\(\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx\) [129]

Optimal result

Integrand size = 18, antiderivative size = 38 \[ \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {738, 212} \[ \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a}} \]

[In]

Int[1/(x*Sqrt[a + b*x + c*x^2]),x]

[Out]

-(ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])]/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 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\sqrt {a}}\)	\(35\)

[In]

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

[Out]

-1/a^(1/2)*ln((2*a+b*x+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 = 111, normalized size of antiderivative = 2.92 \[ \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx=\left [\frac {\log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right )}{a}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx=\frac {2 \, \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx=-\frac {\ln \left (\frac {b}{2}+\frac {a}{x}+\frac {\sqrt {a}\,\sqrt {c\,x^2+b\,x+a}}{x}\right )}{\sqrt {a}} \]

[In]

int(1/(x*(a + b*x + c*x^2)^(1/2)),x)

[Out]

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