3.1680 \(\int \frac{1}{(a+\frac{b}{x})^2 x^{9/2}} \, dx\)

Optimal. Leaf size=69 \[ \frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}+\frac{5 a}{b^3 \sqrt{x}}+\frac{1}{b x^{3/2} (a x+b)}-\frac{5}{3 b^2 x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0242281, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {263, 51, 63, 205} \[ \frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}+\frac{5 a}{b^3 \sqrt{x}}+\frac{1}{b x^{3/2} (a x+b)}-\frac{5}{3 b^2 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^2*x^(9/2)),x]

[Out]

-5/(3*b^2*x^(3/2)) + (5*a)/(b^3*Sqrt[x]) + 1/(b*x^(3/2)*(b + a*x)) + (5*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[
b]])/b^(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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 )^2 x^{9/2}} \, dx &=\int \frac{1}{x^{5/2} (b+a x)^2} \, dx\\ &=\frac{1}{b x^{3/2} (b+a x)}+\frac{5 \int \frac{1}{x^{5/2} (b+a x)} \, dx}{2 b}\\ &=-\frac{5}{3 b^2 x^{3/2}}+\frac{1}{b x^{3/2} (b+a x)}-\frac{(5 a) \int \frac{1}{x^{3/2} (b+a x)} \, dx}{2 b^2}\\ &=-\frac{5}{3 b^2 x^{3/2}}+\frac{5 a}{b^3 \sqrt{x}}+\frac{1}{b x^{3/2} (b+a x)}+\frac{\left (5 a^2\right ) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{2 b^3}\\ &=-\frac{5}{3 b^2 x^{3/2}}+\frac{5 a}{b^3 \sqrt{x}}+\frac{1}{b x^{3/2} (b+a x)}+\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{b^3}\\ &=-\frac{5}{3 b^2 x^{3/2}}+\frac{5 a}{b^3 \sqrt{x}}+\frac{1}{b x^{3/2} (b+a x)}+\frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0045889, size = 27, normalized size = 0.39 \[ -\frac{2 \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};-\frac{a x}{b}\right )}{3 b^2 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^2*x^(9/2)),x]

[Out]

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

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Maple [A]  time = 0.013, size = 60, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}}{{b}^{3} \left ( ax+b \right ) }\sqrt{x}}+5\,{\frac{{a}^{2}}{{b}^{3}\sqrt{ab}}\arctan \left ({\frac{a\sqrt{x}}{\sqrt{ab}}} \right ) }-{\frac{2}{3\,{b}^{2}}{x}^{-{\frac{3}{2}}}}+4\,{\frac{a}{{b}^{3}\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2/b^3*x^(1/2)/(a*x+b)+5*a^2/b^3/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2))-2/3/b^2/x^(3/2)+4*a/b^3/x^(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)^2/x^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.91305, size = 402, normalized size = 5.83 \begin{align*} \left [\frac{15 \,{\left (a^{2} x^{3} + a b x^{2}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{a x + 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - b}{a x + b}\right ) + 2 \,{\left (15 \, a^{2} x^{2} + 10 \, a b x - 2 \, b^{2}\right )} \sqrt{x}}{6 \,{\left (a b^{3} x^{3} + b^{4} x^{2}\right )}}, -\frac{15 \,{\left (a^{2} x^{3} + a b x^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (15 \, a^{2} x^{2} + 10 \, a b x - 2 \, b^{2}\right )} \sqrt{x}}{3 \,{\left (a b^{3} x^{3} + b^{4} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a^2*x^3 + a*b*x^2)*sqrt(-a/b)*log((a*x + 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b)) + 2*(15*a^2*x^2 + 10
*a*b*x - 2*b^2)*sqrt(x))/(a*b^3*x^3 + b^4*x^2), -1/3*(15*(a^2*x^3 + a*b*x^2)*sqrt(a/b)*arctan(b*sqrt(a/b)/(a*s
qrt(x))) - (15*a^2*x^2 + 10*a*b*x - 2*b^2)*sqrt(x))/(a*b^3*x^3 + b^4*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)**2/x**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.13217, size = 78, normalized size = 1.13 \begin{align*} \frac{5 \, a^{2} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} + \frac{a^{2} \sqrt{x}}{{\left (a x + b\right )} b^{3}} + \frac{2 \,{\left (6 \, a x - b\right )}}{3 \, b^{3} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a^2*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + a^2*sqrt(x)/((a*x + b)*b^3) + 2/3*(6*a*x - b)/(b^3*x^(3/2)
)