\(\int \frac {1}{x \sqrt {a x^2+b x^3+c x^4}} \, dx\) [53]

Optimal result

Integrand size = 24, antiderivative size = 77 \[ \int \frac {1}{x \sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {\sqrt {a x^2+b x^3+c x^4}}{a x^2}+\frac {b \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1941, 1918, 212} \[ \int \frac {1}{x \sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {b \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 a^{3/2}}-\frac {\sqrt {a x^2+b x^3+c x^4}}{a x^2} \]

[In]

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

[Out]

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

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 1941

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)/(2*a*(n - q)*(p + 1))), x] - Dist[b/(2*a), Int[x^(m + 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] && EqQ[m + p*q + 1, -2*(n
- q)*(p + 1)]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x \sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {-\sqrt {a} (a+x (b+c x))-b x \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {b x \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )-b x \ln \left (2\right )-2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{2 a^{\frac {3}{2}} x}\)	\(68\)
default	\(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (2 a^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}-b \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) a x \right )}{2 \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, a^{\frac {5}{2}}}\)	\(88\)
risch	\(-\frac {c \,x^{2}+b x +a}{a \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}+\frac {b \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) x \sqrt {c \,x^{2}+b x +a}}{2 a^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.52 \[ \int \frac {1}{x \sqrt {a x^2+b x^3+c x^4}} \, dx=\left [\frac {\sqrt {a} b x^{2} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} a}{4 \, a^{2} x^{2}}, -\frac {\sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} a}{2 \, a^{2} x^{2}}\right ] \]

[In]

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

[Out]

[1/4*(sqrt(a)*b*x^2*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)*a)/(a^2*x^2), -1/2*(sqrt(-a)*b*x^2*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)*a)/(a^2*x^2)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(1/x/(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), x)

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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