3.1793 \(\int \frac{1}{(a+\frac{b}{x})^{3/2} x^{9/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{4 b^{7/2}}-\frac{5 \sqrt{a+\frac{b}{x}}}{2 b^2 x^{3/2}}+\frac{15 a \sqrt{a+\frac{b}{x}}}{4 b^3 \sqrt{x}}+\frac{2}{b x^{5/2} \sqrt{a+\frac{b}{x}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0541592, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {337, 288, 321, 217, 206} \[ -\frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{4 b^{7/2}}-\frac{5 \sqrt{a+\frac{b}{x}}}{2 b^2 x^{3/2}}+\frac{15 a \sqrt{a+\frac{b}{x}}}{4 b^3 \sqrt{x}}+\frac{2}{b x^{5/2} \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(3/2)*x^(9/2)),x]

[Out]

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

Rule 337

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, -Dist[k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x}\right )^{3/2} x^{9/2}} \, dx &=-\left (2 \operatorname{Subst}\left (\int \frac{x^6}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=\frac{2}{b \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{10 \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{b}\\ &=\frac{2}{b \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{5 \sqrt{a+\frac{b}{x}}}{2 b^2 x^{3/2}}+\frac{(15 a) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{2 b^2}\\ &=\frac{2}{b \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{5 \sqrt{a+\frac{b}{x}}}{2 b^2 x^{3/2}}+\frac{15 a \sqrt{a+\frac{b}{x}}}{4 b^3 \sqrt{x}}-\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{4 b^3}\\ &=\frac{2}{b \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{5 \sqrt{a+\frac{b}{x}}}{2 b^2 x^{3/2}}+\frac{15 a \sqrt{a+\frac{b}{x}}}{4 b^3 \sqrt{x}}-\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )}{4 b^3}\\ &=\frac{2}{b \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{5 \sqrt{a+\frac{b}{x}}}{2 b^2 x^{3/2}}+\frac{15 a \sqrt{a+\frac{b}{x}}}{4 b^3 \sqrt{x}}-\frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0164082, size = 56, normalized size = 0.54 \[ -\frac{2 \sqrt{\frac{b}{a x}+1} \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};-\frac{b}{a x}\right )}{7 a x^{7/2} \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(3/2)*x^(9/2)),x]

[Out]

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[3/2, 7/2, 9/2, -(b/(a*x))])/(7*a*Sqrt[a + b/x]*x^(7/2))

________________________________________________________________________________________

Maple [A]  time = 0.016, size = 78, normalized size = 0.8 \begin{align*} -{\frac{1}{4\,ax+4\,b}\sqrt{{\frac{ax+b}{x}}} \left ( 15\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ) \sqrt{ax+b}{x}^{2}{a}^{2}-5\,{b}^{3/2}xa-15\,{a}^{2}{x}^{2}\sqrt{b}+2\,{b}^{5/2} \right ){x}^{-{\frac{3}{2}}}{b}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(3/2)/x^(9/2),x)

[Out]

-1/4*((a*x+b)/x)^(1/2)/x^(3/2)*(15*arctanh((a*x+b)^(1/2)/b^(1/2))*(a*x+b)^(1/2)*x^2*a^2-5*b^(3/2)*x*a-15*a^2*x
^2*b^(1/2)+2*b^(5/2))/(a*x+b)/b^(7/2)

________________________________________________________________________________________

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)^(3/2)/x^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.55056, size = 485, normalized size = 4.66 \begin{align*} \left [\frac{15 \,{\left (a^{3} x^{3} + a^{2} b x^{2}\right )} \sqrt{b} \log \left (\frac{a x - 2 \, \sqrt{b} \sqrt{x} \sqrt{\frac{a x + b}{x}} + 2 \, b}{x}\right ) + 2 \,{\left (15 \, a^{2} b x^{2} + 5 \, a b^{2} x - 2 \, b^{3}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{8 \,{\left (a b^{4} x^{3} + b^{5} x^{2}\right )}}, \frac{15 \,{\left (a^{3} x^{3} + a^{2} b x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (15 \, a^{2} b x^{2} + 5 \, a b^{2} x - 2 \, b^{3}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{4 \,{\left (a b^{4} x^{3} + b^{5} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/x^(9/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(3/2)/x**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.25446, size = 97, normalized size = 0.93 \begin{align*} \frac{1}{4} \, a^{2}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} + \frac{8}{\sqrt{a x + b} b^{3}} + \frac{7 \,{\left (a x + b\right )}^{\frac{3}{2}} - 9 \, \sqrt{a x + b} b}{a^{2} b^{3} x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/x^(9/2),x, algorithm="giac")

[Out]

1/4*a^2*(15*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + 8/(sqrt(a*x + b)*b^3) + (7*(a*x + b)^(3/2) - 9*sqr
t(a*x + b)*b)/(a^2*b^3*x^2))