3.1803 \(\int \frac{1}{(a+\frac{b}{x})^{5/2} x^{11/2}} \, dx\)

Optimal. Leaf size=129 \[ -\frac{35 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{4 b^{9/2}}-\frac{35 \sqrt{a+\frac{b}{x}}}{6 b^3 x^{3/2}}+\frac{14}{3 b^2 x^{5/2} \sqrt{a+\frac{b}{x}}}+\frac{35 a \sqrt{a+\frac{b}{x}}}{4 b^4 \sqrt{x}}+\frac{2}{3 b x^{7/2} \left (a+\frac{b}{x}\right )^{3/2}} \]

[Out]

2/(3*b*(a + b/x)^(3/2)*x^(7/2)) + 14/(3*b^2*Sqrt[a + b/x]*x^(5/2)) - (35*Sqrt[a + b/x])/(6*b^3*x^(3/2)) + (35*
a*Sqrt[a + b/x])/(4*b^4*Sqrt[x]) - (35*a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(4*b^(9/2))

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Rubi [A]  time = 0.0710331, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{35 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{4 b^{9/2}}-\frac{35 \sqrt{a+\frac{b}{x}}}{6 b^3 x^{3/2}}+\frac{14}{3 b^2 x^{5/2} \sqrt{a+\frac{b}{x}}}+\frac{35 a \sqrt{a+\frac{b}{x}}}{4 b^4 \sqrt{x}}+\frac{2}{3 b x^{7/2} \left (a+\frac{b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(5/2)*x^(11/2)),x]

[Out]

2/(3*b*(a + b/x)^(3/2)*x^(7/2)) + 14/(3*b^2*Sqrt[a + b/x]*x^(5/2)) - (35*Sqrt[a + b/x])/(6*b^3*x^(3/2)) + (35*
a*Sqrt[a + b/x])/(4*b^4*Sqrt[x]) - (35*a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(4*b^(9/2))

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 )^{5/2} x^{11/2}} \, dx &=-\left (2 \operatorname{Subst}\left (\int \frac{x^8}{\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^{7/2}}-\frac{14 \operatorname{Subst}\left (\int \frac{x^6}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{3 b}\\ &=\frac{2}{3 b \left (a+\frac{b}{x}\right )^{3/2} x^{7/2}}+\frac{14}{3 b^2 \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{70 \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{3 b^2}\\ &=\frac{2}{3 b \left (a+\frac{b}{x}\right )^{3/2} x^{7/2}}+\frac{14}{3 b^2 \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{35 \sqrt{a+\frac{b}{x}}}{6 b^3 x^{3/2}}+\frac{(35 a) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{2 b^3}\\ &=\frac{2}{3 b \left (a+\frac{b}{x}\right )^{3/2} x^{7/2}}+\frac{14}{3 b^2 \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{35 \sqrt{a+\frac{b}{x}}}{6 b^3 x^{3/2}}+\frac{35 a \sqrt{a+\frac{b}{x}}}{4 b^4 \sqrt{x}}-\frac{\left (35 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{4 b^4}\\ &=\frac{2}{3 b \left (a+\frac{b}{x}\right )^{3/2} x^{7/2}}+\frac{14}{3 b^2 \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{35 \sqrt{a+\frac{b}{x}}}{6 b^3 x^{3/2}}+\frac{35 a \sqrt{a+\frac{b}{x}}}{4 b^4 \sqrt{x}}-\frac{\left (35 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^4}\\ &=\frac{2}{3 b \left (a+\frac{b}{x}\right )^{3/2} x^{7/2}}+\frac{14}{3 b^2 \sqrt{a+\frac{b}{x}} x^{5/2}}-\frac{35 \sqrt{a+\frac{b}{x}}}{6 b^3 x^{3/2}}+\frac{35 a \sqrt{a+\frac{b}{x}}}{4 b^4 \sqrt{x}}-\frac{35 a^2 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0183709, size = 56, normalized size = 0.43 \[ -\frac{2 \sqrt{\frac{b}{a x}+1} \, _2F_1\left (\frac{5}{2},\frac{9}{2};\frac{11}{2};-\frac{b}{a x}\right )}{9 a^2 x^{9/2} \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(5/2)*x^(11/2)),x]

