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

Optimal. Leaf size=28 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {620, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a*x + b*x^2],x]

[Out]

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

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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rubi steps

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

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Mathematica [B]  time = 0.02, size = 57, normalized size = 2.04 \[ \frac {2 \sqrt {a} \sqrt {x} \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {x (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a*x + b*x^2],x]

[Out]

(2*Sqrt[a]*Sqrt[x]*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[b]*Sqrt[x*(a + b*x)])

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fricas [A]  time = 0.46, size = 62, normalized size = 2.21 \[ \left [\frac {\log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right )}{b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b), -2*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x))/b]

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giac [A]  time = 0.47, size = 35, normalized size = 1.25 \[ -\frac {\log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/sqrt(b)

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maple [A]  time = 0.00, size = 29, normalized size = 1.04 \[ \frac {\ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 27, normalized size = 0.96 \[ \frac {\log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b)

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mupad [B]  time = 3.47, size = 28, normalized size = 1.00 \[ \frac {\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x + b*x^2)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a*x + b*x**2), x)

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