3.23.43 \(\int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx\) [2243]

Optimal. Leaf size=80 \[ -\frac {\sqrt {a+b \sqrt {x}}}{a x}+\frac {3 b \sqrt {a+b \sqrt {x}}}{2 a^2 \sqrt {x}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

-3/2*b^2*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2))/a^(5/2)-(a+b*x^(1/2))^(1/2)/a/x+3/2*b*(a+b*x^(1/2))^(1/2)/a^2/x^
(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {272, 44, 65, 214} \begin {gather*} -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{5/2}}+\frac {3 b \sqrt {a+b \sqrt {x}}}{2 a^2 \sqrt {x}}-\frac {\sqrt {a+b \sqrt {x}}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*Sqrt[x]]*x^2),x]

[Out]

-(Sqrt[a + b*Sqrt[x]]/(a*x)) + (3*b*Sqrt[a + b*Sqrt[x]])/(2*a^2*Sqrt[x]) - (3*b^2*ArcTanh[Sqrt[a + b*Sqrt[x]]/
Sqrt[a]])/(2*a^(5/2))

Rule 44

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

Rule 65

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 214

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

Rule 272

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 {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx &=2 \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {\sqrt {a+b \sqrt {x}}}{a x}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )}{2 a}\\ &=-\frac {\sqrt {a+b \sqrt {x}}}{a x}+\frac {3 b \sqrt {a+b \sqrt {x}}}{2 a^2 \sqrt {x}}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )}{4 a^2}\\ &=-\frac {\sqrt {a+b \sqrt {x}}}{a x}+\frac {3 b \sqrt {a+b \sqrt {x}}}{2 a^2 \sqrt {x}}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sqrt {x}}\right )}{2 a^2}\\ &=-\frac {\sqrt {a+b \sqrt {x}}}{a x}+\frac {3 b \sqrt {a+b \sqrt {x}}}{2 a^2 \sqrt {x}}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 68, normalized size = 0.85 \begin {gather*} \frac {\sqrt {a+b \sqrt {x}} \left (-2 a+3 b \sqrt {x}\right )}{2 a^2 x}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*Sqrt[x]]*(-2*a + 3*b*Sqrt[x]))/(2*a^2*x) - (3*b^2*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/(2*a^(5/2)
)

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Maple [A]
time = 0.19, size = 72, normalized size = 0.90

method	result	size
derivativedivides	\(4 b^{2} \left (-\frac {\sqrt {a +b \sqrt {x}}}{4 a \,b^{2} x}-\frac {3 \left (-\frac {\sqrt {a +b \sqrt {x}}}{2 a b \sqrt {x}}+\frac {\arctanh \left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\)	\(72\)
default	\(4 b^{2} \left (-\frac {\sqrt {a +b \sqrt {x}}}{4 a \,b^{2} x}-\frac {3 \left (-\frac {\sqrt {a +b \sqrt {x}}}{2 a b \sqrt {x}}+\frac {\arctanh \left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*b^2*(-1/4*(a+b*x^(1/2))^(1/2)/a/b^2/x-3/4/a*(-1/2*(a+b*x^(1/2))^(1/2)/a/b/x^(1/2)+1/2/a^(3/2)*arctanh((a+b*x
^(1/2))^(1/2)/a^(1/2))))

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Maxima [A]
time = 0.51, size = 104, normalized size = 1.30 \begin {gather*} \frac {3 \, b^{2} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right )}{4 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {b \sqrt {x} + a} a b^{2}}{2 \, {\left ({\left (b \sqrt {x} + a\right )}^{2} a^{2} - 2 \, {\left (b \sqrt {x} + a\right )} a^{3} + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*b^2*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt(a)))/a^(5/2) + 1/2*(3*(b*sqrt(x) + a)^
(3/2)*b^2 - 5*sqrt(b*sqrt(x) + a)*a*b^2)/((b*sqrt(x) + a)^2*a^2 - 2*(b*sqrt(x) + a)*a^3 + a^4)

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Fricas [A]
time = 0.38, size = 137, normalized size = 1.71 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{2} x \log \left (\frac {b x - 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) + 2 \, {\left (3 \, a b \sqrt {x} - 2 \, a^{2}\right )} \sqrt {b \sqrt {x} + a}}{4 \, a^{3} x}, \frac {3 \, \sqrt {-a} b^{2} x \arctan \left (\frac {\sqrt {b \sqrt {x} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b \sqrt {x} - 2 \, a^{2}\right )} \sqrt {b \sqrt {x} + a}}{2 \, a^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*sqrt(a)*b^2*x*log((b*x - 2*sqrt(b*sqrt(x) + a)*sqrt(a)*sqrt(x) + 2*a*sqrt(x))/x) + 2*(3*a*b*sqrt(x) -
2*a^2)*sqrt(b*sqrt(x) + a))/(a^3*x), 1/2*(3*sqrt(-a)*b^2*x*arctan(sqrt(b*sqrt(x) + a)*sqrt(-a)/a) + (3*a*b*sqr
t(x) - 2*a^2)*sqrt(b*sqrt(x) + a))/(a^3*x)]

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Sympy [A]
time = 2.71, size = 110, normalized size = 1.38 \begin {gather*} - \frac {1}{\sqrt {b} x^{\frac {5}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {\sqrt {b}}{2 a x^{\frac {3}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {3 b^{\frac {3}{2}}}{2 a^{2} \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )}}{2 a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(sqrt(b)*x**(5/4)*sqrt(a/(b*sqrt(x)) + 1)) + sqrt(b)/(2*a*x**(3/4)*sqrt(a/(b*sqrt(x)) + 1)) + 3*b**(3/2)/(2
*a**2*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1)) - 3*b**2*asinh(sqrt(a)/(sqrt(b)*x**(1/4)))/(2*a**(5/2))

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Giac [A]
time = 1.51, size = 75, normalized size = 0.94 \begin {gather*} \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} b^{3} - 5 \, \sqrt {b \sqrt {x} + a} a b^{3}}{a^{2} b^{2} x}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*b^3*arctan(sqrt(b*sqrt(x) + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*sqrt(x) + a)^(3/2)*b^3 - 5*sqrt(b*sqrt(
x) + a)*a*b^3)/(a^2*b^2*x))/b

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Mupad [B]
time = 1.51, size = 57, normalized size = 0.71 \begin {gather*} \frac {3\,{\left (a+b\,\sqrt {x}\right )}^{3/2}}{2\,a^2\,x}-\frac {5\,\sqrt {a+b\,\sqrt {x}}}{2\,a\,x}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,\sqrt {x}}}{\sqrt {a}}\right )}{2\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^(1/2))^(1/2)),x)

[Out]

(3*(a + b*x^(1/2))^(3/2))/(2*a^2*x) - (5*(a + b*x^(1/2))^(1/2))/(2*a*x) - (3*b^2*atanh((a + b*x^(1/2))^(1/2)/a
^(1/2)))/(2*a^(5/2))

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