3.1713 \(\int (a+\frac {b}{x})^{5/2} x^2 \, dx\)

Optimal. Leaf size=87 \[ \frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 \sqrt {a}}+\frac {5}{8} b^2 x \sqrt {a+\frac {b}{x}}+\frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{5/2}+\frac {5}{12} b x^2 \left (a+\frac {b}{x}\right )^{3/2} \]

[Out]

5/12*b*(a+b/x)^(3/2)*x^2+1/3*(a+b/x)^(5/2)*x^3+5/8*b^3*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(1/2)+5/8*b^2*x*(a+b/x
)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 87, 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 = {266, 47, 63, 208} \[ \frac {5}{8} b^2 x \sqrt {a+\frac {b}{x}}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 \sqrt {a}}+\frac {5}{12} b x^2 \left (a+\frac {b}{x}\right )^{3/2}+\frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*x^2,x]

[Out]

(5*b^2*Sqrt[a + b/x]*x)/8 + (5*b*(a + b/x)^(3/2)*x^2)/12 + ((a + b/x)^(5/2)*x^3)/3 + (5*b^3*ArcTanh[Sqrt[a + b
/x]/Sqrt[a]])/(8*Sqrt[a])

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 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx &=-\operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} \left (a+\frac {b}{x}\right )^{5/2} x^3-\frac {1}{6} (5 b) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{12} b \left (a+\frac {b}{x}\right )^{3/2} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{5/2} x^3-\frac {1}{8} \left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{8} b^2 \sqrt {a+\frac {b}{x}} x+\frac {5}{12} b \left (a+\frac {b}{x}\right )^{3/2} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{5/2} x^3-\frac {1}{16} \left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {5}{8} b^2 \sqrt {a+\frac {b}{x}} x+\frac {5}{12} b \left (a+\frac {b}{x}\right )^{3/2} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{5/2} x^3-\frac {1}{8} \left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )\\ &=\frac {5}{8} b^2 \sqrt {a+\frac {b}{x}} x+\frac {5}{12} b \left (a+\frac {b}{x}\right )^{3/2} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{5/2} x^3+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 \sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 87, normalized size = 1.00 \[ \frac {x \sqrt {a+\frac {b}{x}} \left (8 a^3 x^3+34 a^2 b x^2+15 b^3 \sqrt {\frac {b}{a x}+1} \tanh ^{-1}\left (\sqrt {\frac {b}{a x}+1}\right )+59 a b^2 x+33 b^3\right )}{24 (a x+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)*x^2,x]

[Out]

(Sqrt[a + b/x]*x*(33*b^3 + 59*a*b^2*x + 34*a^2*b*x^2 + 8*a^3*x^3 + 15*b^3*Sqrt[1 + b/(a*x)]*ArcTanh[Sqrt[1 + b
/(a*x)]]))/(24*(b + a*x))

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fricas [A]  time = 1.21, size = 152, normalized size = 1.75 \[ \left [\frac {15 \, \sqrt {a} b^{3} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (8 \, a^{3} x^{3} + 26 \, a^{2} b x^{2} + 33 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{48 \, a}, -\frac {15 \, \sqrt {-a} b^{3} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (8 \, a^{3} x^{3} + 26 \, a^{2} b x^{2} + 33 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{24 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*sqrt(a)*b^3*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(8*a^3*x^3 + 26*a^2*b*x^2 + 33*a*b^2*
x)*sqrt((a*x + b)/x))/a, -1/24*(15*sqrt(-a)*b^3*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (8*a^3*x^3 + 26*a^2*b*x
^2 + 33*a*b^2*x)*sqrt((a*x + b)/x))/a]

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giac [A]  time = 0.25, size = 93, normalized size = 1.07 \[ -\frac {5 \, b^{3} \log \left ({\left | -2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} - b \right |}\right ) \mathrm {sgn}\relax (x)}{16 \, \sqrt {a}} + \frac {5 \, b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{16 \, \sqrt {a}} + \frac {1}{24} \, \sqrt {a x^{2} + b x} {\left (33 \, b^{2} \mathrm {sgn}\relax (x) + 2 \, {\left (4 \, a^{2} x \mathrm {sgn}\relax (x) + 13 \, a b \mathrm {sgn}\relax (x)\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/16*b^3*log(abs(-2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) - b))*sgn(x)/sqrt(a) + 5/16*b^3*log(abs(b))*sgn(x
)/sqrt(a) + 1/24*sqrt(a*x^2 + b*x)*(33*b^2*sgn(x) + 2*(4*a^2*x*sgn(x) + 13*a*b*sgn(x))*x)

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maple [A]  time = 0.01, size = 115, normalized size = 1.32 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (15 a \,b^{3} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+36 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b x +66 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{2}+16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}}\right ) x}{48 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*((a*x+b)/x)^(1/2)*x*(16*(a*x^2+b*x)^(3/2)*a^(5/2)+36*(a*x^2+b*x)^(1/2)*a^(5/2)*b*x+66*(a*x^2+b*x)^(1/2)*a
^(3/2)*b^2+15*a*b^3*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2)))/((a*x+b)*x)^(1/2)/a^(3/2)

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maxima [A]  time = 2.31, size = 131, normalized size = 1.51 \[ -\frac {5 \, b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{16 \, \sqrt {a}} + \frac {33 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{3} - 40 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a b^{3} + 15 \, \sqrt {a + \frac {b}{x}} a^{2} b^{3}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{3} - 3 \, {\left (a + \frac {b}{x}\right )}^{2} a + 3 \, {\left (a + \frac {b}{x}\right )} a^{2} - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/16*b^3*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a) + 1/24*(33*(a + b/x)^(5/2)*b^3 - 40
*(a + b/x)^(3/2)*a*b^3 + 15*sqrt(a + b/x)*a^2*b^3)/((a + b/x)^3 - 3*(a + b/x)^2*a + 3*(a + b/x)*a^2 - a^3)

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mupad [B]  time = 1.35, size = 72, normalized size = 0.83 \[ \frac {11\,x^3\,{\left (a+\frac {b}{x}\right )}^{5/2}}{8}-\frac {5\,a\,x^3\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3}+\frac {5\,a^2\,x^3\,\sqrt {a+\frac {b}{x}}}{8}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{8\,\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b/x)^(5/2),x)

[Out]

(11*x^3*(a + b/x)^(5/2))/8 - (b^3*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*5i)/(8*a^(1/2)) - (5*a*x^3*(a + b/x)^(3/2
))/3 + (5*a^2*x^3*(a + b/x)^(1/2))/8

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sympy [A]  time = 5.58, size = 102, normalized size = 1.17 \[ \frac {a^{2} \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a x}{b} + 1}}{3} + \frac {13 a b^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {\frac {a x}{b} + 1}}{12} + \frac {11 b^{\frac {5}{2}} \sqrt {x} \sqrt {\frac {a x}{b} + 1}}{8} + \frac {5 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{8 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*sqrt(b)*x**(5/2)*sqrt(a*x/b + 1)/3 + 13*a*b**(3/2)*x**(3/2)*sqrt(a*x/b + 1)/12 + 11*b**(5/2)*sqrt(x)*sqrt
(a*x/b + 1)/8 + 5*b**3*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(8*sqrt(a))

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