)

input
int((c*x^4+b*x^2+a)^(3/2)/x^5,x,method=_RETURNVERBOSE)
 
output
-3/4/a^(1/2)*(x^4*(a*c+1/4*b^2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^(1 
/2))/x^2)-ln((2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/2)+b)/c^(1/2))*b*x^4*a^ 
(1/2)*c^(1/2)+1/3*(a^(3/2)+(-2*c*x^4+5/2*b*x^2)*a^(1/2))*(c*x^4+b*x^2+a)^( 
1/2)+ln(2)*b*x^4*a^(1/2)*c^(1/2))/x^4
 
3.10.42.5 Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 713, normalized size of antiderivative = 4.72 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\left [\frac {12 \, a b \sqrt {c} x^{4} \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 ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {a} x^{4} \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 (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, a x^{4}}, -\frac {24 \, a b \sqrt {-c} x^{4} \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 ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {a} x^{4} \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 (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, a x^{4}}, \frac {6 \, a b \sqrt {c} x^{4} \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 ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-a} x^{4} \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 (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, a x^{4}}, -\frac {12 \, a b \sqrt {-c} x^{4} \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 ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-a} x^{4} \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 (4 \, a c x^{4} - 5 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, a x^{4}}\right ] \]

input
integrate((c*x^4+b*x^2+a)^(3/2)/x^5,x, algorithm="fricas")
 
output
[1/32*(12*a*b*sqrt(c)*x^4*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) + 3*(b^2 + 4*a*c)*sqrt(a)*x^4* 
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*(4*a*c*x^4 - 5*a*b*x^2 - 2*a^2)*sqrt(c*x^4 + 
 b*x^2 + a))/(a*x^4), -1/32*(24*a*b*sqrt(-c)*x^4*arctan(1/2*sqrt(c*x^4 + b 
*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) - 3*(b^2 + 4*a 
*c)*sqrt(a)*x^4*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*(4*a*c*x^4 - 5*a*b*x^2 - 2*a 
^2)*sqrt(c*x^4 + b*x^2 + a))/(a*x^4), 1/16*(6*a*b*sqrt(c)*x^4*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) + 3*(b^2 + 4*a*c)*sqrt(-a)*x^4*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*(4*a*c*x^4 - 5*a*b*x^2 
 - 2*a^2)*sqrt(c*x^4 + b*x^2 + a))/(a*x^4), -1/16*(12*a*b*sqrt(-c)*x^4*arc 
tan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 
+ a*c)) - 3*(b^2 + 4*a*c)*sqrt(-a)*x^4*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*(4*a*c*x^4 - 5*a*b*x 
^2 - 2*a^2)*sqrt(c*x^4 + b*x^2 + a))/(a*x^4)]
 
3.10.42.6 Sympy [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x^{5}}\, dx \]

input
integrate((c*x**4+b*x**2+a)**(3/2)/x**5,x)
 
output
Integral((a + b*x**2 + c*x**4)**(3/2)/x**5, x)
 
3.10.42.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\text {Exception raised: ValueError} \]

input
integrate((c*x^4+b*x^2+a)^(3/2)/x^5,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
 
3.10.42.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (125) = 250\).

Time = 0.37 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.01 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx=-\frac {3}{4} \, b \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right ) + \frac {1}{2} \, \sqrt {c x^{4} + b x^{2} + a} c + \frac {3 \, {\left (b^{2} + 4 \, a c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a}} + \frac {5 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a c + 16 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a b \sqrt {c} - 3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} c - 8 \, a^{2} b \sqrt {c}}{8 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{2}} \]

input
integrate((c*x^4+b*x^2+a)^(3/2)/x^5,x, algorithm="giac")
 
output
-3/4*b*sqrt(c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) 
- b)) + 1/2*sqrt(c*x^4 + b*x^2 + a)*c + 3/8*(b^2 + 4*a*c)*arctan(-(sqrt(c) 
*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/sqrt(-a) + 1/8*(5*(sqrt(c)*x^2 - 
 sqrt(c*x^4 + b*x^2 + a))^3*b^2 + 4*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a) 
)^3*a*c + 16*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a*b*sqrt(c) - 3*(sq 
rt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a*b^2 + 4*(sqrt(c)*x^2 - sqrt(c*x^4 + 
 b*x^2 + a))*a^2*c - 8*a^2*b*sqrt(c))/((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + 
 a))^2 - a)^2
 
3.10.42.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^5} \, dx=\int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x^5} \,d x \]

input
int((a + b*x^2 + c*x^4)^(3/2)/x^5,x)
 
output
int((a + b*x^2 + c*x^4)^(3/2)/x^5, x)