3.18.23 \(\int \frac {x^2}{\sqrt {a+\frac {b}{x}}} \, dx\) [1723]

Optimal. Leaf size=96 \[ \frac {5 b^2 \sqrt {a+\frac {b}{x}} x}{8 a^3}-\frac {5 b \sqrt {a+\frac {b}{x}} x^2}{12 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^3}{3 a}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{7/2}} \]

[Out]

-5/8*b^3*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)+5/8*b^2*x*(a+b/x)^(1/2)/a^3-5/12*b*x^2*(a+b/x)^(1/2)/a^2+1/3*x
^3*(a+b/x)^(1/2)/a

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Rubi [A]
time = 0.03, antiderivative size = 96, 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 = {272, 44, 65, 214} \begin {gather*} -\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{7/2}}+\frac {5 b^2 x \sqrt {a+\frac {b}{x}}}{8 a^3}-\frac {5 b x^2 \sqrt {a+\frac {b}{x}}}{12 a^2}+\frac {x^3 \sqrt {a+\frac {b}{x}}}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b/x],x]

[Out]

(5*b^2*Sqrt[a + b/x]*x)/(8*a^3) - (5*b*Sqrt[a + b/x]*x^2)/(12*a^2) + (Sqrt[a + b/x]*x^3)/(3*a) - (5*b^3*ArcTan
h[Sqrt[a + b/x]/Sqrt[a]])/(8*a^(7/2))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

Rule 272

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 \frac {x^2}{\sqrt {a+\frac {b}{x}}} \, dx &=-\text {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\sqrt {a+\frac {b}{x}} x^3}{3 a}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{6 a}\\ &=-\frac {5 b \sqrt {a+\frac {b}{x}} x^2}{12 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^3}{3 a}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{8 a^2}\\ &=\frac {5 b^2 \sqrt {a+\frac {b}{x}} x}{8 a^3}-\frac {5 b \sqrt {a+\frac {b}{x}} x^2}{12 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^3}{3 a}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{16 a^3}\\ &=\frac {5 b^2 \sqrt {a+\frac {b}{x}} x}{8 a^3}-\frac {5 b \sqrt {a+\frac {b}{x}} x^2}{12 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^3}{3 a}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{8 a^3}\\ &=\frac {5 b^2 \sqrt {a+\frac {b}{x}} x}{8 a^3}-\frac {5 b \sqrt {a+\frac {b}{x}} x^2}{12 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^3}{3 a}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 69, normalized size = 0.72 \begin {gather*} \frac {\sqrt {a+\frac {b}{x}} x \left (15 b^2-10 a b x+8 a^2 x^2\right )}{24 a^3}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + b/x],x]

[Out]

(Sqrt[a + b/x]*x*(15*b^2 - 10*a*b*x + 8*a^2*x^2))/(24*a^3) - (5*b^3*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(8*a^(7/2)
)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(163\) vs. \(2(76)=152\).
time = 0.04, size = 164, normalized size = 1.71

method	result	size
risch	\(\frac {\left (8 a^{2} x^{2}-10 a b x +15 b^{2}\right ) \left (a x +b \right )}{24 a^{3} \sqrt {\frac {a x +b}{x}}}-\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {x \left (a x +b \right )}}{16 a^{\frac {7}{2}} x \sqrt {\frac {a x +b}{x}}}\)	\(97\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}}-36 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b x +48 \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}} b^{2}-18 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{2}-24 a \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{3}+9 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{3}\right )}{48 a^{\frac {9}{2}} \sqrt {x \left (a x +b \right )}}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+1/x*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/48*((a*x+b)/x)^(1/2)*x/a^(9/2)*(16*(a*x^2+b*x)^(3/2)*a^(5/2)-36*(a*x^2+b*x)^(1/2)*a^(5/2)*b*x+48*(x*(a*x+b))
^(1/2)*a^(3/2)*b^2-18*(a*x^2+b*x)^(1/2)*a^(3/2)*b^2-24*a*ln(1/2*(2*(x*(a*x+b))^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))
*b^3+9*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*b^3)/(x*(a*x+b))^(1/2)

