3.1681 \(\int \frac{1}{(a+\frac{b}{x})^2 x^{11/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{7 a^2}{b^4 \sqrt{x}}-\frac{7 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{9/2}}+\frac{7 a}{3 b^3 x^{3/2}}+\frac{1}{b x^{5/2} (a x+b)}-\frac{7}{5 b^2 x^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.032364, antiderivative size = 84, 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} \[ -\frac{7 a^2}{b^4 \sqrt{x}}-\frac{7 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{b^{9/2}}+\frac{7 a}{3 b^3 x^{3/2}}+\frac{1}{b x^{5/2} (a x+b)}-\frac{7}{5 b^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^2*x^(11/2)),x]

[Out]

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

Mathematica [C]  time = 0.0052615, size = 27, normalized size = 0.32 \[ -\frac{2 \, _2F_1\left (-\frac{5}{2},2;-\frac{3}{2};-\frac{a x}{b}\right )}{5 b^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^2*x^(11/2)),x]

[Out]

(-2*Hypergeometric2F1[-5/2, 2, -3/2, -((a*x)/b)])/(5*b^2*x^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.014, size = 72, normalized size = 0.9 \begin{align*} -{\frac{{a}^{3}}{{b}^{4} \left ( ax+b \right ) }\sqrt{x}}-7\,{\frac{{a}^{3}}{{b}^{4}\sqrt{ab}}\arctan \left ({\frac{a\sqrt{x}}{\sqrt{ab}}} \right ) }-{\frac{2}{5\,{b}^{2}}{x}^{-{\frac{5}{2}}}}-6\,{\frac{{a}^{2}}{{b}^{4}\sqrt{x}}}+{\frac{4\,a}{3\,{b}^{3}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3/b^4*x^(1/2)/(a*x+b)-7*a^3/b^4/(a*b)^(1/2)*arctan(a*x^(1/2)/(a*b)^(1/2))-2/5/b^2/x^(5/2)-6*a^2/b^4/x^(1/2)
+4/3*a/b^3/x^(3/2)

________________________________________________________________________________________

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(1/(a+b/x)^2/x^(11/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.83907, size = 460, normalized size = 5.48 \begin{align*} \left [\frac{105 \,{\left (a^{3} x^{4} + a^{2} b x^{3}\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} + 70 \, a^{2} b x^{2} - 14 \, a b^{2} x + 6 \, b^{3}\right )} \sqrt{x}}{30 \,{\left (a b^{4} x^{4} + b^{5} x^{3}\right )}}, \frac{105 \,{\left (a^{3} x^{4} + a^{2} b x^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (105 \, a^{3} x^{3} + 70 \, a^{2} b x^{2} - 14 \, a b^{2} x + 6 \, b^{3}\right )} \sqrt{x}}{15 \,{\left (a b^{4} x^{4} + b^{5} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(105*(a^3*x^4 + a^2*b*x^3)*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b)) - 2*(105*a^3*x^3
+ 70*a^2*b*x^2 - 14*a*b^2*x + 6*b^3)*sqrt(x))/(a*b^4*x^4 + b^5*x^3), 1/15*(105*(a^3*x^4 + a^2*b*x^3)*sqrt(a/b
)*arctan(b*sqrt(a/b)/(a*sqrt(x))) - (105*a^3*x^3 + 70*a^2*b*x^2 - 14*a*b^2*x + 6*b^3)*sqrt(x))/(a*b^4*x^4 + b^
5*x^3)]

________________________________________________________________________________________

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)**2/x**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.09854, size = 95, normalized size = 1.13 \begin{align*} -\frac{7 \, a^{3} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{4}} - \frac{a^{3} \sqrt{x}}{{\left (a x + b\right )} b^{4}} - \frac{2 \,{\left (45 \, a^{2} x^{2} - 10 \, a b x + 3 \, b^{2}\right )}}{15 \, b^{4} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*a^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - a^3*sqrt(x)/((a*x + b)*b^4) - 2/15*(45*a^2*x^2 - 10*a*b*x
+ 3*b^2)/(b^4*x^(5/2))