3.2236 \(\int \frac {\sqrt {a+b \sqrt {x}}}{x} \, dx\)

Optimal. Leaf size=43 \[ 4 \sqrt {a+b \sqrt {x}}-4 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right ) \]

[Out]

-4*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2))*a^(1/2)+4*(a+b*x^(1/2))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {266, 50, 63, 208} \[ 4 \sqrt {a+b \sqrt {x}}-4 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[x]]/x,x]

[Out]

4*Sqrt[a + b*Sqrt[x]] - 4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \sqrt {x}}}{x} \, dx &=2 \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sqrt {x}\right )\\ &=4 \sqrt {a+b \sqrt {x}}+(2 a) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )\\ &=4 \sqrt {a+b \sqrt {x}}+\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sqrt {x}}\right )}{b}\\ &=4 \sqrt {a+b \sqrt {x}}-4 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.00 \[ 4 \sqrt {a+b \sqrt {x}}-4 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[x]]/x,x]

[Out]

4*Sqrt[a + b*Sqrt[x]] - 4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]]

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fricas [A]  time = 0.90, size = 88, normalized size = 2.05 \[ \left [2 \, \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) + 4 \, \sqrt {b \sqrt {x} + a}, 4 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b \sqrt {x} + a} \sqrt {-a}}{a}\right ) + 4 \, \sqrt {b \sqrt {x} + a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2*sqrt(a)*log((b*x - 2*sqrt(b*sqrt(x) + a)*sqrt(a)*sqrt(x) + 2*a*sqrt(x))/x) + 4*sqrt(b*sqrt(x) + a), 4*sqrt(
-a)*arctan(sqrt(b*sqrt(x) + a)*sqrt(-a)/a) + 4*sqrt(b*sqrt(x) + a)]

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giac [A]  time = 0.17, size = 41, normalized size = 0.95 \[ \frac {4 \, {\left (\frac {a b \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \sqrt {b \sqrt {x} + a} b\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(a*b*arctan(sqrt(b*sqrt(x) + a)/sqrt(-a))/sqrt(-a) + sqrt(b*sqrt(x) + a)*b)/b

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maple [A]  time = 0.00, size = 32, normalized size = 0.74 \[ -4 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b \sqrt {x}+a}}{\sqrt {a}}\right )+4 \sqrt {b \sqrt {x}+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*arctanh((b*x^(1/2)+a)^(1/2)/a^(1/2))*a^(1/2)+4*(b*x^(1/2)+a)^(1/2)

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maxima [A]  time = 2.04, size = 49, normalized size = 1.14 \[ 2 \, \sqrt {a} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right ) + 4 \, \sqrt {b \sqrt {x} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(a)*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt(a))) + 4*sqrt(b*sqrt(x) + a)

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mupad [B]  time = 1.89, size = 31, normalized size = 0.72 \[ 4\,\sqrt {a+b\,\sqrt {x}}-4\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,\sqrt {x}}}{\sqrt {a}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/2))^(1/2)/x,x)

[Out]

4*(a + b*x^(1/2))^(1/2) - 4*a^(1/2)*atanh((a + b*x^(1/2))^(1/2)/a^(1/2))

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sympy [B]  time = 1.57, size = 75, normalized size = 1.74 \[ - 4 \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )} + \frac {4 a}{\sqrt {b} \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {4 \sqrt {b} \sqrt [4]{x}}{\sqrt {\frac {a}{b \sqrt {x}} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*sqrt(a)*asinh(sqrt(a)/(sqrt(b)*x**(1/4))) + 4*a/(sqrt(b)*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1)) + 4*sqrt(b)*x**(
1/4)/sqrt(a/(b*sqrt(x)) + 1)

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