[Out]

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[5/2, 9/2, 11/2, -(b/(a*x))])/(9*a^2*Sqrt[a + b/x]*x^(9/2))

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Maple [A]  time = 0.022, size = 117, normalized size = 0.9 \begin{align*} -{\frac{1}{12\, \left ( ax+b \right ) ^{2}}\sqrt{{\frac{ax+b}{x}}} \left ( 105\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ) \sqrt{ax+b}{x}^{3}{a}^{3}+105\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ){x}^{2}{a}^{2}b\sqrt{ax+b}-105\,{a}^{3}{x}^{3}\sqrt{b}-140\,{b}^{3/2}{x}^{2}{a}^{2}-21\,{b}^{5/2}xa+6\,{b}^{7/2} \right ){x}^{-{\frac{3}{2}}}{b}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(5/2)/x^(11/2),x)

[Out]

-1/12*((a*x+b)/x)^(1/2)/x^(3/2)*(105*arctanh((a*x+b)^(1/2)/b^(1/2))*(a*x+b)^(1/2)*x^3*a^3+105*arctanh((a*x+b)^
(1/2)/b^(1/2))*x^2*a^2*b*(a*x+b)^(1/2)-105*a^3*x^3*b^(1/2)-140*b^(3/2)*x^2*a^2-21*b^(5/2)*x*a+6*b^(7/2))/(a*x+
b)^2/b^(9/2)

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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^(11/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53939, size = 630, normalized size = 4.88 \begin{align*} \left [\frac{105 \,{\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} 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 (105 \, a^{3} b x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a b^{3} x - 6 \, b^{4}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{24 \,{\left (a^{2} b^{5} x^{4} + 2 \, a b^{6} x^{3} + b^{7} x^{2}\right )}}, \frac{105 \,{\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (105 \, a^{3} b x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a b^{3} x - 6 \, b^{4}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{12 \,{\left (a^{2} b^{5} x^{4} + 2 \, a b^{6} x^{3} + b^{7} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x^(11/2),x, algorithm="fricas")

[Out]

[1/24*(105*(a^4*x^4 + 2*a^3*b*x^3 + a^2*b^2*x^2)*sqrt(b)*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)
/x) + 2*(105*a^3*b*x^3 + 140*a^2*b^2*x^2 + 21*a*b^3*x - 6*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(a^2*b^5*x^4 + 2*a*b
^6*x^3 + b^7*x^2), 1/12*(105*(a^4*x^4 + 2*a^3*b*x^3 + a^2*b^2*x^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x
+ b)/x)/b) + (105*a^3*b*x^3 + 140*a^2*b^2*x^2 + 21*a*b^3*x - 6*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(a^2*b^5*x^4 +
2*a*b^6*x^3 + b^7*x^2)]

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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**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.29731, size = 109, normalized size = 0.84 \begin{align*} \frac{1}{12} \, a^{2}{\left (\frac{105 \, \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{4}} + \frac{8 \,{\left (9 \, a x + 10 \, b\right )}}{{\left (a x + b\right )}^{\frac{3}{2}} b^{4}} + \frac{3 \,{\left (11 \,{\left (a x + b\right )}^{\frac{3}{2}} - 13 \, \sqrt{a x + b} b\right )}}{a^{2} b^{4} x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2)/x^(11/2),x, algorithm="giac")

[Out]

1/12*a^2*(105*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^4) + 8*(9*a*x + 10*b)/((a*x + b)^(3/2)*b^4) + 3*(11*(
a*x + b)^(3/2) - 13*sqrt(a*x + b)*b)/(a^2*b^4*x^2))