\(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^3} \, dx\) [969]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3}{8} \left (3 b+2 c x^2\right ) \sqrt {a+b x^2+c x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{2 x^2}-\frac {3}{4} \sqrt {a} b \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )+\frac {3 \left (b^2+4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{16 \sqrt {c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{16} \left (\frac {2 \sqrt {a+b x^2+c x^4} \left (-4 a+5 b x^2+2 c x^4\right )}{x^2}+24 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )-\frac {3 \left (b^2+4 a c\right ) \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{\sqrt {c}}\right ) \] Input:

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

Output:

((2*Sqrt[a + b*x^2 + c*x^4]*(-4*a + 5*b*x^2 + 2*c*x^4))/x^2 + 24*Sqrt[a]*b 
*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]] - (3*(b^2 + 4*a* 
c)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/Sqrt[c])/16
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1434, 1161, 1231, 25, 27, 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.

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1161

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

\(\Big \downarrow \) 1231

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 1231

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 x^{2}}-\frac {3 \sqrt {a}\, b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}+\frac {c \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}+\frac {3 a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}\)	\(170\)
risch	\(\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 x^{2}}-\frac {3 \sqrt {a}\, b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}+\frac {c \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}+\frac {3 a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}\)	\(170\)
elliptic	\(\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{16 \sqrt {c}}-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 x^{2}}-\frac {3 \sqrt {a}\, b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4}+\frac {c \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{4}+\frac {5 b \sqrt {c \,x^{4}+b \,x^{2}+a}}{8}+\frac {3 a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4}\)	\(170\)
pseudoelliptic	\(\frac {4 c^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{4}-12 \sqrt {a}\, b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) x^{2} \sqrt {c}+10 b \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2} \sqrt {c}+12 a c \,x^{2} \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )+3 b^{2} x^{2} \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )-12 a c \,x^{2} \ln \left (2\right )-3 b^{2} x^{2} \ln \left (2\right )-8 a \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}}{16 x^{2} \sqrt {c}}\)	\(218\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 713, normalized size of antiderivative = 4.75 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/32*(12*sqrt(a)*b*c*x^2*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) + 3*(b^2 + 4*a*c)*sqrt 
(c)*x^2*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*c^2*x^4 + 5*b*c*x^2 - 4*a*c)*sqrt(c*x^4 + 
 b*x^2 + a))/(c*x^2), 1/16*(6*sqrt(a)*b*c*x^2*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) - 
3*(b^2 + 4*a*c)*sqrt(-c)*x^2*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)) + 2*(2*c^2*x^4 + 5*b*c*x^2 - 4*a*c 
)*sqrt(c*x^4 + b*x^2 + a))/(c*x^2), 1/32*(24*sqrt(-a)*b*c*x^2*arctan(1/2*s 
qrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 
 3*(b^2 + 4*a*c)*sqrt(c)*x^2*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*c^2*x^4 + 5*b*c*x^2 
- 4*a*c)*sqrt(c*x^4 + b*x^2 + a))/(c*x^2), 1/16*(12*sqrt(-a)*b*c*x^2*arcta 
n(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)) - 3*(b^2 + 4*a*c)*sqrt(-c)*x^2*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)) + 2*(2*c^2*x^4 + 5*b*c*x^2 
 - 4*a*c)*sqrt(c*x^4 + b*x^2 + a))/(c*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((c*x^4+b*x^2+a)^(3/2)/x^3,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 = 0.17 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3 \, a b \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a}} + \frac {1}{8} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + 5 \, b\right )} - \frac {3 \, {\left (b^{2} + 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, \sqrt {c}} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a b + 2 \, a^{2} \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}} \] Input:

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

Output:

3/2*a*b*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/sqrt(-a) 
 + 1/8*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + 5*b) - 3/16*(b^2 + 4*a*c)*log(ab 
s(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/sqrt(c) + 1/2*(( 
sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a*b + 2*a^2*sqrt(c))/((sqrt(c)*x^2 
- sqrt(c*x^4 + b*x^2 + a))^2 - a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {24 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+b +2 c \,x^{2}}{\sqrt {4 a c -b^{2}}}\right ) a c \,x^{2}+6 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+b +2 c \,x^{2}}{\sqrt {4 a c -b^{2}}}\right ) b^{2} x^{2}-8 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, a b -16 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, a c \,x^{2}+10 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2} x^{2}+24 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, b c \,x^{4}+8 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2} x^{6}+48 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{5}+b \,x^{3}+a x}d x \right ) a b c \,x^{2}+24 \sqrt {c}\, \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{5}+b \,x^{3}+a x}d x \right ) a \,b^{2} x^{2}+48 \sqrt {c}\, \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{5}+b \,x^{3}+a x}d x \right ) a b c \,x^{4}+12 \,\mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+b +2 c \,x^{2}}{\sqrt {4 a c -b^{2}}}\right ) a b c \,x^{2}+24 \,\mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+b +2 c \,x^{2}}{\sqrt {4 a c -b^{2}}}\right ) a \,c^{2} x^{4}+3 \,\mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+b +2 c \,x^{2}}{\sqrt {4 a c -b^{2}}}\right ) b^{3} x^{2}+6 \,\mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+b +2 c \,x^{2}}{\sqrt {4 a c -b^{2}}}\right ) b^{2} c \,x^{4}-16 a^{2} c +4 a b c \,x^{2}-8 a \,c^{2} x^{4}+20 b^{2} c \,x^{4}+28 b \,c^{2} x^{6}+8 c^{3} x^{8}}{16 x^{2} \left (2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c +\sqrt {c}\, b +2 \sqrt {c}\, c \,x^{2}\right )} \] Input:

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

Output:

(24*sqrt(c)*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*c*x**2 + 6*sqrt(c)*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))*b**2*x**2 - 8*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*a*b - 16* 
sqrt(c)*sqrt(a + b*x**2 + c*x**4)*a*c*x**2 + 10*sqrt(c)*sqrt(a + b*x**2 + 
c*x**4)*b**2*x**2 + 24*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*b*c*x**4 + 8*sqrt 
(c)*sqrt(a + b*x**2 + c*x**4)*c**2*x**6 + 48*sqrt(a + b*x**2 + c*x**4)*int 
(sqrt(a + b*x**2 + c*x**4)/(a*x + b*x**3 + c*x**5),x)*a*b*c*x**2 + 24*sqrt 
(c)*int(sqrt(a + b*x**2 + c*x**4)/(a*x + b*x**3 + c*x**5),x)*a*b**2*x**2 + 
 48*sqrt(c)*int(sqrt(a + b*x**2 + c*x**4)/(a*x + b*x**3 + c*x**5),x)*a*b*c 
*x**4 + 12*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4 
*a*c - b**2))*a*b*c*x**2 + 24*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b 
 + 2*c*x**2)/sqrt(4*a*c - b**2))*a*c**2*x**4 + 3*log((2*sqrt(c)*sqrt(a + b 
*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**3*x**2 + 6*log((2*s 
qrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*b**2* 
c*x**4 - 16*a**2*c + 4*a*b*c*x**2 - 8*a*c**2*x**4 + 20*b**2*c*x**4 + 28*b* 
c**2*x**6 + 8*c**3*x**8)/(16*x**2*(2*sqrt(a + b*x**2 + c*x**4)*c + sqrt(c) 
*b + 2*sqrt(c)*c*x**2))