3.1672 \(\int \frac{1}{(a+\frac{b}{x}) x^{9/2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac{2 a^2}{b^3 \sqrt{x}}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}+\frac{2 a}{3 b^2 x^{3/2}}-\frac{2}{5 b x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0253934, antiderivative size = 68, 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} \[ -\frac{2 a^2}{b^3 \sqrt{x}}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{7/2}}+\frac{2 a}{3 b^2 x^{3/2}}-\frac{2}{5 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)*x^(9/2)),x]

[Out]

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

Mathematica [C]  time = 0.0049082, size = 27, normalized size = 0.4 \[ -\frac{2 \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};-\frac{a x}{b}\right )}{5 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)*x^(9/2)),x]

[Out]

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

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Maple [A]  time = 0.008, size = 54, normalized size = 0.8 \begin{align*} -2\,{\frac{{a}^{3}}{{b}^{3}\sqrt{ab}}\arctan \left ({\frac{a\sqrt{x}}{\sqrt{ab}}} \right ) }-{\frac{2}{5\,b}{x}^{-{\frac{5}{2}}}}-2\,{\frac{{a}^{2}}{{b}^{3}\sqrt{x}}}+{\frac{2\,a}{3\,{b}^{2}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/x^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.75661, size = 336, normalized size = 4.94 \begin{align*} \left [\frac{15 \, a^{2} x^{3} \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} - 5 \, a b x + 3 \, b^{2}\right )} \sqrt{x}}{15 \, b^{3} x^{3}}, \frac{2 \,{\left (15 \, a^{2} x^{3} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (15 \, a^{2} x^{2} - 5 \, a b x + 3 \, b^{2}\right )} \sqrt{x}\right )}}{15 \, b^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 103.053, size = 139, normalized size = 2.04 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{x^{\frac{5}{2}}} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{2}{5 b x^{\frac{5}{2}}} & \text{for}\: a = 0 \\- \frac{2}{7 a x^{\frac{7}{2}}} & \text{for}\: b = 0 \\- \frac{2 a^{2}}{b^{3} \sqrt{x}} + \frac{i a^{2} \log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} + \sqrt{x} \right )}}{b^{\frac{7}{2}} \sqrt{\frac{1}{a}}} - \frac{i a^{2} \log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} + \sqrt{x} \right )}}{b^{\frac{7}{2}} \sqrt{\frac{1}{a}}} + \frac{2 a}{3 b^{2} x^{\frac{3}{2}}} - \frac{2}{5 b x^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)/x**(9/2),x)

[Out]

Piecewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(a, 0)), (-2/(7*a*x**(7/2)), Eq(b, 0)), (
-2*a**2/(b**3*sqrt(x)) + I*a**2*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(b**(7/2)*sqrt(1/a)) - I*a**2*log(I*sqrt(b
)*sqrt(1/a) + sqrt(x))/(b**(7/2)*sqrt(1/a)) + 2*a/(3*b**2*x**(3/2)) - 2/(5*b*x**(5/2)), True))

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Giac [A]  time = 1.11328, size = 70, normalized size = 1.03 \begin{align*} -\frac{2 \, a^{3} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} - \frac{2 \,{\left (15 \, a^{2} x^{2} - 5 \, a b x + 3 \, b^{2}\right )}}{15 \, b^{3} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 2/15*(15*a^2*x^2 - 5*a*b*x + 3*b^2)/(b^3*x^(5/2))