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]

(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)

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Rubi [A]  time = 0.0192307, antiderivative size = 57, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [C]  time = 0.0037448, 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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Maple [A]  time = 0.01, size = 47, normalized size = 0.8 \begin{align*} 2\,{\frac{\sqrt{x}}{{a}^{2}}}+{\frac{b}{{a}^{2} \left ( ax+b \right ) }\sqrt{x}}-3\,{\frac{b}{{a}^{2}\sqrt{ab}}\arctan \left ({\frac{a\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81404, size = 300, normalized size = 5.26 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 10.3269, size = 411, normalized size = 7.21 \begin{align*} \begin{cases} \tilde{\infty } x^{\frac{5}{2}} & \text{for}\: a = 0 \wedge b = 0 \\\frac{2 x^{\frac{5}{2}}}{5 b^{2}} & \text{for}\: a = 0 \\\frac{2 \sqrt{x}}{a^{2}} & \text{for}\: b = 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} \end{align*}

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*x**(5/2)/(5*b**2), Eq(a, 0)), (2*sqrt(x)/a**2, Eq(b, 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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Giac [A]  time = 1.11867, size = 62, normalized size = 1.09 \begin{align*} -\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}} \end{align*}

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)