3.1690 \(\int \frac{1}{(a+\frac{b}{x})^3 x^{13/2}} \, dx\)

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

[Out]

-63/(20*b^3*x^(5/2)) + (21*a)/(4*b^4*x^(3/2)) - (63*a^2)/(4*b^5*Sqrt[x]) + 1/(2*b*x^(5/2)*(b + a*x)^2) + 9/(4*
b^2*x^(5/2)*(b + a*x)) - (63*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(11/2))

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Rubi [A]  time = 0.0430688, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {263, 51, 63, 205} \[ -\frac{63 a^2}{4 b^5 \sqrt{x}}-\frac{63 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{11/2}}+\frac{21 a}{4 b^4 x^{3/2}}+\frac{9}{4 b^2 x^{5/2} (a x+b)}+\frac{1}{2 b x^{5/2} (a x+b)^2}-\frac{63}{20 b^3 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^3*x^(13/2)),x]

[Out]

-63/(20*b^3*x^(5/2)) + (21*a)/(4*b^4*x^(3/2)) - (63*a^2)/(4*b^5*Sqrt[x]) + 1/(2*b*x^(5/2)*(b + a*x)^2) + 9/(4*
b^2*x^(5/2)*(b + a*x)) - (63*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[x])/Sqrt[b]])/(4*b^(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 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^{13/2}} \, dx &=\int \frac{1}{x^{7/2} (b+a x)^3} \, dx\\ &=\frac{1}{2 b x^{5/2} (b+a x)^2}+\frac{9 \int \frac{1}{x^{7/2} (b+a x)^2} \, dx}{4 b}\\ &=\frac{1}{2 b x^{5/2} (b+a x)^2}+\frac{9}{4 b^2 x^{5/2} (b+a x)}+\frac{63 \int \frac{1}{x^{7/2} (b+a x)} \, dx}{8 b^2}\\ &=-\frac{63}{20 b^3 x^{5/2}}+\frac{1}{2 b x^{5/2} (b+a x)^2}+\frac{9}{4 b^2 x^{5/2} (b+a x)}-\frac{(63 a) \int \frac{1}{x^{5/2} (b+a x)} \, dx}{8 b^3}\\ &=-\frac{63}{20 b^3 x^{5/2}}+\frac{21 a}{4 b^4 x^{3/2}}+\frac{1}{2 b x^{5/2} (b+a x)^2}+\frac{9}{4 b^2 x^{5/2} (b+a x)}+\frac{\left (63 a^2\right ) \int \frac{1}{x^{3/2} (b+a x)} \, dx}{8 b^4}\\ &=-\frac{63}{20 b^3 x^{5/2}}+\frac{21 a}{4 b^4 x^{3/2}}-\frac{63 a^2}{4 b^5 \sqrt{x}}+\frac{1}{2 b x^{5/2} (b+a x)^2}+\frac{9}{4 b^2 x^{5/2} (b+a x)}-\frac{\left (63 a^3\right ) \int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 b^5}\\ &=-\frac{63}{20 b^3 x^{5/2}}+\frac{21 a}{4 b^4 x^{3/2}}-\frac{63 a^2}{4 b^5 \sqrt{x}}+\frac{1}{2 b x^{5/2} (b+a x)^2}+\frac{9}{4 b^2 x^{5/2} (b+a x)}-\frac{\left (63 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 b^5}\\ &=-\frac{63}{20 b^3 x^{5/2}}+\frac{21 a}{4 b^4 x^{3/2}}-\frac{63 a^2}{4 b^5 \sqrt{x}}+\frac{1}{2 b x^{5/2} (b+a x)^2}+\frac{9}{4 b^2 x^{5/2} (b+a x)}-\frac{63 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0054438, size = 27, normalized size = 0.25 \[ -\frac{2 \, _2F_1\left (-\frac{5}{2},3;-\frac{3}{2};-\frac{a x}{b}\right )}{5 b^3 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^3*x^(13/2)),x]

[Out]

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

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Maple [A]  time = 0.015, size = 90, normalized size = 0.8 \begin{align*} -{\frac{15\,{a}^{4}}{4\,{b}^{5} \left ( ax+b \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{17\,{a}^{3}}{4\,{b}^{4} \left ( ax+b \right ) ^{2}}\sqrt{x}}-{\frac{63\,{a}^{3}}{4\,{b}^{5}}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{2}{5\,{b}^{3}}{x}^{-{\frac{5}{2}}}}-12\,{\frac{{a}^{2}}{{b}^{5}\sqrt{x}}}+2\,{\frac{a}{{b}^{4}{x}^{3/2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^3/x^(13/2),x)

[Out]

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

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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^(13/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83845, size = 598, normalized size = 5.44 \begin{align*} \left [\frac{315 \,{\left (a^{4} x^{5} + 2 \, a^{3} b x^{4} + a^{2} b^{2} 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 (315 \, a^{4} x^{4} + 525 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a b^{3} x + 8 \, b^{4}\right )} \sqrt{x}}{40 \,{\left (a^{2} b^{5} x^{5} + 2 \, a b^{6} x^{4} + b^{7} x^{3}\right )}}, \frac{315 \,{\left (a^{4} x^{5} + 2 \, a^{3} b x^{4} + a^{2} b^{2} x^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a \sqrt{x}}\right ) -{\left (315 \, a^{4} x^{4} + 525 \, a^{3} b x^{3} + 168 \, a^{2} b^{2} x^{2} - 24 \, a b^{3} x + 8 \, b^{4}\right )} \sqrt{x}}{20 \,{\left (a^{2} b^{5} x^{5} + 2 \, a b^{6} x^{4} + b^{7} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(13/2),x, algorithm="fricas")

[Out]

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

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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**(13/2),x)

[Out]

Timed out

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Giac [A]  time = 1.1058, size = 108, normalized size = 0.98 \begin{align*} -\frac{63 \, a^{3} \arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{5}} - \frac{15 \, a^{4} x^{\frac{3}{2}} + 17 \, a^{3} b \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} b^{5}} - \frac{2 \,{\left (30 \, a^{2} x^{2} - 5 \, a b x + b^{2}\right )}}{5 \, b^{5} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(13/2),x, algorithm="giac")

[Out]

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