3.1682 \(\int \frac{x^{3/2}}{(a+\frac{b}{x})^3} \, dx\)

Optimal. Leaf size=110 \[ \frac{63 b^2 \sqrt{x}}{4 a^5}-\frac{63 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{11/2}}-\frac{9 x^{7/2}}{4 a^2 (a x+b)}-\frac{21 b x^{3/2}}{4 a^4}+\frac{63 x^{5/2}}{20 a^3}-\frac{x^{9/2}}{2 a (a x+b)^2} \]

[Out]

(63*b^2*Sqrt[x])/(4*a^5) - (21*b*x^(3/2))/(4*a^4) + (63*x^(5/2))/(20*a^3) - x^(9/2)/(2*a*(b + a*x)^2) - (9*x^(
7/2))/(4*a^2*(b + a*x)) - (63*b^(5/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*a^(11/2))

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Rubi [A]  time = 0.0413503, antiderivative size = 110, 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} \[ \frac{63 b^2 \sqrt{x}}{4 a^5}-\frac{63 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{11/2}}-\frac{9 x^{7/2}}{4 a^2 (a x+b)}-\frac{21 b x^{3/2}}{4 a^4}+\frac{63 x^{5/2}}{20 a^3}-\frac{x^{9/2}}{2 a (a x+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + b/x)^3,x]

[Out]

(63*b^2*Sqrt[x])/(4*a^5) - (21*b*x^(3/2))/(4*a^4) + (63*x^(5/2))/(20*a^3) - x^(9/2)/(2*a*(b + a*x)^2) - (9*x^(
7/2))/(4*a^2*(b + a*x)) - (63*b^(5/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*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^{3/2}}{\left (a+\frac{b}{x}\right )^3} \, dx &=\int \frac{x^{9/2}}{(b+a x)^3} \, dx\\ &=-\frac{x^{9/2}}{2 a (b+a x)^2}+\frac{9 \int \frac{x^{7/2}}{(b+a x)^2} \, dx}{4 a}\\ &=-\frac{x^{9/2}}{2 a (b+a x)^2}-\frac{9 x^{7/2}}{4 a^2 (b+a x)}+\frac{63 \int \frac{x^{5/2}}{b+a x} \, dx}{8 a^2}\\ &=\frac{63 x^{5/2}}{20 a^3}-\frac{x^{9/2}}{2 a (b+a x)^2}-\frac{9 x^{7/2}}{4 a^2 (b+a x)}-\frac{(63 b) \int \frac{x^{3/2}}{b+a x} \, dx}{8 a^3}\\ &=-\frac{21 b x^{3/2}}{4 a^4}+\frac{63 x^{5/2}}{20 a^3}-\frac{x^{9/2}}{2 a (b+a x)^2}-\frac{9 x^{7/2}}{4 a^2 (b+a x)}+\frac{\left (63 b^2\right ) \int \frac{\sqrt{x}}{b+a x} \, dx}{8 a^4}\\ &=\frac{63 b^2 \sqrt{x}}{4 a^5}-\frac{21 b x^{3/2}}{4 a^4}+\frac{63 x^{5/2}}{20 a^3}-\frac{x^{9/2}}{2 a (b+a x)^2}-\frac{9 x^{7/2}}{4 a^2 (b+a x)}-\frac{\left (63 b^3\right ) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 a^5}\\ &=\frac{63 b^2 \sqrt{x}}{4 a^5}-\frac{21 b x^{3/2}}{4 a^4}+\frac{63 x^{5/2}}{20 a^3}-\frac{x^{9/2}}{2 a (b+a x)^2}-\frac{9 x^{7/2}}{4 a^2 (b+a x)}-\frac{\left (63 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 a^5}\\ &=\frac{63 b^2 \sqrt{x}}{4 a^5}-\frac{21 b x^{3/2}}{4 a^4}+\frac{63 x^{5/2}}{20 a^3}-\frac{x^{9/2}}{2 a (b+a x)^2}-\frac{9 x^{7/2}}{4 a^2 (b+a x)}-\frac{63 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0059522, size = 27, normalized size = 0.25 \[ \frac{2 x^{11/2} \, _2F_1\left (3,\frac{11}{2};\frac{13}{2};-\frac{a x}{b}\right )}{11 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + b/x)^3,x]

