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

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

[Out]

-2/5/b/x^(5/2)+2/3*a/b^2/x^(3/2)-2*a^(5/2)*arctan(a^(1/2)*x^(1/2)/b^(1/2))/b^(7/2)-2*a^2/b^3/x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.40 \[ -\frac {2 \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\frac {a x}{b}\right )}{5 b x^{5/2}} \]

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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fricas [A]  time = 0.79, size = 144, normalized size = 2.12 \[ \left [\frac {15 \, a^{2} x^{3} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (15 \, a^{2} x^{2} - 5 \, a b x + 3 \, b^{2}\right )} \sqrt {x}}{15 \, b^{3} x^{3}}, \frac {2 \, {\left (15 \, a^{2} x^{3} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (15 \, a^{2} x^{2} - 5 \, a b x + 3 \, b^{2}\right )} \sqrt {x}\right )}}{15 \, b^{3} x^{3}}\right ] \]

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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giac [A]  time = 0.16, size = 52, normalized size = 0.76 \[ -\frac {2 \, a^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {2 \, {\left (15 \, a^{2} x^{2} - 5 \, a b x + 3 \, b^{2}\right )}}{15 \, b^{3} x^{\frac {5}{2}}} \]

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))

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maple [A]  time = 0.01, size = 54, normalized size = 0.79 \[ -\frac {2 a^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}-\frac {2 a^{2}}{b^{3} \sqrt {x}}+\frac {2 a}{3 b^{2} x^{\frac {3}{2}}}-\frac {2}{5 b \,x^{\frac {5}{2}}} \]

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(1/(a*b)^(1/2)*a*x^(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 [A]  time = 2.29, size = 54, normalized size = 0.79 \[ \frac {2 \, a^{3} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b^{3}} - \frac {2 \, {\left (\frac {15 \, a^{2}}{\sqrt {x}} - \frac {5 \, a b}{x^{\frac {3}{2}}} + \frac {3 \, b^{2}}{x^{\frac {5}{2}}}\right )}}{15 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 49, normalized size = 0.72 \[ -\frac {\frac {2}{5\,b}+\frac {2\,a^2\,x^2}{b^3}-\frac {2\,a\,x}{3\,b^2}}{x^{5/2}}-\frac {2\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(9/2)*(a + b/x)),x)

[Out]

- (2/(5*b) + (2*a^2*x^2)/b^3 - (2*a*x)/(3*b^2))/x^(5/2) - (2*a^(5/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/b^(7/2)

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sympy [A]  time = 149.24, size = 139, normalized size = 2.04 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{7 a x^{\frac {7}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {2 a^{2}}{b^{3} \sqrt {x}} + \frac {i a^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{b^{\frac {7}{2}} \sqrt {\frac {1}{a}}} - \frac {i a^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{b^{\frac {7}{2}} \sqrt {\frac {1}{a}}} + \frac {2 a}{3 b^{2} x^{\frac {3}{2}}} - \frac {2}{5 b x^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

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/(7*a*x**(7/2)), Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(a, 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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