3.1673 \(\int \frac{x^{5/2}}{(a+\frac{b}{x})^2} \, dx\)

Optimal. Leaf size=98 \[ \frac{3 b^2 x^{3/2}}{a^4}-\frac{9 b^3 \sqrt{x}}{a^5}+\frac{9 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{a^{11/2}}-\frac{9 b x^{5/2}}{5 a^3}+\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (a x+b)} \]

[Out]

(-9*b^3*Sqrt[x])/a^5 + (3*b^2*x^(3/2))/a^4 - (9*b*x^(5/2))/(5*a^3) + (9*x^(7/2))/(7*a^2) - x^(9/2)/(a*(b + a*x
)) + (9*b^(7/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(11/2)

________________________________________________________________________________________

Rubi [A]  time = 0.0406906, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {263, 47, 50, 63, 205} \[ \frac{3 b^2 x^{3/2}}{a^4}-\frac{9 b^3 \sqrt{x}}{a^5}+\frac{9 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{a^{11/2}}-\frac{9 b x^{5/2}}{5 a^3}+\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (a x+b)} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b/x)^2,x]

[Out]

(-9*b^3*Sqrt[x])/a^5 + (3*b^2*x^(3/2))/a^4 - (9*b*x^(5/2))/(5*a^3) + (9*x^(7/2))/(7*a^2) - x^(9/2)/(a*(b + a*x
)) + (9*b^(7/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(11/2)

Rule 263

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 205

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

Rubi steps

\begin{align*} \int \frac{x^{5/2}}{\left (a+\frac{b}{x}\right )^2} \, dx &=\int \frac{x^{9/2}}{(b+a x)^2} \, dx\\ &=-\frac{x^{9/2}}{a (b+a x)}+\frac{9 \int \frac{x^{7/2}}{b+a x} \, dx}{2 a}\\ &=\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (b+a x)}-\frac{(9 b) \int \frac{x^{5/2}}{b+a x} \, dx}{2 a^2}\\ &=-\frac{9 b x^{5/2}}{5 a^3}+\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (b+a x)}+\frac{\left (9 b^2\right ) \int \frac{x^{3/2}}{b+a x} \, dx}{2 a^3}\\ &=\frac{3 b^2 x^{3/2}}{a^4}-\frac{9 b x^{5/2}}{5 a^3}+\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (b+a x)}-\frac{\left (9 b^3\right ) \int \frac{\sqrt{x}}{b+a x} \, dx}{2 a^4}\\ &=-\frac{9 b^3 \sqrt{x}}{a^5}+\frac{3 b^2 x^{3/2}}{a^4}-\frac{9 b x^{5/2}}{5 a^3}+\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (b+a x)}+\frac{\left (9 b^4\right ) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{2 a^5}\\ &=-\frac{9 b^3 \sqrt{x}}{a^5}+\frac{3 b^2 x^{3/2}}{a^4}-\frac{9 b x^{5/2}}{5 a^3}+\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (b+a x)}+\frac{\left (9 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{a^5}\\ &=-\frac{9 b^3 \sqrt{x}}{a^5}+\frac{3 b^2 x^{3/2}}{a^4}-\frac{9 b x^{5/2}}{5 a^3}+\frac{9 x^{7/2}}{7 a^2}-\frac{x^{9/2}}{a (b+a x)}+\frac{9 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{a^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0048274, size = 27, normalized size = 0.28 \[ \frac{2 x^{11/2} \, _2F_1\left (2,\frac{11}{2};\frac{13}{2};-\frac{a x}{b}\right )}{11 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b/x)^2,x]

[Out]

(2*x^(11/2)*Hypergeometric2F1[2, 11/2, 13/2, -((a*x)/b)])/(11*b^2)

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 83, normalized size = 0.9 \begin{align*}{\frac{2}{7\,{a}^{2}}{x}^{{\frac{7}{2}}}}-{\frac{4\,b}{5\,{a}^{3}}{x}^{{\frac{5}{2}}}}+2\,{\frac{{b}^{2}{x}^{3/2}}{{a}^{4}}}-8\,{\frac{{b}^{3}\sqrt{x}}{{a}^{5}}}-{\frac{{b}^{4}}{{a}^{5} \left ( ax+b \right ) }\sqrt{x}}+9\,{\frac{{b}^{4}}{{a}^{5}\sqrt{ab}}\arctan \left ({\frac{a\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*x^(7/2)/a^2-4/5*b*x^(5/2)/a^3+2*b^2*x^(3/2)/a^4-8*b^3*x^(1/2)/a^5-1/a^5*b^4*x^(1/2)/(a*x+b)+9/a^5*b^4/(a*b
)^(1/2)*arctan(a*x^(1/2)/(a*b)^(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^(5/2)/(a+b/x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.84736, size = 477, normalized size = 4.87 \begin{align*} \left [\frac{315 \,{\left (a b^{3} x + b^{4}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{a x + 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - b}{a x + b}\right ) + 2 \,{\left (10 \, a^{4} x^{4} - 18 \, a^{3} b x^{3} + 42 \, a^{2} b^{2} x^{2} - 210 \, a b^{3} x - 315 \, b^{4}\right )} \sqrt{x}}{70 \,{\left (a^{6} x + a^{5} b\right )}}, \frac{315 \,{\left (a b^{3} x + b^{4}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{x} \sqrt{\frac{b}{a}}}{b}\right ) +{\left (10 \, a^{4} x^{4} - 18 \, a^{3} b x^{3} + 42 \, a^{2} b^{2} x^{2} - 210 \, a b^{3} x - 315 \, b^{4}\right )} \sqrt{x}}{35 \,{\left (a^{6} x + a^{5} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/70*(315*(a*b^3*x + b^4)*sqrt(-b/a)*log((a*x + 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(10*a^4*x^4 - 18*a
^3*b*x^3 + 42*a^2*b^2*x^2 - 210*a*b^3*x - 315*b^4)*sqrt(x))/(a^6*x + a^5*b), 1/35*(315*(a*b^3*x + b^4)*sqrt(b/
a)*arctan(a*sqrt(x)*sqrt(b/a)/b) + (10*a^4*x^4 - 18*a^3*b*x^3 + 42*a^2*b^2*x^2 - 210*a*b^3*x - 315*b^4)*sqrt(x
))/(a^6*x + a^5*b)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.11506, size = 119, normalized size = 1.21 \begin{align*} \frac{9 \, b^{4} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{5}} - \frac{b^{4} \sqrt{x}}{{\left (a x + b\right )} a^{5}} + \frac{2 \,{\left (5 \, a^{12} x^{\frac{7}{2}} - 14 \, a^{11} b x^{\frac{5}{2}} + 35 \, a^{10} b^{2} x^{\frac{3}{2}} - 140 \, a^{9} b^{3} \sqrt{x}\right )}}{35 \, a^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9*b^4*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) - b^4*sqrt(x)/((a*x + b)*a^5) + 2/35*(5*a^12*x^(7/2) - 14*a^
11*b*x^(5/2) + 35*a^10*b^2*x^(3/2) - 140*a^9*b^3*sqrt(x))/a^14