[Out]

(2*x^(11/2)*Hypergeometric2F1[3, 11/2, 13/2, -((a*x)/b)])/(11*b^3)

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Maple [A]  time = 0.013, size = 90, normalized size = 0.8 \begin{align*}{\frac{2}{5\,{a}^{3}}{x}^{{\frac{5}{2}}}}-2\,{\frac{b{x}^{3/2}}{{a}^{4}}}+12\,{\frac{{b}^{2}\sqrt{x}}{{a}^{5}}}+{\frac{17\,{b}^{3}}{4\,{a}^{4} \left ( ax+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{15\,{b}^{4}}{4\,{a}^{5} \left ( ax+b \right ) ^{2}}\sqrt{x}}-{\frac{63\,{b}^{3}}{4\,{a}^{5}}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^(5/2)/a^3-2*b*x^(3/2)/a^4+12*b^2*x^(1/2)/a^5+17/4/a^4*b^3/(a*x+b)^2*x^(3/2)+15/4/a^5*b^4/(a*x+b)^2*x^(1/
2)-63/4/a^5*b^3/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86749, size = 564, normalized size = 5.13 \begin{align*} \left [\frac{315 \,{\left (a^{2} b^{2} x^{2} + 2 \, 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 (8 \, a^{4} x^{4} - 24 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} + 525 \, a b^{3} x + 315 \, b^{4}\right )} \sqrt{x}}{40 \,{\left (a^{7} x^{2} + 2 \, a^{6} b x + a^{5} b^{2}\right )}}, -\frac{315 \,{\left (a^{2} b^{2} x^{2} + 2 \, a b^{3} x + b^{4}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{x} \sqrt{\frac{b}{a}}}{b}\right ) -{\left (8 \, a^{4} x^{4} - 24 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} + 525 \, a b^{3} x + 315 \, b^{4}\right )} \sqrt{x}}{20 \,{\left (a^{7} x^{2} + 2 \, a^{6} b x + a^{5} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/40*(315*(a^2*b^2*x^2 + 2*a*b^3*x + b^4)*sqrt(-b/a)*log((a*x - 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(8
*a^4*x^4 - 24*a^3*b*x^3 + 168*a^2*b^2*x^2 + 525*a*b^3*x + 315*b^4)*sqrt(x))/(a^7*x^2 + 2*a^6*b*x + a^5*b^2), -
1/20*(315*(a^2*b^2*x^2 + 2*a*b^3*x + b^4)*sqrt(b/a)*arctan(a*sqrt(x)*sqrt(b/a)/b) - (8*a^4*x^4 - 24*a^3*b*x^3
+ 168*a^2*b^2*x^2 + 525*a*b^3*x + 315*b^4)*sqrt(x))/(a^7*x^2 + 2*a^6*b*x + a^5*b^2)]

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

[Out]

Timed out

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Giac [A]  time = 1.08539, size = 119, normalized size = 1.08 \begin{align*} -\frac{63 \, b^{3} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{5}} + \frac{17 \, a b^{3} x^{\frac{3}{2}} + 15 \, b^{4} \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} a^{5}} + \frac{2 \,{\left (a^{12} x^{\frac{5}{2}} - 5 \, a^{11} b x^{\frac{3}{2}} + 30 \, a^{10} b^{2} \sqrt{x}\right )}}{5 \, a^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-63/4*b^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/4*(17*a*b^3*x^(3/2) + 15*b^4*sqrt(x))/((a*x + b)^2*a
^5) + 2/5*(a^12*x^(5/2) - 5*a^11*b*x^(3/2) + 30*a^10*b^2*sqrt(x))/a^15