3.1801 \(\int \frac{1}{(a+\frac{b}{x})^{5/2} x^{7/2}} \, dx\)

Optimal. Leaf size=75 \[ \frac{2}{b^2 \sqrt{x} \sqrt{a+\frac{b}{x}}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{b^{5/2}}+\frac{2}{3 b x^{3/2} \left (a+\frac{b}{x}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0404249, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {337, 288, 217, 206} \[ \frac{2}{b^2 \sqrt{x} \sqrt{a+\frac{b}{x}}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{b^{5/2}}+\frac{2}{3 b x^{3/2} \left (a+\frac{b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(5/2)*x^(7/2)),x]

[Out]

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

Mathematica [A]  time = 0.16973, size = 94, normalized size = 1.25 \[ \frac{2 \left (\sqrt{b} \sqrt{x} (3 a x+4 b)-3 \sqrt{a} x \sqrt{\frac{b}{a x}+1} (a x+b) \sinh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}}\right )\right )}{3 b^{5/2} x \sqrt{a+\frac{b}{x}} (a x+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(5/2)*x^(7/2)),x]

[Out]

(2*(Sqrt[b]*Sqrt[x]*(4*b + 3*a*x) - 3*Sqrt[a]*Sqrt[1 + b/(a*x)]*x*(b + a*x)*ArcSinh[Sqrt[b]/(Sqrt[a]*Sqrt[x])]
))/(3*b^(5/2)*Sqrt[a + b/x]*x*(b + a*x))

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Maple [A]  time = 0.014, size = 85, normalized size = 1.1 \begin{align*}{\frac{2}{3\, \left ( ax+b \right ) ^{2}}\sqrt{{\frac{ax+b}{x}}}\sqrt{x} \left ( -3\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ) \sqrt{ax+b}xa+4\,{b}^{3/2}+3\,ax\sqrt{b}-3\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ) b\sqrt{ax+b} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(5/2)/x^(7/2),x)

[Out]

2/3*((a*x+b)/x)^(1/2)*x^(1/2)*(-3*arctanh((a*x+b)^(1/2)/b^(1/2))*(a*x+b)^(1/2)*x*a+4*b^(3/2)+3*a*x*b^(1/2)-3*a
rctanh((a*x+b)^(1/2)/b^(1/2))*b*(a*x+b)^(1/2))/b^(5/2)/(a*x+b)^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)^(5/2)/x^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45265, size = 474, normalized size = 6.32 \begin{align*} \left [\frac{3 \,{\left (a^{2} x^{2} + 2 \, a b x + b^{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 (3 \, a b x + 4 \, b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{3 \,{\left (a^{2} b^{3} x^{2} + 2 \, a b^{4} x + b^{5}\right )}}, \frac{2 \,{\left (3 \,{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (3 \, a b x + 4 \, b^{2}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}\right )}}{3 \,{\left (a^{2} b^{3} x^{2} + 2 \, a b^{4} x + b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x^(7/2),x, algorithm="fricas")

[Out]

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

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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)**(5/2)/x**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.19328, size = 105, normalized size = 1.4 \begin{align*} \frac{2 \, \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} - \frac{2 \,{\left (3 \, \sqrt{b} \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) + 4 \, \sqrt{-b}\right )}}{3 \, \sqrt{-b} b^{\frac{5}{2}}} + \frac{2 \,{\left (3 \, a x + 4 \, b\right )}}{3 \,{\left (a x + b\right )}^{\frac{3}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x^(7/2),x, algorithm="giac")

[Out]

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