3.1689 \(\int \frac{1}{(a+\frac{b}{x})^3 x^{11/2}} \, dx\)

Optimal. Leaf size=95 \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{9/2}}+\frac{7}{4 b^2 x^{3/2} (a x+b)}+\frac{35 a}{4 b^4 \sqrt{x}}+\frac{1}{2 b x^{3/2} (a x+b)^2}-\frac{35}{12 b^3 x^{3/2}} \]

[Out]

-35/(12*b^3*x^(3/2)) + (35*a)/(4*b^4*Sqrt[x]) + 1/(2*b*x^(3/2)*(b + a*x)^2) + 7/(4*b^2*x^(3/2)*(b + a*x)) + (3
5*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(9/2))

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Rubi [A]  time = 0.0324622, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {263, 51, 63, 205} \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{9/2}}+\frac{7}{4 b^2 x^{3/2} (a x+b)}+\frac{35 a}{4 b^4 \sqrt{x}}+\frac{1}{2 b x^{3/2} (a x+b)^2}-\frac{35}{12 b^3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^3*x^(11/2)),x]

[Out]

-35/(12*b^3*x^(3/2)) + (35*a)/(4*b^4*Sqrt[x]) + 1/(2*b*x^(3/2)*(b + a*x)^2) + 7/(4*b^2*x^(3/2)*(b + a*x)) + (3
5*a^(3/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(9/2))

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 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 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{1}{\left (a+\frac{b}{x}\right )^3 x^{11/2}} \, dx &=\int \frac{1}{x^{5/2} (b+a x)^3} \, dx\\ &=\frac{1}{2 b x^{3/2} (b+a x)^2}+\frac{7 \int \frac{1}{x^{5/2} (b+a x)^2} \, dx}{4 b}\\ &=\frac{1}{2 b x^{3/2} (b+a x)^2}+\frac{7}{4 b^2 x^{3/2} (b+a x)}+\frac{35 \int \frac{1}{x^{5/2} (b+a x)} \, dx}{8 b^2}\\ &=-\frac{35}{12 b^3 x^{3/2}}+\frac{1}{2 b x^{3/2} (b+a x)^2}+\frac{7}{4 b^2 x^{3/2} (b+a x)}-\frac{(35 a) \int \frac{1}{x^{3/2} (b+a x)} \, dx}{8 b^3}\\ &=-\frac{35}{12 b^3 x^{3/2}}+\frac{35 a}{4 b^4 \sqrt{x}}+\frac{1}{2 b x^{3/2} (b+a x)^2}+\frac{7}{4 b^2 x^{3/2} (b+a x)}+\frac{\left (35 a^2\right ) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 b^4}\\ &=-\frac{35}{12 b^3 x^{3/2}}+\frac{35 a}{4 b^4 \sqrt{x}}+\frac{1}{2 b x^{3/2} (b+a x)^2}+\frac{7}{4 b^2 x^{3/2} (b+a x)}+\frac{\left (35 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 b^4}\\ &=-\frac{35}{12 b^3 x^{3/2}}+\frac{35 a}{4 b^4 \sqrt{x}}+\frac{1}{2 b x^{3/2} (b+a x)^2}+\frac{7}{4 b^2 x^{3/2} (b+a x)}+\frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0054574, size = 27, normalized size = 0.28 \[ -\frac{2 \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};-\frac{a x}{b}\right )}{3 b^3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^3*x^(11/2)),x]

[Out]

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

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Maple [A]  time = 0.015, size = 79, normalized size = 0.8 \begin{align*}{\frac{11\,{a}^{3}}{4\,{b}^{4} \left ( ax+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{13\,{a}^{2}}{4\,{b}^{3} \left ( ax+b \right ) ^{2}}\sqrt{x}}+{\frac{35\,{a}^{2}}{4\,{b}^{4}}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{2}{3\,{b}^{3}}{x}^{-{\frac{3}{2}}}}+6\,{\frac{a}{{b}^{4}\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^3/x^(11/2),x)

[Out]

11/4/b^4*a^3/(a*x+b)^2*x^(3/2)+13/4/b^3*a^2/(a*x+b)^2*x^(1/2)+35/4/b^4*a^2/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^
(1/2))-2/3/b^3/x^(3/2)+6*a/b^4/x^(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(1/(a+b/x)^3/x^(11/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9296, size = 545, normalized size = 5.74 \begin{align*} \left [\frac{105 \,{\left (a^{3} x^{4} + 2 \, a^{2} b x^{3} + a b^{2} x^{2}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{a x + 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - b}{a x + b}\right ) + 2 \,{\left (105 \, a^{3} x^{3} + 175 \, a^{2} b x^{2} + 56 \, a b^{2} x - 8 \, b^{3}\right )} \sqrt{x}}{24 \,{\left (a^{2} b^{4} x^{4} + 2 \, a b^{5} x^{3} + b^{6} x^{2}\right )}}, -\frac{105 \,{\left (a^{3} x^{4} + 2 \, a^{2} b x^{3} + a b^{2} x^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (105 \, a^{3} x^{3} + 175 \, a^{2} b x^{2} + 56 \, a b^{2} x - 8 \, b^{3}\right )} \sqrt{x}}{12 \,{\left (a^{2} b^{4} x^{4} + 2 \, a b^{5} x^{3} + b^{6} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(11/2),x, algorithm="fricas")

[Out]

[1/24*(105*(a^3*x^4 + 2*a^2*b*x^3 + a*b^2*x^2)*sqrt(-a/b)*log((a*x + 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b)) +
2*(105*a^3*x^3 + 175*a^2*b*x^2 + 56*a*b^2*x - 8*b^3)*sqrt(x))/(a^2*b^4*x^4 + 2*a*b^5*x^3 + b^6*x^2), -1/12*(10
5*(a^3*x^4 + 2*a^2*b*x^3 + a*b^2*x^2)*sqrt(a/b)*arctan(b*sqrt(a/b)/(a*sqrt(x))) - (105*a^3*x^3 + 175*a^2*b*x^2
+ 56*a*b^2*x - 8*b^3)*sqrt(x))/(a^2*b^4*x^4 + 2*a*b^5*x^3 + b^6*x^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(1/(a+b/x)**3/x**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11787, size = 96, normalized size = 1.01 \begin{align*} \frac{35 \, a^{2} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{4}} + \frac{2 \,{\left (9 \, a x - b\right )}}{3 \, b^{4} x^{\frac{3}{2}}} + \frac{11 \, a^{3} x^{\frac{3}{2}} + 13 \, a^{2} b \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(11/2),x, algorithm="giac")

[Out]

35/4*a^2*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 2/3*(9*a*x - b)/(b^4*x^(3/2)) + 1/4*(11*a^3*x^(3/2) + 1
3*a^2*b*sqrt(x))/((a*x + b)^2*b^4)