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

Optimal. Leaf size=45 \[ -\frac {\tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{\sqrt {a}} \]

[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)

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Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1996, 1904, 206} \[ -\frac {\tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{\sqrt {a}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[x^2*(a + b*x + c*x^2)],x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1904

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 1996

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedTrinomialQ[u, x] &&  !Gen
eralizedTrinomialMatchQ[u, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {x^2 \left (a+b x+c x^2\right )}} \, dx &=\int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx\\ &=-\left (2 \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 70, normalized size = 1.56 \[ -\frac {x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} \sqrt {x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 130, normalized size = 2.89 \[ \left [\frac {\log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right )}{a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.47, size = 59, normalized size = 1.31 \[ -\frac {2 \, \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-a}} + \frac {2 \, \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.01, size = 64, normalized size = 1.42 \[ -\frac {\sqrt {c \,x^{2}+b x +a}\, x \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{\sqrt {\left (c \,x^{2}+b x +a \right ) x^{2}}\, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {{\left (c x^{2} + b x + a\right )} x^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {x^2\,\left (c\,x^2+b\,x+a\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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