3.1675 \(\int \frac {\sqrt {x}}{(a+\frac {b}{x})^2} \, dx\)

Optimal. Leaf size=70 \[ \frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{7/2}}-\frac {5 b \sqrt {x}}{a^3}+\frac {5 x^{3/2}}{3 a^2}-\frac {x^{5/2}}{a (a x+b)} \]

[Out]

5/3*x^(3/2)/a^2-x^(5/2)/a/(a*x+b)+5*b^(3/2)*arctan(a^(1/2)*x^(1/2)/b^(1/2))/a^(7/2)-5*b*x^(1/2)/a^3

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{7/2}}-\frac {5 b \sqrt {x}}{a^3}+\frac {5 x^{3/2}}{3 a^2}-\frac {x^{5/2}}{a (a x+b)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b/x)^2,x]

[Out]

(-5*b*Sqrt[x])/a^3 + (5*x^(3/2))/(3*a^2) - x^(5/2)/(a*(b + a*x)) + (5*b^(3/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]
])/a^(7/2)

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 {\sqrt {x}}{\left (a+\frac {b}{x}\right )^2} \, dx &=\int \frac {x^{5/2}}{(b+a x)^2} \, dx\\ &=-\frac {x^{5/2}}{a (b+a x)}+\frac {5 \int \frac {x^{3/2}}{b+a x} \, dx}{2 a}\\ &=\frac {5 x^{3/2}}{3 a^2}-\frac {x^{5/2}}{a (b+a x)}-\frac {(5 b) \int \frac {\sqrt {x}}{b+a x} \, dx}{2 a^2}\\ &=-\frac {5 b \sqrt {x}}{a^3}+\frac {5 x^{3/2}}{3 a^2}-\frac {x^{5/2}}{a (b+a x)}+\frac {\left (5 b^2\right ) \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{2 a^3}\\ &=-\frac {5 b \sqrt {x}}{a^3}+\frac {5 x^{3/2}}{3 a^2}-\frac {x^{5/2}}{a (b+a x)}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{a^3}\\ &=-\frac {5 b \sqrt {x}}{a^3}+\frac {5 x^{3/2}}{3 a^2}-\frac {x^{5/2}}{a (b+a x)}+\frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.39 \[ \frac {2 x^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {a x}{b}\right )}{7 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a + b/x)^2,x]

[Out]

(2*x^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, -((a*x)/b)])/(7*b^2)

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fricas [A]  time = 0.92, size = 161, normalized size = 2.30 \[ \left [\frac {15 \, {\left (a b x + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (2 \, a^{2} x^{2} - 10 \, a b x - 15 \, b^{2}\right )} \sqrt {x}}{6 \, {\left (a^{4} x + a^{3} b\right )}}, \frac {15 \, {\left (a b x + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) + {\left (2 \, a^{2} x^{2} - 10 \, a b x - 15 \, b^{2}\right )} \sqrt {x}}{3 \, {\left (a^{4} x + a^{3} b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b*x + b^2)*sqrt(-b/a)*log((a*x + 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(2*a^2*x^2 - 10*a*b*x
- 15*b^2)*sqrt(x))/(a^4*x + a^3*b), 1/3*(15*(a*b*x + b^2)*sqrt(b/a)*arctan(a*sqrt(x)*sqrt(b/a)/b) + (2*a^2*x^2
- 10*a*b*x - 15*b^2)*sqrt(x))/(a^4*x + a^3*b)]

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giac [A]  time = 0.18, size = 65, normalized size = 0.93 \[ \frac {5 \, b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {b^{2} \sqrt {x}}{{\left (a x + b\right )} a^{3}} + \frac {2 \, {\left (a^{4} x^{\frac {3}{2}} - 6 \, a^{3} b \sqrt {x}\right )}}{3 \, a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*b^2*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - b^2*sqrt(x)/((a*x + b)*a^3) + 2/3*(a^4*x^(3/2) - 6*a^3*b*s
qrt(x))/a^6

