3.1792 \(\int \frac {1}{(a+\frac {b}{x})^{3/2} x^{7/2}} \, dx\)

Optimal. Leaf size=74 \[ \frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{b^{5/2}}-\frac {3 \sqrt {a+\frac {b}{x}}}{b^2 \sqrt {x}}+\frac {2}{b x^{3/2} \sqrt {a+\frac {b}{x}}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {337, 288, 321, 217, 206} \[ -\frac {3 \sqrt {a+\frac {b}{x}}}{b^2 \sqrt {x}}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{b^{5/2}}+\frac {2}{b x^{3/2} \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(3/2)*x^(7/2)),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 337

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, -Dist[k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{7/2}} \, dx &=-\left (2 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=\frac {2}{b \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {6 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{b}\\ &=\frac {2}{b \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {3 \sqrt {a+\frac {b}{x}}}{b^2 \sqrt {x}}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{b^2}\\ &=\frac {2}{b \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {3 \sqrt {a+\frac {b}{x}}}{b^2 \sqrt {x}}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{b^2}\\ &=\frac {2}{b \sqrt {a+\frac {b}{x}} x^{3/2}}-\frac {3 \sqrt {a+\frac {b}{x}}}{b^2 \sqrt {x}}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{b^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.76 \[ -\frac {2 \sqrt {\frac {b}{a x}+1} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {b}{a x}\right )}{5 a x^{5/2} \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(3/2)*x^(7/2)),x]

[Out]

(-2*Sqrt[1 + b/(a*x)]*Hypergeometric2F1[3/2, 5/2, 7/2, -(b/(a*x))])/(5*a*Sqrt[a + b/x]*x^(5/2))

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fricas [A]  time = 1.07, size = 179, normalized size = 2.42 \[ \left [\frac {3 \, {\left (a^{2} x^{2} + a b x\right )} \sqrt {b} \log \left (\frac {a x + 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, {\left (3 \, a b x + b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{2 \, {\left (a b^{3} x^{2} + b^{4} x\right )}}, -\frac {3 \, {\left (a^{2} x^{2} + a b x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (3 \, a b x + b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{a b^{3} x^{2} + b^{4} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/x^(7/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 64, normalized size = 0.86 \[ -\frac {3 \, a \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} - \frac {3 \, {\left (a x + b\right )} a - 2 \, a b}{{\left ({\left (a x + b\right )}^{\frac {3}{2}} - \sqrt {a x + b} b\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(3/2)/x^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 61, normalized size = 0.82 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-3 \sqrt {a x +b}\, a x \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )+3 a \sqrt {b}\, x +b^{\frac {3}{2}}\right )}{\left (a x +b \right ) b^{\frac {5}{2}} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(3/2)/x^(7/2),x)

[Out]

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

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maxima [A]  time = 2.40, size = 101, normalized size = 1.36 \[ -\frac {3 \, {\left (a + \frac {b}{x}\right )} a x - 2 \, a b}{{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{2} x^{\frac {3}{2}} - \sqrt {a + \frac {b}{x}} b^{3} \sqrt {x}} - \frac {3 \, a \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{2 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^{7/2}\,{\left (a+\frac {b}{x}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(a + b/x)^(3/2)),x)

[Out]

int(1/(x^(7/2)*(a + b/x)^(3/2)), x)

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sympy [A]  time = 38.02, size = 73, normalized size = 0.99 \[ - \frac {3 \sqrt {a}}{b^{2} \sqrt {x} \sqrt {1 + \frac {b}{a x}}} + \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{b^{\frac {5}{2}}} - \frac {1}{\sqrt {a} b x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(3/2)/x**(7/2),x)

[Out]

-3*sqrt(a)/(b**2*sqrt(x)*sqrt(1 + b/(a*x))) + 3*a*asinh(sqrt(b)/(sqrt(a)*sqrt(x)))/b**(5/2) - 1/(sqrt(a)*b*x**
(3/2)*sqrt(1 + b/(a*x)))

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