3.2.5 \(\int \frac {1}{\sqrt {b \sqrt {x}+a x}} \, dx\) [105]

Optimal. Leaf size=56 \[ \frac {2 \sqrt {b \sqrt {x}+a x}}{a}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{a^{3/2}} \]

[Out]

-2*b*arctanh(a^(1/2)*x^(1/2)/(b*x^(1/2)+a*x)^(1/2))/a^(3/2)+2*(b*x^(1/2)+a*x)^(1/2)/a

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Rubi [A]
time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2035, 2038, 634, 212} \begin {gather*} \frac {2 \sqrt {a x+b \sqrt {x}}}{a}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

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

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]

Rule 2035

Int[1/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[-2*(Sqrt[a*x^j + b*x^n]/(b*(n - 2)*x^(n - 1
))), x] - Dist[a*((2*n - j - 2)/(b*(n - 2))), Int[1/(x^(n - j)*Sqrt[a*x^j + b*x^n]), x], x] /; FreeQ[{a, b}, x
] && LtQ[2*(n - 1), j, n]

Rule 2038

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(a*x^Simplify[j/n]
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && IntegerQ[Simplify[j
/n]] && EqQ[Simplify[m - n + 1], 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {b \sqrt {x}+a x}} \, dx &=\frac {2 \sqrt {b \sqrt {x}+a x}}{a}-\frac {b \int \frac {1}{\sqrt {x} \sqrt {b \sqrt {x}+a x}} \, dx}{2 a}\\ &=\frac {2 \sqrt {b \sqrt {x}+a x}}{a}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {2 \sqrt {b \sqrt {x}+a x}}{a}-\frac {(2 b) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{a}\\ &=\frac {2 \sqrt {b \sqrt {x}+a x}}{a}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{a^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 65, normalized size = 1.16 \begin {gather*} \frac {2 \sqrt {b \sqrt {x}+a x}}{a}+\frac {b \log \left (a b+2 a^2 \sqrt {x}-2 a^{3/2} \sqrt {b \sqrt {x}+a x}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(2*Sqrt[b*Sqrt[x] + a*x])/a + (b*Log[a*b + 2*a^2*Sqrt[x] - 2*a^(3/2)*Sqrt[b*Sqrt[x] + a*x]])/a^(3/2)

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Maple [A]
time = 0.38, size = 83, normalized size = 1.48

method	result	size
derivativedivides	\(\frac {2 \sqrt {b \sqrt {x}+a x}}{a}-\frac {b \ln \left (\frac {\frac {b}{2}+a \sqrt {x}}{\sqrt {a}}+\sqrt {b \sqrt {x}+a x}\right )}{a^{\frac {3}{2}}}\)	\(50\)
default	\(-\frac {\sqrt {b \sqrt {x}+a x}\, \left (b \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right )-2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}\right )}{\sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {3}{2}}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x^(1/2)+a*x)^(1/2)*(b*ln(1/2*(2*a*x^(1/2)+2*(x^(1/2)*(a*x^(1/2)+b))^(1/2)*a^(1/2)+b)/a^(1/2))-2*(x^(1/2)*(
a*x^(1/2)+b))^(1/2)*a^(1/2))/(x^(1/2)*(a*x^(1/2)+b))^(1/2)/a^(3/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(a*x + b*sqrt(x)), x)

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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/(b*x^(1/2)+a*x)^(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 x + b \sqrt {x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a*x + b*sqrt(x)), x)

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Giac [A]
time = 1.20, size = 54, normalized size = 0.96 \begin {gather*} \frac {b \log \left ({\left | -2 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} - b \right |}\right )}{a^{\frac {3}{2}}} + \frac {2 \, \sqrt {a x + b \sqrt {x}}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(abs(-2*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(3/2) + 2*sqrt(a*x + b*sqrt(x))/a

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Mupad [B]
time = 5.24, size = 72, normalized size = 1.29 \begin {gather*} \frac {4\,x\,\left (\frac {3\,\sqrt {b}\,\sqrt {b+a\,\sqrt {x}}}{2\,a\,\sqrt {x}}+\frac {b^{3/2}\,\mathrm {asin}\left (\frac {\sqrt {a}\,x^{1/4}\,1{}\mathrm {i}}{\sqrt {b}}\right )\,3{}\mathrm {i}}{2\,a^{3/2}\,x^{3/4}}\right )\,\sqrt {\frac {a\,\sqrt {x}}{b}+1}}{3\,\sqrt {a\,x+b\,\sqrt {x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x + b*x^(1/2))^(1/2),x)

[Out]

(4*x*((3*b^(1/2)*(b + a*x^(1/2))^(1/2))/(2*a*x^(1/2)) + (b^(3/2)*asin((a^(1/2)*x^(1/4)*1i)/b^(1/2))*3i)/(2*a^(
3/2)*x^(3/4)))*((a*x^(1/2))/b + 1)^(1/2))/(3*(a*x + b*x^(1/2))^(1/2))

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