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Maxima [A]
time = 0.52, size = 137, normalized size = 1.43 \begin {gather*} \frac {5 \, b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{16 \, a^{\frac {7}{2}}} + \frac {15 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{3} - 40 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a b^{3} + 33 \, \sqrt {a + \frac {b}{x}} a^{2} b^{3}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{3} a^{3} - 3 \, {\left (a + \frac {b}{x}\right )}^{2} a^{4} + 3 \, {\left (a + \frac {b}{x}\right )} a^{5} - a^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*b^3*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2) + 1/24*(15*(a + b/x)^(5/2)*b^3 - 40*
(a + b/x)^(3/2)*a*b^3 + 33*sqrt(a + b/x)*a^2*b^3)/((a + b/x)^3*a^3 - 3*(a + b/x)^2*a^4 + 3*(a + b/x)*a^5 - a^6
)

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Fricas [A]
time = 0.37, size = 151, normalized size = 1.57 \begin {gather*} \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} - 10 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{48 \, a^{4}}, \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} - 10 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{24 \, a^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b/x)^(1/2),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 - 10*a^2*b*x^2 + 15*a*b^2*
x)*sqrt((a*x + b)/x))/a^4, 1/24*(15*sqrt(-a)*b^3*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (8*a^3*x^3 - 10*a^2*b*
x^2 + 15*a*b^2*x)*sqrt((a*x + b)/x))/a^4]

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Sympy [A]
time = 7.40, size = 128, normalized size = 1.33 \begin {gather*} \frac {x^{\frac {7}{2}}}{3 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {\sqrt {b} x^{\frac {5}{2}}}{12 a \sqrt {\frac {a x}{b} + 1}} + \frac {5 b^{\frac {3}{2}} x^{\frac {3}{2}}}{24 a^{2} \sqrt {\frac {a x}{b} + 1}} + \frac {5 b^{\frac {5}{2}} \sqrt {x}}{8 a^{3} \sqrt {\frac {a x}{b} + 1}} - \frac {5 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{8 a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**(7/2)/(3*sqrt(b)*sqrt(a*x/b + 1)) - sqrt(b)*x**(5/2)/(12*a*sqrt(a*x/b + 1)) + 5*b**(3/2)*x**(3/2)/(24*a**2*
sqrt(a*x/b + 1)) + 5*b**(5/2)*sqrt(x)/(8*a**3*sqrt(a*x/b + 1)) - 5*b**3*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(8*a**(
7/2))

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Giac [A]
time = 0.51, size = 106, normalized size = 1.10 \begin {gather*} \frac {1}{24} \, \sqrt {a x^{2} + b x} {\left (2 \, x {\left (\frac {4 \, x}{a \mathrm {sgn}\left (x\right )} - \frac {5 \, b}{a^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {15 \, b^{2}}{a^{3} \mathrm {sgn}\left (x\right )}\right )} - \frac {5 \, b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {7}{2}}} + \frac {5 \, b^{3} \log \left ({\left | -2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} - b \right |}\right )}{16 \, a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(a*x^2 + b*x)*(2*x*(4*x/(a*sgn(x)) - 5*b/(a^2*sgn(x))) + 15*b^2/(a^3*sgn(x))) - 5/16*b^3*log(abs(b))*
sgn(x)/a^(7/2) + 5/16*b^3*log(abs(-2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) - b))/(a^(7/2)*sgn(x))

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Mupad [B]
time = 1.40, size = 77, normalized size = 0.80 \begin {gather*} \frac {11\,x^3\,\sqrt {a+\frac {b}{x}}}{8\,a}-\frac {5\,x^3\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3\,a^2}+\frac {5\,x^3\,{\left (a+\frac {b}{x}\right )}^{5/2}}{8\,a^3}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{8\,a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b/x)^(1/2),x)

[Out]

(b^3*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*5i)/(8*a^(7/2)) + (11*x^3*(a + b/x)^(1/2))/(8*a) - (5*x^3*(a + b/x)^(3
/2))/(3*a^2) + (5*x^3*(a + b/x)^(5/2))/(8*a^3)

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