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maple [A]  time = 0.01, size = 61, normalized size = 0.87 \[ \frac {5 b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}+\frac {2 x^{\frac {3}{2}}}{3 a^{2}}-\frac {b^{2} \sqrt {x}}{\left (a x +b \right ) a^{3}}-\frac {4 b \sqrt {x}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.35, size = 66, normalized size = 0.94 \[ \frac {2 \, a^{2} - \frac {10 \, a b}{x} - \frac {15 \, b^{2}}{x^{2}}}{3 \, {\left (\frac {a^{4}}{x^{\frac {3}{2}}} + \frac {a^{3} b}{x^{\frac {5}{2}}}\right )}} - \frac {5 \, b^{2} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*a^2 - 10*a*b/x - 15*b^2/x^2)/(a^4/x^(3/2) + a^3*b/x^(5/2)) - 5*b^2*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(
a*b)*a^3)

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mupad [B]  time = 0.06, size = 58, normalized size = 0.83 \[ \frac {2\,x^{3/2}}{3\,a^2}-\frac {4\,b\,\sqrt {x}}{a^3}-\frac {b^2\,\sqrt {x}}{x\,a^4+b\,a^3}+\frac {5\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a + b/x)^2,x)

[Out]

(2*x^(3/2))/(3*a^2) - (4*b*x^(1/2))/a^3 - (b^2*x^(1/2))/(a^3*b + a^4*x) + (5*b^(3/2)*atan((a^(1/2)*x^(1/2))/b^
(1/2)))/a^(7/2)

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sympy [A]  time = 9.64, size = 479, normalized size = 6.84 \[ \begin {cases} \tilde {\infty } x^{\frac {7}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 b^{2}} & \text {for}\: a = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a^{2}} & \text {for}\: b = 0 \\\frac {4 i a^{3} \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {1}{a}}}{6 i a^{5} \sqrt {b} x \sqrt {\frac {1}{a}} + 6 i a^{4} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {20 i a^{2} b^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{a}}}{6 i a^{5} \sqrt {b} x \sqrt {\frac {1}{a}} + 6 i a^{4} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {30 i a b^{\frac {5}{2}} \sqrt {x} \sqrt {\frac {1}{a}}}{6 i a^{5} \sqrt {b} x \sqrt {\frac {1}{a}} + 6 i a^{4} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {15 a b^{2} x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{6 i a^{5} \sqrt {b} x \sqrt {\frac {1}{a}} + 6 i a^{4} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {15 a b^{2} x \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{6 i a^{5} \sqrt {b} x \sqrt {\frac {1}{a}} + 6 i a^{4} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {15 b^{3} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{6 i a^{5} \sqrt {b} x \sqrt {\frac {1}{a}} + 6 i a^{4} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {15 b^{3} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{6 i a^{5} \sqrt {b} x \sqrt {\frac {1}{a}} + 6 i a^{4} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(7/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*b**2), Eq(a, 0)), (2*x**(3/2)/(3*a**2), Eq(b, 0)
), (4*I*a**3*sqrt(b)*x**(5/2)*sqrt(1/a)/(6*I*a**5*sqrt(b)*x*sqrt(1/a) + 6*I*a**4*b**(3/2)*sqrt(1/a)) - 20*I*a*
*2*b**(3/2)*x**(3/2)*sqrt(1/a)/(6*I*a**5*sqrt(b)*x*sqrt(1/a) + 6*I*a**4*b**(3/2)*sqrt(1/a)) - 30*I*a*b**(5/2)*
sqrt(x)*sqrt(1/a)/(6*I*a**5*sqrt(b)*x*sqrt(1/a) + 6*I*a**4*b**(3/2)*sqrt(1/a)) + 15*a*b**2*x*log(-I*sqrt(b)*sq
rt(1/a) + sqrt(x))/(6*I*a**5*sqrt(b)*x*sqrt(1/a) + 6*I*a**4*b**(3/2)*sqrt(1/a)) - 15*a*b**2*x*log(I*sqrt(b)*sq
rt(1/a) + sqrt(x))/(6*I*a**5*sqrt(b)*x*sqrt(1/a) + 6*I*a**4*b**(3/2)*sqrt(1/a)) + 15*b**3*log(-I*sqrt(b)*sqrt(
1/a) + sqrt(x))/(6*I*a**5*sqrt(b)*x*sqrt(1/a) + 6*I*a**4*b**(3/2)*sqrt(1/a)) - 15*b**3*log(I*sqrt(b)*sqrt(1/a)
+ sqrt(x))/(6*I*a**5*sqrt(b)*x*sqrt(1/a) + 6*I*a**4*b**(3/2)*sqrt(1/a)), True))

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