3.1.26 \(\int \frac {1}{\sqrt {b x+b^2 x^2}} \, dx\) [26]

Optimal. Leaf size=24 \[ \frac {2 \tanh ^{-1}\left (\frac {b x}{\sqrt {b x+b^2 x^2}}\right )}{b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {634, 212} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {b x}{\sqrt {b^2 x^2+b x}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*x + b^2*x^2],x]

[Out]

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

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 634

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(24)=48\).
time = 0.04, size = 56, normalized size = 2.33 \begin {gather*} -\frac {2 \sqrt {x} \sqrt {1+b x} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {1+b x}\right )}{\sqrt {b} \sqrt {b x (1+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*x + b^2*x^2],x]

[Out]

(-2*Sqrt[x]*Sqrt[1 + b*x]*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[1 + b*x]])/(Sqrt[b]*Sqrt[b*x*(1 + b*x)])

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Maple [A]
time = 0.37, size = 37, normalized size = 1.54

method	result	size
default	\(\frac {\ln \left (\frac {\frac {1}{2} b +b^{2} x}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+b x}\right )}{\sqrt {b^{2}}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 29, normalized size = 1.21 \begin {gather*} \frac {\log \left (2 \, b^{2} x + 2 \, \sqrt {b^{2} x^{2} + b x} b + b\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.56, size = 27, normalized size = 1.12 \begin {gather*} -\frac {\log \left (-2 \, b x + 2 \, \sqrt {b^{2} x^{2} + b x} - 1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b^{2} x^{2} + b x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b**2*x**2 + b*x), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (22) = 44\).
time = 2.22, size = 59, normalized size = 2.46 \begin {gather*} \frac {1}{4} \, \sqrt {b^{2} x^{2} + b x} {\left (2 \, x + \frac {1}{b}\right )} + \frac {\log \left ({\left | -2 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + b x}\right )} {\left | b \right |} - b \right |}\right )}{8 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(b^2*x^2 + b*x)*(2*x + 1/b) + 1/8*log(abs(-2*(x*abs(b) - sqrt(b^2*x^2 + b*x))*abs(b) - b))/abs(b)

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Mupad [B]
time = 0.23, size = 36, normalized size = 1.50 \begin {gather*} \frac {\ln \left (\frac {x\,b^2+\frac {b}{2}}{\sqrt {b^2}}+\sqrt {b^2\,x^2+b\,x}\right )}{\sqrt {b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + b^2*x^2)^(1/2),x)

[Out]

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

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