3.1723 \(\int \frac{x^2}{\sqrt{a+\frac{b}{x}}} \, dx\)

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.039083, antiderivative size = 96, normalized size of antiderivative = 1., 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, 51, 63, 208} \[ \frac{5 b^2 x \sqrt{a+\frac{b}{x}}}{8 a^3}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{8 a^{7/2}}-\frac{5 b x^2 \sqrt{a+\frac{b}{x}}}{12 a^2}+\frac{x^3 \sqrt{a+\frac{b}{x}}}{3 a} \]

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 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]]

Rule 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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]

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{a+\frac{b}{x}}} \, dx &=-\operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.0091067, size = 37, normalized size = 0.39 \[ -\frac{2 b^3 \sqrt{a+\frac{b}{x}} \, _2F_1\left (\frac{1}{2},4;\frac{3}{2};\frac{b}{a x}+1\right )}{a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + b/x],x]

[Out]

(-2*b^3*Sqrt[a + b/x]*Hypergeometric2F1[1/2, 4, 3/2, 1 + b/(a*x)])/a^4

________________________________________________________________________________________

Maple [B]  time = 0.008, size = 164, normalized size = 1.7 \begin{align*}{\frac{x}{48}\sqrt{{\frac{ax+b}{x}}} \left ( 16\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{5/2}-36\,\sqrt{a{x}^{2}+bx}{a}^{5/2}xb+48\,\sqrt{ \left ( ax+b \right ) x}{a}^{3/2}{b}^{2}-18\,\sqrt{a{x}^{2}+bx}{a}^{3/2}{b}^{2}-24\,a\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){b}^{3}+9\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) a{b}^{3} \right ){a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*x*b+48*((a*x+b)*x)
^(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*((a*x+b)*x)^(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)/((a*x+b)*x)^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.76019, size = 358, normalized size = 3.73 \begin{align*} \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{align*}

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]

________________________________________________________________________________________

Sympy [A]  time = 5.80324, size = 128, normalized size = 1.33 \begin{align*} \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{align*}

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))

________________________________________________________________________________________

Giac [A]  time = 1.16718, size = 169, normalized size = 1.76 \begin{align*} \frac{1}{24} \, b{\left (\frac{15 \, b^{2} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} - \frac{33 \, a^{2} b^{2} \sqrt{\frac{a x + b}{x}} - \frac{40 \,{\left (a x + b\right )} a b^{2} \sqrt{\frac{a x + b}{x}}}{x} + \frac{15 \,{\left (a x + b\right )}^{2} b^{2} \sqrt{\frac{a x + b}{x}}}{x^{2}}}{{\left (a - \frac{a x + b}{x}\right )}^{3} a^{3}}\right )} \end{align*}

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*b*(15*b^2*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a^3) - (33*a^2*b^2*sqrt((a*x + b)/x) - 40*(a*x + b
)*a*b^2*sqrt((a*x + b)/x)/x + 15*(a*x + b)^2*b^2*sqrt((a*x + b)/x)/x^2)/((a - (a*x + b)/x)^3*a^3))