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

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

[Out]

-15/4*arctan(a^(1/2)*x^(1/2)/b^(1/2))*a^(1/2)/b^(7/2)-15/4/b^3/x^(1/2)+1/2/b/(a*x+b)^2/x^(1/2)+5/4/b^2/(a*x+b)
/x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 82, 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} \[ \frac {5}{4 b^2 \sqrt {x} (a x+b)}-\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{7/2}}+\frac {1}{2 b \sqrt {x} (a x+b)^2}-\frac {15}{4 b^3 \sqrt {x}} \]

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 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 )^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*}

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Mathematica [C]  time = 0.00, size = 25, normalized size = 0.30 \[ -\frac {2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};-\frac {a x}{b}\right )}{b^3 \sqrt {x}} \]

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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fricas [A]  time = 1.09, size = 214, normalized size = 2.61 \[ \left [\frac {15 \, {\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, {\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}, \frac {15 \, {\left (a^{2} x^{3} + 2 \, a b x^{2} + b^{2} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - {\left (15 \, a^{2} x^{2} + 25 \, a b x + 8 \, b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{3} x^{3} + 2 \, a b^{4} x^{2} + b^{5} x\right )}}\right ] \]

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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giac [A]  time = 0.16, size = 59, normalized size = 0.72 \[ -\frac {15 \, a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {2}{b^{3} \sqrt {x}} - \frac {7 \, a^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} b^{3}} \]

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)

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maple [A]  time = 0.02, size = 66, normalized size = 0.80 \[ -\frac {7 a^{2} x^{\frac {3}{2}}}{4 \left (a x +b \right )^{2} b^{3}}-\frac {9 a \sqrt {x}}{4 \left (a x +b \right )^{2} b^{2}}-\frac {15 a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{3}}-\frac {2}{b^{3} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.42, size = 75, normalized size = 0.91 \[ -\frac {\frac {7 \, a^{2}}{\sqrt {x}} + \frac {9 \, a b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{2} b^{3} + \frac {2 \, a b^{4}}{x} + \frac {b^{5}}{x^{2}}\right )}} + \frac {15 \, a \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {2}{b^{3} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(7*a^2/sqrt(x) + 9*a*b/x^(3/2))/(a^2*b^3 + 2*a*b^4/x + b^5/x^2) + 15/4*a*arctan(b/(sqrt(a*b)*sqrt(x)))/(s
qrt(a*b)*b^3) - 2/(b^3*sqrt(x))

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mupad [B]  time = 1.13, size = 70, normalized size = 0.85 \[ -\frac {\frac {2}{b}+\frac {15\,a^2\,x^2}{4\,b^3}+\frac {25\,a\,x}{4\,b^2}}{a^2\,x^{5/2}+b^2\,\sqrt {x}+2\,a\,b\,x^{3/2}}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(9/2)*(a + b/x)^3),x)

[Out]

- (2/b + (15*a^2*x^2)/(4*b^3) + (25*a*x)/(4*b^2))/(a^2*x^(5/2) + b^2*x^(1/2) + 2*a*b*x^(3/2)) - (15*a^(1/2)*at
an((a^(1/2)*x^(1/2))/b^(1/2)))/(4*b^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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