3.1684 \(\int \frac{1}{(a+\frac{b}{x})^3 \sqrt{x}} \, dx\)

Optimal. Leaf size=82 \[ -\frac{5 x^{3/2}}{4 a^2 (a x+b)}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{7/2}}+\frac{15 \sqrt{x}}{4 a^3}-\frac{x^{5/2}}{2 a (a x+b)^2} \]

[Out]

(15*Sqrt[x])/(4*a^3) - x^(5/2)/(2*a*(b + a*x)^2) - (5*x^(3/2))/(4*a^2*(b + a*x)) - (15*Sqrt[b]*ArcTan[(Sqrt[a]
*Sqrt[x])/Sqrt[b]])/(4*a^(7/2))

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Rubi [A]  time = 0.0250705, antiderivative size = 82, normalized size of antiderivative = 1., 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} \[ -\frac{5 x^{3/2}}{4 a^2 (a x+b)}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{7/2}}+\frac{15 \sqrt{x}}{4 a^3}-\frac{x^{5/2}}{2 a (a x+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^3*Sqrt[x]),x]

[Out]

(15*Sqrt[x])/(4*a^3) - x^(5/2)/(2*a*(b + a*x)^2) - (5*x^(3/2))/(4*a^2*(b + a*x)) - (15*Sqrt[b]*ArcTan[(Sqrt[a]
*Sqrt[x])/Sqrt[b]])/(4*a^(7/2))

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 )^3 \sqrt{x}} \, dx &=\int \frac{x^{5/2}}{(b+a x)^3} \, dx\\ &=-\frac{x^{5/2}}{2 a (b+a x)^2}+\frac{5 \int \frac{x^{3/2}}{(b+a x)^2} \, dx}{4 a}\\ &=-\frac{x^{5/2}}{2 a (b+a x)^2}-\frac{5 x^{3/2}}{4 a^2 (b+a x)}+\frac{15 \int \frac{\sqrt{x}}{b+a x} \, dx}{8 a^2}\\ &=\frac{15 \sqrt{x}}{4 a^3}-\frac{x^{5/2}}{2 a (b+a x)^2}-\frac{5 x^{3/2}}{4 a^2 (b+a x)}-\frac{(15 b) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 a^3}\\ &=\frac{15 \sqrt{x}}{4 a^3}-\frac{x^{5/2}}{2 a (b+a x)^2}-\frac{5 x^{3/2}}{4 a^2 (b+a x)}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 a^3}\\ &=\frac{15 \sqrt{x}}{4 a^3}-\frac{x^{5/2}}{2 a (b+a x)^2}-\frac{5 x^{3/2}}{4 a^2 (b+a x)}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0047981, size = 27, normalized size = 0.33 \[ \frac{2 x^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};-\frac{a x}{b}\right )}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^3*Sqrt[x]),x]

[Out]

(2*x^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, -((a*x)/b)])/(7*b^3)

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Maple [A]  time = 0.011, size = 66, normalized size = 0.8 \begin{align*} 2\,{\frac{\sqrt{x}}{{a}^{3}}}+{\frac{9\,b}{4\,{a}^{2} \left ( ax+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{7\,{b}^{2}}{4\,{a}^{3} \left ( ax+b \right ) ^{2}}\sqrt{x}}-{\frac{15\,b}{4\,{a}^{3}}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^3/x^(1/2),x)

[Out]

2*x^(1/2)/a^3+9/4/a^2*b/(a*x+b)^2*x^(3/2)+7/4/a^3*b^2/(a*x+b)^2*x^(1/2)-15/4/a^3*b/(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)^3/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.90873, size = 443, normalized size = 5.4 \begin{align*} \left [\frac{15 \,{\left (a^{2} x^{2} + 2 \, 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 (8 \, a^{2} x^{2} + 25 \, a b x + 15 \, b^{2}\right )} \sqrt{x}}{8 \,{\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}, -\frac{15 \,{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{x} \sqrt{\frac{b}{a}}}{b}\right ) -{\left (8 \, a^{2} x^{2} + 25 \, a b x + 15 \, b^{2}\right )} \sqrt{x}}{4 \,{\left (a^{5} x^{2} + 2 \, a^{4} b x + a^{3} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(1/2),x, algorithm="fricas")

[Out]

[1/8*(15*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(-b/a)*log((a*x - 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(8*a^2*x^2
+ 25*a*b*x + 15*b^2)*sqrt(x))/(a^5*x^2 + 2*a^4*b*x + a^3*b^2), -1/4*(15*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(b/a)*a
rctan(a*sqrt(x)*sqrt(b/a)/b) - (8*a^2*x^2 + 25*a*b*x + 15*b^2)*sqrt(x))/(a^5*x^2 + 2*a^4*b*x + a^3*b^2)]

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Sympy [A]  time = 45.3965, size = 816, normalized size = 9.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**3/x**(1/2),x)

[Out]

Piecewise((zoo*x**(7/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*b**3), Eq(a, 0)), (2*sqrt(x)/a**3, Eq(b, 0)), (1
6*I*a**3*sqrt(b)*x**(5/2)*sqrt(1/a)/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a*
*4*b**(5/2)*sqrt(1/a)) + 50*I*a**2*b**(3/2)*x**(3/2)*sqrt(1/a)/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b*
*(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) - 15*a**2*b*x**2*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a*
*6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) + 15*a**2*b*x**2*log
(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b
**(5/2)*sqrt(1/a)) + 30*I*a*b**(5/2)*sqrt(x)*sqrt(1/a)/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x
*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) - 30*a*b**2*x*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*
x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) + 30*a*b**2*x*log(I*sqrt(b)*sqr
t(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1
/a)) - 15*b**3*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**5*b**(3/2)*x*sqr
t(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)) + 15*b**3*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**6*sqrt(b)*x**2*sqrt
(1/a) + 16*I*a**5*b**(3/2)*x*sqrt(1/a) + 8*I*a**4*b**(5/2)*sqrt(1/a)), True))

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Giac [A]  time = 1.11098, size = 80, normalized size = 0.98 \begin{align*} -\frac{15 \, b \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{3}} + \frac{2 \, \sqrt{x}}{a^{3}} + \frac{9 \, a b x^{\frac{3}{2}} + 7 \, b^{2} \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(1/2),x, algorithm="giac")

[Out]

-15/4*b*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 2*sqrt(x)/a^3 + 1/4*(9*a*b*x^(3/2) + 7*b^2*sqrt(x))/((a*
x + b)^2*a^3)