3.1685 \(\int \frac {1}{(a+\frac {b}{x})^3 x^{3/2}} \, dx\)

Optimal. Leaf size=70 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{5/2} \sqrt {b}}-\frac {3 \sqrt {x}}{4 a^2 (a x+b)}-\frac {x^{3/2}}{2 a (a x+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, 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, 47, 63, 205} \[ -\frac {3 \sqrt {x}}{4 a^2 (a x+b)}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{5/2} \sqrt {b}}-\frac {x^{3/2}}{2 a (a x+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^3*x^(3/2)),x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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^{3/2}} \, dx &=\int \frac {x^{3/2}}{(b+a x)^3} \, dx\\ &=-\frac {x^{3/2}}{2 a (b+a x)^2}+\frac {3 \int \frac {\sqrt {x}}{(b+a x)^2} \, dx}{4 a}\\ &=-\frac {x^{3/2}}{2 a (b+a x)^2}-\frac {3 \sqrt {x}}{4 a^2 (b+a x)}+\frac {3 \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{8 a^2}\\ &=-\frac {x^{3/2}}{2 a (b+a x)^2}-\frac {3 \sqrt {x}}{4 a^2 (b+a x)}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{4 a^2}\\ &=-\frac {x^{3/2}}{2 a (b+a x)^2}-\frac {3 \sqrt {x}}{4 a^2 (b+a x)}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{5/2} \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 59, normalized size = 0.84 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{4 a^{5/2} \sqrt {b}}-\frac {\sqrt {x} (5 a x+3 b)}{4 a^2 (a x+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^3*x^(3/2)),x]

[Out]

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

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fricas [A]  time = 1.01, size = 185, normalized size = 2.64 \[ \left [-\frac {3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-a b} \log \left (\frac {a x - b - 2 \, \sqrt {-a b} \sqrt {x}}{a x + b}\right ) + 2 \, {\left (5 \, a^{2} b x + 3 \, a b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{5} b x^{2} + 2 \, a^{4} b^{2} x + a^{3} b^{3}\right )}}, -\frac {3 \, {\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{a \sqrt {x}}\right ) + {\left (5 \, a^{2} b x + 3 \, a b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{5} b x^{2} + 2 \, a^{4} b^{2} x + a^{3} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(-a*b)*log((a*x - b - 2*sqrt(-a*b)*sqrt(x))/(a*x + b)) + 2*(5*a^2*b*x +
3*a*b^2)*sqrt(x))/(a^5*b*x^2 + 2*a^4*b^2*x + a^3*b^3), -1/4*(3*(a^2*x^2 + 2*a*b*x + b^2)*sqrt(a*b)*arctan(sqr
t(a*b)/(a*sqrt(x))) + (5*a^2*b*x + 3*a*b^2)*sqrt(x))/(a^5*b*x^2 + 2*a^4*b^2*x + a^3*b^3)]

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giac [A]  time = 0.16, size = 47, normalized size = 0.67 \[ \frac {3 \, \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{2}} - \frac {5 \, a x^{\frac {3}{2}} + 3 \, b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 50, normalized size = 0.71 \[ \frac {3 \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{2}}+\frac {-\frac {5 x^{\frac {3}{2}}}{4 a}-\frac {3 b \sqrt {x}}{4 a^{2}}}{\left (a x +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-5/8/a*x^(3/2)-3/8/a^2*b*x^(1/2))/(a*x+b)^2+3/4/a^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*a*x^(1/2))

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maxima [A]  time = 2.33, size = 62, normalized size = 0.89 \[ -\frac {\frac {5 \, a}{\sqrt {x}} + \frac {3 \, b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{4} + \frac {2 \, a^{3} b}{x} + \frac {a^{2} b^{2}}{x^{2}}\right )}} - \frac {3 \, \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.12, size = 58, normalized size = 0.83 \[ \frac {3\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,a^{5/2}\,\sqrt {b}}-\frac {\frac {5\,x^{3/2}}{4\,a}+\frac {3\,b\,\sqrt {x}}{4\,a^2}}{a^2\,x^2+2\,a\,b\,x+b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(3/2)*(a + b/x)^3),x)

[Out]

(3*atan((a^(1/2)*x^(1/2))/b^(1/2)))/(4*a^(5/2)*b^(1/2)) - ((5*x^(3/2))/(4*a) + (3*b*x^(1/2))/(4*a^2))/(b^2 + a
^2*x^2 + 2*a*b*x)

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sympy [A]  time = 110.84, size = 726, normalized size = 10.37 \[ \begin {cases} \tilde {\infty } x^{\frac {5}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b^{3}} & \text {for}\: a = 0 \\- \frac {2}{a^{3} \sqrt {x}} & \text {for}\: b = 0 \\- \frac {10 i a^{2} \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {1}{a}}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {3 a^{2} x^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {3 a^{2} x^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {6 i a b^{\frac {3}{2}} \sqrt {x} \sqrt {\frac {1}{a}}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {6 a b x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {6 a b x \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {3 b^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {3 b^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{8 i a^{5} \sqrt {b} x^{2} \sqrt {\frac {1}{a}} + 16 i a^{4} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 8 i a^{3} b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**3/x**(3/2),x)

[Out]

Piecewise((zoo*x**(5/2), Eq(a, 0) & Eq(b, 0)), (2*x**(5/2)/(5*b**3), Eq(a, 0)), (-2/(a**3*sqrt(x)), Eq(b, 0)),
(-10*I*a**2*sqrt(b)*x**(3/2)*sqrt(1/a)/(8*I*a**5*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**4*b**(3/2)*x*sqrt(1/a) + 8*
I*a**3*b**(5/2)*sqrt(1/a)) + 3*a**2*x**2*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**5*sqrt(b)*x**2*sqrt(1/a)
+ 16*I*a**4*b**(3/2)*x*sqrt(1/a) + 8*I*a**3*b**(5/2)*sqrt(1/a)) - 3*a**2*x**2*log(I*sqrt(b)*sqrt(1/a) + sqrt(x
))/(8*I*a**5*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**4*b**(3/2)*x*sqrt(1/a) + 8*I*a**3*b**(5/2)*sqrt(1/a)) - 6*I*a*b*
*(3/2)*sqrt(x)*sqrt(1/a)/(8*I*a**5*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**4*b**(3/2)*x*sqrt(1/a) + 8*I*a**3*b**(5/2)
*sqrt(1/a)) + 6*a*b*x*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**5*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**4*b**(3/2
)*x*sqrt(1/a) + 8*I*a**3*b**(5/2)*sqrt(1/a)) - 6*a*b*x*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**5*sqrt(b)*x*
*2*sqrt(1/a) + 16*I*a**4*b**(3/2)*x*sqrt(1/a) + 8*I*a**3*b**(5/2)*sqrt(1/a)) + 3*b**2*log(-I*sqrt(b)*sqrt(1/a)
+ sqrt(x))/(8*I*a**5*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**4*b**(3/2)*x*sqrt(1/a) + 8*I*a**3*b**(5/2)*sqrt(1/a)) -
3*b**2*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(8*I*a**5*sqrt(b)*x**2*sqrt(1/a) + 16*I*a**4*b**(3/2)*x*sqrt(1/a) +
8*I*a**3*b**(5/2)*sqrt(1/a)), True))

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