\(\int \frac {1}{x^3 (a+b x^2+c x^4)^{3/2}} \, dx\) [1014]

Optimal result

Integrand size = 20, antiderivative size = 139 \[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x^2+c x^4}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {3 b \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{5/2}} \] Output:

(b*c*x^2-2*a*c+b^2)/a/(-4*a*c+b^2)/x^2/(c*x^4+b*x^2+a)^(1/2)-1/2*(-8*a*c+3 
*b^2)*(c*x^4+b*x^2+a)^(1/2)/a^2/(-4*a*c+b^2)/x^2+3/4*b*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 = 0.50 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {-4 a^2 c+3 b^2 x^2 \left (b+c x^2\right )+a \left (b^2-10 b c x^2-8 c^2 x^4\right )}{2 a^2 \left (-b^2+4 a c\right ) x^2 \sqrt {a+b x^2+c x^4}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{2 a^{5/2}} \] Input:

Integrate[1/(x^3*(a + b*x^2 + c*x^4)^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1434, 1165, 27, 1228, 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.

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1165

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1228

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

Int[1/(x^3*(a + b*x^2 + c*x^4)^(3/2)),x]
 

Output:

((2*(b^2 - 2*a*c + b*c*x^2))/(a*(b^2 - 4*a*c)*x^2*Sqrt[a + b*x^2 + c*x^4]) 
 + (-(((3*b^2 - 8*a*c)*Sqrt[a + b*x^2 + c*x^4])/(a*x^2)) + (3*b*(b^2 - 4*a 
*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*a^(3/2) 
))/(a*(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 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 1165

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

rule 1228

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + 
 b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e 
*f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x 
] && EqQ[Simplify[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 = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(-\frac {-\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (a c -\frac {b^{2}}{4}\right ) a b \,x^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2}+a^{\frac {3}{2}} \left (c \,a^{2}+\left (2 c^{2} x^{4}+\frac {5}{2} b c \,x^{2}-\frac {1}{4} b^{2}\right ) a -\frac {3 b^{2} x^{2} \left (c \,x^{2}+b \right )}{4}\right )}{2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (a c -\frac {b^{2}}{4}\right ) a^{\frac {7}{2}} x^{2}}\)	\(146\)
default	\(-\frac {1}{2 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (2 c \,x^{2}+b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{a^{\frac {3}{2}}}\right )}{4 a}-\frac {2 c \left (2 c \,x^{2}+b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}\)	\(167\)
elliptic	\(-\frac {1}{2 a \,x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {b \left (2 c \,x^{2}+b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{a^{\frac {3}{2}}}\right )}{4 a}-\frac {2 c \left (2 c \,x^{2}+b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}\)	\(167\)
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a^{2} x^{2}}+\frac {b^{2} x^{2} c}{a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {b^{3}}{4 a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {2 c^{2} x^{2}}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {3 b}{4 a^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {3 b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\)	\(191\)

Input:

int(1/x^3/(c*x^4+b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 485, normalized size of antiderivative = 3.49 \[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{6} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{4} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{8 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{6} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{4} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}, -\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{6} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{4} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{4 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{6} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{4} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}\right ] \] Input:

integrate(1/x^3/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[1/8*(3*((b^3*c - 4*a*b*c^2)*x^6 + (b^4 - 4*a*b^2*c)*x^4 + (a*b^3 - 4*a^2* 
b*c)*x^2)*sqrt(a)*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^2*c - 8*a^2*c^2)*x 
^4 + a^2*b^2 - 4*a^3*c + (3*a*b^3 - 10*a^2*b*c)*x^2)*sqrt(c*x^4 + b*x^2 + 
a))/((a^3*b^2*c - 4*a^4*c^2)*x^6 + (a^3*b^3 - 4*a^4*b*c)*x^4 + (a^4*b^2 - 
4*a^5*c)*x^2), -1/4*(3*((b^3*c - 4*a*b*c^2)*x^6 + (b^4 - 4*a*b^2*c)*x^4 + 
(a*b^3 - 4*a^2*b*c)*x^2)*sqrt(-a)*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^2*c - 8*a^2*c^2)* 
x^4 + a^2*b^2 - 4*a^3*c + (3*a*b^3 - 10*a^2*b*c)*x^2)*sqrt(c*x^4 + b*x^2 + 
 a))/((a^3*b^2*c - 4*a^4*c^2)*x^6 + (a^3*b^3 - 4*a^4*b*c)*x^4 + (a^4*b^2 - 
 4*a^5*c)*x^2)]
 

Sympy [F]

\[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/x**3/(c*x**4+b*x**2+a)**(3/2),x)
 

Output:

Integral(1/(x**3*(a + b*x**2 + c*x**4)**(3/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/x^3/(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 = 200, normalized size of antiderivative = 1.44 \[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\frac {{\left (a^{2} b^{2} c - 2 \, a^{3} c^{2}\right )} x^{2}}{a^{4} b^{2} - 4 \, a^{5} c} + \frac {a^{2} b^{3} - 3 \, a^{3} b c}{a^{4} b^{2} - 4 \, a^{5} c}}{\sqrt {c x^{4} + b x^{2} + a}} - \frac {3 \, b \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{2}} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} b + 2 \, a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \] Input:

integrate(1/x^3/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

-((a^2*b^2*c - 2*a^3*c^2)*x^2/(a^4*b^2 - 4*a^5*c) + (a^2*b^3 - 3*a^3*b*c)/ 
(a^4*b^2 - 4*a^5*c))/sqrt(c*x^4 + b*x^2 + a) - 3/2*b*arctan(-(sqrt(c)*x^2 
- sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) + 1/2*((sqrt(c)*x^2 - 
sqrt(c*x^4 + b*x^2 + a))*b + 2*a*sqrt(c))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b* 
x^2 + a))^2 - a)*a^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int(1/(x^3*(a + b*x^2 + c*x^4)^(3/2)),x)
 

Output:

int(1/(x^3*(a + b*x^2 + c*x^4)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,x^{3}+\sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,x^{5}+\sqrt {c \,x^{4}+b \,x^{2}+a}\, c \,x^{7}}d x \] Input:

int(1/x^3/(c*x^4+b*x^2+a)^(3/2),x)
 

Output:

int(1/(sqrt(a + b*x**2 + c*x**4)*a*x**3 + sqrt(a + b*x**2 + c*x**4)*b*x**5 
 + sqrt(a + b*x**2 + c*x**4)*c*x**7),x)