3.1676 \(\int \frac {1}{(a+\frac {b}{x})^2 \sqrt {x}} \, dx\)

Optimal. Leaf size=57 \[ -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}}+\frac {3 \sqrt {x}}{a^2}-\frac {x^{3/2}}{a (a x+b)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {263, 47, 50, 63, 205} \[ -\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}}+\frac {3 \sqrt {x}}{a^2}-\frac {x^{3/2}}{a (a x+b)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^2*Sqrt[x]),x]

[Out]

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

Rule 47

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

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 205

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

Rule 263

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

Rubi steps

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

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.47 \[ \frac {2 x^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {a x}{b}\right )}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^2*Sqrt[x]),x]

[Out]

(2*x^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((a*x)/b)])/(5*b^2)

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fricas [A]  time = 0.91, size = 134, normalized size = 2.35 \[ \left [\frac {3 \, {\left (a x + b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (2 \, a x + 3 \, b\right )} \sqrt {x}}{2 \, {\left (a^{3} x + a^{2} b\right )}}, -\frac {3 \, {\left (a x + b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - {\left (2 \, a x + 3 \, b\right )} \sqrt {x}}{a^{3} x + a^{2} b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 46, normalized size = 0.81 \[ -\frac {3 \, b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, \sqrt {x}}{a^{2}} + \frac {b \sqrt {x}}{{\left (a x + b\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 47, normalized size = 0.82 \[ -\frac {3 b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}+\frac {b \sqrt {x}}{\left (a x +b \right ) a^{2}}+\frac {2 \sqrt {x}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.35, size = 52, normalized size = 0.91 \[ \frac {2 \, a + \frac {3 \, b}{x}}{\frac {a^{3}}{\sqrt {x}} + \frac {a^{2} b}{x^{\frac {3}{2}}}} + \frac {3 \, b \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.10, size = 46, normalized size = 0.81 \[ \frac {2\,\sqrt {x}}{a^2}+\frac {b\,\sqrt {x}}{x\,a^3+b\,a^2}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(a + b/x)^2),x)

[Out]

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

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sympy [A]  time = 12.34, size = 411, normalized size = 7.21 \[ \begin {cases} \tilde {\infty } x^{\frac {5}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 \sqrt {x}}{a^{2}} & \text {for}\: b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: a = 0 \\\frac {4 i a^{2} \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {1}{a}}}{2 i a^{4} \sqrt {b} x \sqrt {\frac {1}{a}} + 2 i a^{3} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {6 i a b^{\frac {3}{2}} \sqrt {x} \sqrt {\frac {1}{a}}}{2 i a^{4} \sqrt {b} x \sqrt {\frac {1}{a}} + 2 i a^{3} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {3 a b x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{4} \sqrt {b} x \sqrt {\frac {1}{a}} + 2 i a^{3} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {3 a b x \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{4} \sqrt {b} x \sqrt {\frac {1}{a}} + 2 i a^{3} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {3 b^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{4} \sqrt {b} x \sqrt {\frac {1}{a}} + 2 i a^{3} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {3 b^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{4} \sqrt {b} x \sqrt {\frac {1}{a}} + 2 i a^{3} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(5/2), Eq(a, 0) & Eq(b, 0)), (2*sqrt(x)/a**2, Eq(b, 0)), (2*x**(5/2)/(5*b**2), Eq(a, 0)), (4
*I*a**2*sqrt(b)*x**(3/2)*sqrt(1/a)/(2*I*a**4*sqrt(b)*x*sqrt(1/a) + 2*I*a**3*b**(3/2)*sqrt(1/a)) + 6*I*a*b**(3/
2)*sqrt(x)*sqrt(1/a)/(2*I*a**4*sqrt(b)*x*sqrt(1/a) + 2*I*a**3*b**(3/2)*sqrt(1/a)) - 3*a*b*x*log(-I*sqrt(b)*sqr
t(1/a) + sqrt(x))/(2*I*a**4*sqrt(b)*x*sqrt(1/a) + 2*I*a**3*b**(3/2)*sqrt(1/a)) + 3*a*b*x*log(I*sqrt(b)*sqrt(1/
a) + sqrt(x))/(2*I*a**4*sqrt(b)*x*sqrt(1/a) + 2*I*a**3*b**(3/2)*sqrt(1/a)) - 3*b**2*log(-I*sqrt(b)*sqrt(1/a) +
sqrt(x))/(2*I*a**4*sqrt(b)*x*sqrt(1/a) + 2*I*a**3*b**(3/2)*sqrt(1/a)) + 3*b**2*log(I*sqrt(b)*sqrt(1/a) + sqrt
(x))/(2*I*a**4*sqrt(b)*x*sqrt(1/a) + 2*I*a**3*b**(3/2)*sqrt(1/a)), True))

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