3.1671 \(\int \frac{1}{(a+\frac{b}{x}) x^{7/2}} \, dx\)

Optimal. Leaf size=53 \[ \frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{5/2}}+\frac{2 a}{b^2 \sqrt{x}}-\frac{2}{3 b x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0187194, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {263, 51, 63, 205} \[ \frac{2 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{5/2}}+\frac{2 a}{b^2 \sqrt{x}}-\frac{2}{3 b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)*x^(7/2)),x]

[Out]

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

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

Mathematica [C]  time = 0.0051113, size = 27, normalized size = 0.51 \[ -\frac{2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{a x}{b}\right )}{3 b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)*x^(7/2)),x]

[Out]

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

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Maple [A]  time = 0.009, size = 43, normalized size = 0.8 \begin{align*} 2\,{\frac{{a}^{2}}{{b}^{2}\sqrt{ab}}\arctan \left ({\frac{a\sqrt{x}}{\sqrt{ab}}} \right ) }-{\frac{2}{3\,b}{x}^{-{\frac{3}{2}}}}+2\,{\frac{a}{{b}^{2}\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77515, size = 275, normalized size = 5.19 \begin{align*} \left [\frac{3 \, a x^{2} \sqrt{-\frac{a}{b}} \log \left (\frac{a x + 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - b}{a x + b}\right ) + 2 \,{\left (3 \, a x - b\right )} \sqrt{x}}{3 \, b^{2} x^{2}}, -\frac{2 \,{\left (3 \, a x^{2} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (3 \, a x - b\right )} \sqrt{x}\right )}}{3 \, b^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 29.1598, size = 121, normalized size = 2.28 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{x^{\frac{3}{2}}} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{2}{3 b x^{\frac{3}{2}}} & \text{for}\: a = 0 \\- \frac{2}{5 a x^{\frac{5}{2}}} & \text{for}\: b = 0 \\\frac{2 a}{b^{2} \sqrt{x}} - \frac{i a \log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} + \sqrt{x} \right )}}{b^{\frac{5}{2}} \sqrt{\frac{1}{a}}} + \frac{i a \log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} + \sqrt{x} \right )}}{b^{\frac{5}{2}} \sqrt{\frac{1}{a}}} - \frac{2}{3 b x^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)/x**(7/2),x)

[Out]

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

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Giac [A]  time = 1.61717, size = 55, normalized size = 1.04 \begin{align*} \frac{2 \, a^{2} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{2}} + \frac{2 \,{\left (3 \, a x - b\right )}}{3 \, b^{2} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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