3.1792 \(\int \frac{1}{(a+\frac{b}{x})^{3/2} x^{7/2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{3 \sqrt{a+\frac{b}{x}}}{b^2 \sqrt{x}}+\frac{3 a \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{b^{5/2}}+\frac{2}{b x^{3/2} \sqrt{a+\frac{b}{x}}} \]

[Out]

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

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Rubi [A]  time = 0.0399742, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ -\frac{3 \sqrt{a+\frac{b}{x}}}{b^2 \sqrt{x}}+\frac{3 a \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{b^{5/2}}+\frac{2}{b x^{3/2} \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(3/2)*x^(7/2)),x]

[Out]

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

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

Mathematica [C]  time = 0.0164479, size = 56, normalized size = 0.76 \[ -\frac{2 \sqrt{\frac{b}{a x}+1} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};-\frac{b}{a x}\right )}{5 a x^{5/2} \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(3/2)*x^(7/2)),x]

[Out]

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[3/2, 5/2, 7/2, -(b/(a*x))])/(5*a*Sqrt[a + b/x]*x^(5/2))

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Maple [A]  time = 0.018, size = 61, normalized size = 0.8 \begin{align*} -{\frac{1}{ax+b}\sqrt{{\frac{ax+b}{x}}} \left ( -3\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ) \sqrt{ax+b}xa+3\,ax\sqrt{b}+{b}^{{\frac{3}{2}}} \right ){\frac{1}{\sqrt{x}}}{b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(3/2)/x^(7/2),x)

[Out]

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

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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)^(3/2)/x^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56506, size = 410, normalized size = 5.54 \begin{align*} \left [\frac{3 \,{\left (a^{2} x^{2} + a b x\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 (3 \, a b x + b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{2 \,{\left (a b^{3} x^{2} + b^{4} x\right )}}, -\frac{3 \,{\left (a^{2} x^{2} + a b x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (3 \, a b x + b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{a b^{3} x^{2} + b^{4} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/x^(7/2),x, algorithm="fricas")

[Out]

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

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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**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.3323, size = 78, normalized size = 1.05 \begin{align*} -a{\left (\frac{3 \, \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} + \frac{3 \, a x + b}{{\left ({\left (a x + b\right )}^{\frac{3}{2}} - \sqrt{a x + b} b\right )} b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/x^(7/2),x, algorithm="giac")

[Out]

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