3.1772 \(\int (a+\frac {b}{x})^{5/2} x^{3/2} \, dx\)

Optimal. Leaf size=93 \[ -2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )+2 b^2 \sqrt {x} \sqrt {a+\frac {b}{x}}+\frac {2}{3} b x^{3/2} \left (a+\frac {b}{x}\right )^{3/2}+\frac {2}{5} x^{5/2} \left (a+\frac {b}{x}\right )^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {337, 277, 217, 206} \[ 2 b^2 \sqrt {x} \sqrt {a+\frac {b}{x}}-2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )+\frac {2}{3} b x^{3/2} \left (a+\frac {b}{x}\right )^{3/2}+\frac {2}{5} x^{5/2} \left (a+\frac {b}{x}\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*x^(3/2),x]

[Out]

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

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 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 \left (a+\frac {b}{x}\right )^{5/2} x^{3/2} \, dx &=-\left (2 \operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{5/2}}{x^6} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}-(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x^4} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=\frac {2}{3} b \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}+\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}-\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{x^2} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=2 b^2 \sqrt {a+\frac {b}{x}} \sqrt {x}+\frac {2}{3} b \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}+\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}-\left (2 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=2 b^2 \sqrt {a+\frac {b}{x}} \sqrt {x}+\frac {2}{3} b \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}+\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}-\left (2 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )\\ &=2 b^2 \sqrt {a+\frac {b}{x}} \sqrt {x}+\frac {2}{3} b \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}+\frac {2}{5} \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}-2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.60 \[ \frac {2 a^2 x^{5/2} \sqrt {a+\frac {b}{x}} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};-\frac {b}{a x}\right )}{5 \sqrt {\frac {b}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)*x^(3/2),x]

[Out]

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

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fricas [A]  time = 0.90, size = 142, normalized size = 1.53 \[ \left [b^{\frac {5}{2}} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + \frac {2}{15} \, {\left (3 \, a^{2} x^{2} + 11 \, a b x + 23 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}, 2 \, \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + \frac {2}{15} \, {\left (3 \, a^{2} x^{2} + 11 \, a b x + 23 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^(3/2),x, algorithm="fricas")

[Out]

[b^(5/2)*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) + 2/15*(3*a^2*x^2 + 11*a*b*x + 23*b^2)*sqrt(
x)*sqrt((a*x + b)/x), 2*sqrt(-b)*b^2*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + 2/15*(3*a^2*x^2 + 11*a*b*x
+ 23*b^2)*sqrt(x)*sqrt((a*x + b)/x)]

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giac [A]  time = 0.26, size = 56, normalized size = 0.60 \[ \frac {2 \, b^{3} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + \frac {2}{5} \, {\left (a x + b\right )}^{\frac {5}{2}} + \frac {2}{3} \, {\left (a x + b\right )}^{\frac {3}{2}} b + 2 \, \sqrt {a x + b} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 81, normalized size = 0.87 \[ -\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (-3 \sqrt {a x +b}\, a^{2} x^{2}+15 b^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )-11 \sqrt {a x +b}\, a b x -23 \sqrt {a x +b}\, b^{2}\right ) \sqrt {x}}{15 \sqrt {a x +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)*x^(3/2),x)

[Out]

-2/15*((a*x+b)/x)^(1/2)*x^(1/2)*(-3*x^2*a^2*(a*x+b)^(1/2)+15*b^(5/2)*arctanh((a*x+b)^(1/2)/b^(1/2))-11*x*a*b*(
a*x+b)^(1/2)-23*(a*x+b)^(1/2)*b^2)/(a*x+b)^(1/2)

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maxima [A]  time = 2.34, size = 91, normalized size = 0.98 \[ \frac {2}{5} \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b x^{\frac {3}{2}} + b^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right ) + 2 \, \sqrt {a + \frac {b}{x}} b^{2} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)*x^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{3/2}\,{\left (a+\frac {b}{x}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b/x)^(5/2),x)

[Out]

int(x^(3/2)*(a + b/x)^(5/2), x)

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sympy [A]  time = 26.95, size = 97, normalized size = 1.04 \[ \frac {2 a^{2} \sqrt {b} x^{2} \sqrt {\frac {a x}{b} + 1}}{5} + \frac {22 a b^{\frac {3}{2}} x \sqrt {\frac {a x}{b} + 1}}{15} + \frac {46 b^{\frac {5}{2}} \sqrt {\frac {a x}{b} + 1}}{15} + b^{\frac {5}{2}} \log {\left (\frac {a x}{b} \right )} - 2 b^{\frac {5}{2}} \log {\left (\sqrt {\frac {a x}{b} + 1} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)*x**(3/2),x)

[Out]

2*a**2*sqrt(b)*x**2*sqrt(a*x/b + 1)/5 + 22*a*b**(3/2)*x*sqrt(a*x/b + 1)/15 + 46*b**(5/2)*sqrt(a*x/b + 1)/15 +
b**(5/2)*log(a*x/b) - 2*b**(5/2)*log(sqrt(a*x/b + 1) + 1)

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