\(\int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx\) [54]

Optimal result

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

[Out]

-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)-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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1943, 1965, 12, 1918, 212} \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {\left (3 b^2-4 a c\right ) \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{8 a^{5/2}}+\frac {3 b \sqrt {a x^2+b x^3+c x^4}}{4 a^2 x^2}-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 a x^3} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1918

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-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 1943

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] - Dist[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] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c,
0] && IGtQ[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && LtQ[m + p*q + 1, 0]

Rule 1965

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> 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] + Dist[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] && Eq
Q[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {a x^2+b x^3+c x^4}}{2 a x^3}+\frac {\int \frac {-\frac {3 b}{2}-c x}{x \sqrt {a x^2+b x^3+c x^4}} \, dx}{2 a} \\ & = -\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 {\int \frac {-\frac {3 b^2}{4}+a c}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{2 a^2} \\ & = -\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 ) \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{8 a^2} \\ & = -\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 {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {x (2 a+b x)}{\sqrt {a x^2+b x^3+c x^4}}\right )}{4 a^2} \\ & = -\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 ) \tanh ^{-1}\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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {-\sqrt {a} (2 a-3 b x) (a+x (b+c x))+\left (3 b^2-4 a c\right ) x^2 \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{4 a^{5/2} x \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (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	\(\frac {4 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right ) a c \,x^{2}-3 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right ) b^{2} x^{2}-4 \ln \left (2\right ) a c \,x^{2}+3 \ln \left (2\right ) b^{2} x^{2}-4 a^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}+6 b x \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{8 a^{\frac {5}{2}} x^{2}}\)	\(144\)
default	\(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (-6 a^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b x -4 c \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) a^{2} x^{2}+3 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) a \,b^{2} x^{2}+4 a^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\right )}{8 x \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, a^{\frac {7}{2}}}\)	\(152\)

[In]

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

[Out]

-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)

none

Time = 0.31 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.95 \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\left [-\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {a} x^{3} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{16 \, a^{3} x^{3}}, \frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (3 \, a b x - 2 \, a^{2}\right )}}{8 \, a^{3} x^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Not invertible Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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