3.5.61 \(\int \frac {1}{\sqrt {a^2+2 a b \sqrt {x}+b^2 x}} \, dx\) [461]

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

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Rubi [A]
time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 = {1355, 654, 622, 31} \begin {gather*} \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 {gather*}

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 31

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

Rule 622

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 654

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 1355

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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a^2+2 a b \sqrt {x}+b^2 x}} \, dx &=2 \text {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) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.02, size = 50, normalized size = 0.67 \begin {gather*} \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}} \end {gather*}

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 = 0.06, size = 50, normalized size = 0.67

method	result	size
derivativedivides	\(-\frac {2 \left (a +b \sqrt {x}\right ) \left (a \ln \left (a +b \sqrt {x}\right )-b \sqrt {x}\right )}{\sqrt {\left (a +b \sqrt {x}\right )^{2}}\, b^{2}}\)	\(41\)
default	\(\frac {2 \sqrt {a^{2}+b^{2} x +2 a b \sqrt {x}}\, \left (b \sqrt {x}-a \ln \left (a +b \sqrt {x}\right )\right )}{\left (a +b \sqrt {x}\right ) b^{2}}\)	\(50\)

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,method=_RETURNVERBOSE)

[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 = 0.34, size = 23, normalized size = 0.31 \begin {gather*} -\frac {2 \, a \log \left (b \sqrt {x} + a\right )}{b^{2}} + \frac {2 \, \sqrt {x}}{b} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a^{2} + 2 a b \sqrt {x} + b^{2} x}}\, dx \end {gather*}

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 = 4.73, size = 45, normalized size = 0.60 \begin {gather*} -\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 \, \sqrt {b^{2} x}}{b^{2} \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )} \end {gather*}

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*sqrt(b^2*x)/(b^2*sgn(a)*sgn(b))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {b^2\,x+a^2+2\,a\,b\,\sqrt {x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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