\(\int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx\) [52]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {\text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{\sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1918, 212} \[ \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {\text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{\sqrt {a}} \]

[In]

Int[1/Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {2 x \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(\frac {\ln \left (2\right )-\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )}{\sqrt {a}}\)	\(42\)
default	\(-\frac {x \sqrt {c \,x^{2}+b x +a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \sqrt {a}}\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.89 \[ \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\left [\frac {\log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right )}{a}\right ] \]

[In]

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

[Out]

[1/2*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)/s
qrt(a), sqrt(-a)*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))/a]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(1/(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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {2 \, \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} + \frac {2 \, \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-2*arctan(sqrt(a)/sqrt(-a))*sgn(x)/sqrt(-a) + 2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a
)*sgn(x))

Mupad [F(-1)]

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

[In]

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

[Out]

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