3.1688 \(\int \frac{1}{(a+\frac{b}{x})^3 x^{9/2}} \, dx\)

Optimal. Leaf size=82 \[ \frac{5}{4 b^2 \sqrt{x} (a x+b)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2}}+\frac{1}{2 b \sqrt{x} (a x+b)^2}-\frac{15}{4 b^3 \sqrt{x}} \]

[Out]

-15/(4*b^3*Sqrt[x]) + 1/(2*b*Sqrt[x]*(b + a*x)^2) + 5/(4*b^2*Sqrt[x]*(b + a*x)) - (15*Sqrt[a]*ArcTan[(Sqrt[a]*
Sqrt[x])/Sqrt[b]])/(4*b^(7/2))

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Rubi [A]  time = 0.0286377, antiderivative size = 82, 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} \[ \frac{5}{4 b^2 \sqrt{x} (a x+b)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2}}+\frac{1}{2 b \sqrt{x} (a x+b)^2}-\frac{15}{4 b^3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^3*x^(9/2)),x]

[Out]

-15/(4*b^3*Sqrt[x]) + 1/(2*b*Sqrt[x]*(b + a*x)^2) + 5/(4*b^2*Sqrt[x]*(b + a*x)) - (15*Sqrt[a]*ArcTan[(Sqrt[a]*
Sqrt[x])/Sqrt[b]])/(4*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 )^3 x^{9/2}} \, dx &=\int \frac{1}{x^{3/2} (b+a x)^3} \, dx\\ &=\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5 \int \frac{1}{x^{3/2} (b+a x)^2} \, dx}{4 b}\\ &=\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}+\frac{15 \int \frac{1}{x^{3/2} (b+a x)} \, dx}{8 b^2}\\ &=-\frac{15}{4 b^3 \sqrt{x}}+\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}-\frac{(15 a) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 b^3}\\ &=-\frac{15}{4 b^3 \sqrt{x}}+\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}-\frac{(15 a) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 b^3}\\ &=-\frac{15}{4 b^3 \sqrt{x}}+\frac{1}{2 b \sqrt{x} (b+a x)^2}+\frac{5}{4 b^2 \sqrt{x} (b+a x)}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0042387, size = 25, normalized size = 0.3 \[ -\frac{2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\frac{a x}{b}\right )}{b^3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^3*x^(9/2)),x]

[Out]

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

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Maple [A]  time = 0.012, size = 66, normalized size = 0.8 \begin{align*} -{\frac{7\,{a}^{2}}{4\,{b}^{3} \left ( ax+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{9\,a}{4\,{b}^{2} \left ( ax+b \right ) ^{2}}\sqrt{x}}-{\frac{15\,a}{4\,{b}^{3}}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-2\,{\frac{1}{{b}^{3}\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/4/b^3*a^2/(a*x+b)^2*x^(3/2)-9/4/b^2*a/(a*x+b)^2*x^(1/2)-15/4/b^3*a/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2)
)-2/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)^3/x^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78374, size = 466, normalized size = 5.68 \begin{align*} \left [\frac{15 \,{\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\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} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt{x}}{8 \,{\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}, \frac{15 \,{\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt{x}}{4 \,{\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(a^2*x^3 + 2*a*b*x^2 + b^2*x)*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b)) - 2*(15*a^
2*x^2 + 25*a*b*x + 8*b^2)*sqrt(x))/(a^2*b^3*x^3 + 2*a*b^4*x^2 + b^5*x), 1/4*(15*(a^2*x^3 + 2*a*b*x^2 + b^2*x)*
sqrt(a/b)*arctan(b*sqrt(a/b)/(a*sqrt(x))) - (15*a^2*x^2 + 25*a*b*x + 8*b^2)*sqrt(x))/(a^2*b^3*x^3 + 2*a*b^4*x^
2 + b^5*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/x**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.09443, size = 80, normalized size = 0.98 \begin{align*} -\frac{15 \, a \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{3}} - \frac{2}{b^{3} \sqrt{x}} - \frac{7 \, a^{2} x^{\frac{3}{2}} + 9 \, a b \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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