\(\int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx\) [71]

Optimal result

Integrand size = 24, antiderivative size = 119 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 a x^3}+\frac {3 b \sqrt {a x^2+b x^3+c x^4}}{4 a^2 x^2}-\frac {\left (3 b^2-4 a c\right ) \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{8 a^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {-\sqrt {a} (2 a-3 b x) (a+x (b+c x))+\left (3 b^2-4 a c\right ) x^2 \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{4 a^{5/2} x \sqrt {x^2 (a+x (b+c x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1976, 27, 1998, 27, 1951, 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.

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

\(\Big \downarrow \) 1976

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1998

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1951

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 1951

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] : 
> Simp[-2/(n - 2)   Subst[Int[1/(4*a - x^2), x], x, x*((2*a + b*x^(n - 2))/ 
Sqrt[a*x^2 + b*x^n + c*x^r])], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r, 2* 
n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]
 

rule 1976

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[x^(m - q + 1)*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1) 
/(a*(m + p*q + 1))), x] - Simp[1/(a*(m + p*q + 1))   Int[x^(m + n - q)*(b*( 
m + p*q + (n - q)*(p + 1) + 1) + c*(m + p*q + 2*(n - q)*(p + 1) + 1)*x^(n - 
 q))*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c}, x] && Eq 
Q[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGt 
Q[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && LtQ[m + p*q + 1, 0 
]
 

rule 1998

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ 
.)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[A*x^(m - q + 1)*((a*x^q + b 
*x^n + c*x^(2*n - q))^(p + 1)/(a*(m + p*q + 1))), x] + Simp[1/(a*(m + p*q + 
 1))   Int[x^(m + n - q)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p 
 + 1) + 1) - A*c*(m + p*q + 2*(n - q)*(p + 1) + 1)*x^(n - q), x]*(a*x^q + b 
*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - 
q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] 
 && RationalQ[m, p, q] && ((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - 
q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*q + 1, 0]
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {\left (c \,x^{2}+b x +a \right ) \left (-3 b x +2 a \right )}{4 a^{2} x \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}+\frac {\left (4 a c -3 b^{2}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) x \sqrt {c \,x^{2}+b x +a}}{8 a^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(117\)
pseudoelliptic	\(\frac {4 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right ) a c \,x^{2}-3 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right ) b^{2} x^{2}-4 \ln \left (2\right ) a c \,x^{2}+3 \ln \left (2\right ) b^{2} x^{2}+6 b x \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-4 a^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}}{8 a^{\frac {5}{2}} x^{2}}\)	\(144\)
default	\(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (-6 a^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b x -4 c \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) a^{2} x^{2}+3 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) a \,b^{2} x^{2}+4 a^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\right )}{8 x \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, a^{\frac {7}{2}}}\)	\(152\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.95 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\left [-\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {a} x^{3} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{16 \, a^{3} x^{3}}, \frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{8 \, a^{3} x^{3}}\right ] \] Input:

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

Output:

[-1/16*((3*b^2 - 4*a*c)*sqrt(a)*x^3*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 
8*a^2*x + 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) - 4*sqrt 
(c*x^4 + b*x^3 + a*x^2)*(3*a*b*x - 2*a^2))/(a^3*x^3), 1/8*((3*b^2 - 4*a*c) 
*sqrt(-a)*x^3*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(-a)/ 
(a*c*x^3 + a*b*x^2 + a^2*x)) + 2*sqrt(c*x^4 + b*x^3 + a*x^2)*(3*a*b*x - 2* 
a^2))/(a^3*x^3)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{3} + a x^{2}} x^{2}} \,d x } \] Input:

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {-4 \sqrt {c \,x^{2}+b x +a}\, a^{2}+6 \sqrt {c \,x^{2}+b x +a}\, a b x +4 \sqrt {a}\, \mathrm {log}\left (-2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a c \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (-2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) b^{2} x^{2}-4 \sqrt {a}\, \mathrm {log}\left (x \right ) a c \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (x \right ) b^{2} x^{2}}{8 a^{3} x^{2}} \] Input:

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

Output:

( - 4*sqrt(a + b*x + c*x**2)*a**2 + 6*sqrt(a + b*x + c*x**2)*a*b*x + 4*sqr 
t(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a*c*x**2 - 3*sqr 
t(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*b**2*x**2 - 4*sq 
rt(a)*log(x)*a*c*x**2 + 3*sqrt(a)*log(x)*b**2*x**2)/(8*a**3*x**2)