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

Optimal. Leaf size=75 \[ \frac{2 \sqrt{a^2+2 a b \sqrt{x}+b^2 x}}{b^2}-\frac{2 a \left (a+b \sqrt{x}\right ) \log \left (a+b \sqrt{x}\right )}{b^2 \sqrt{a^2+2 a b \sqrt{x}+b^2 x}} \]

[Out]

(2*Sqrt[a^2 + 2*a*b*Sqrt[x] + b^2*x])/b^2 - (2*a*(a + b*Sqrt[x])*Log[a + b*Sqrt[x]])/(b^2*Sqrt[a^2 + 2*a*b*Sqr
t[x] + b^2*x])

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Rubi [A]  time = 0.039791, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1341, 640, 608, 31} \[ \frac{2 \sqrt{a^2+2 a b \sqrt{x}+b^2 x}}{b^2}-\frac{2 a \left (a+b \sqrt{x}\right ) \log \left (a+b \sqrt{x}\right )}{b^2 \sqrt{a^2+2 a b \sqrt{x}+b^2 x}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a^2 + 2*a*b*Sqrt[x] + b^2*x],x]

[Out]

(2*Sqrt[a^2 + 2*a*b*Sqrt[x] + b^2*x])/b^2 - (2*a*(a + b*Sqrt[x])*Log[a + b*Sqrt[x]])/(b^2*Sqrt[a^2 + 2*a*b*Sqr
t[x] + b^2*x])

Rule 1341

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0299355, size = 50, normalized size = 0.67 \[ \frac{2 \left (a+b \sqrt{x}\right ) \left (b \sqrt{x}-a \log \left (a+b \sqrt{x}\right )\right )}{b^2 \sqrt{\left (a+b \sqrt{x}\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 50, normalized size = 0.7 \begin{align*} 2\,{\frac{\sqrt{{a}^{2}+{b}^{2}x+2\,ab\sqrt{x}} \left ( b\sqrt{x}-a\ln \left ( a+b\sqrt{x} \right ) \right ) }{ \left ( a+b\sqrt{x} \right ){b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03503, size = 31, normalized size = 0.41 \begin{align*} -\frac{2 \, a \log \left (b \sqrt{x} + a\right )}{b^{2}} + \frac{2 \, \sqrt{x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16299, size = 74, normalized size = 0.99 \begin{align*} -\frac{2 \,{\left | a \right |} \log \left ({\left | \sqrt{b^{2} x} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) +{\left | a \right |} \right |}\right )}{b^{2}} + \frac{2 \,{\left | a \right |} \log \left ({\left | a \right |}\right )}{b^{2}} + \frac{2 \, \sqrt{b^{2} x}}{b^{2} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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