3.262 \(\int \frac{1}{(a+\frac{b}{x})^{5/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac{5 b}{a^3 \sqrt{a+\frac{b}{x}}}+\frac{5 b}{3 a^2 \left (a+\frac{b}{x}\right )^{3/2}}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{x}{a \left (a+\frac{b}{x}\right )^{3/2}} \]

[Out]

(5*b)/(3*a^2*(a + b/x)^(3/2)) + (5*b)/(a^3*Sqrt[a + b/x]) + x/(a*(a + b/x)^(3/2)) - (5*b*ArcTanh[Sqrt[a + b/x]
/Sqrt[a]])/a^(7/2)

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Rubi [A]  time = 0.0377736, antiderivative size = 82, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {242, 51, 63, 208} \[ \frac{5 x \sqrt{a+\frac{b}{x}}}{a^3}-\frac{10 x}{3 a^2 \sqrt{a+\frac{b}{x}}}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}-\frac{2 x}{3 a \left (a+\frac{b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(-5/2),x]

[Out]

(-2*x)/(3*a*(a + b/x)^(3/2)) - (10*x)/(3*a^2*Sqrt[a + b/x]) + (5*Sqrt[a + b/x]*x)/a^3 - (5*b*ArcTanh[Sqrt[a +
b/x]/Sqrt[a]])/a^(7/2)

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

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 208

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x}\right )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{5/2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 x}{3 a \left (a+\frac{b}{x}\right )^{3/2}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac{1}{x}\right )}{3 a}\\ &=-\frac{2 x}{3 a \left (a+\frac{b}{x}\right )^{3/2}}-\frac{10 x}{3 a^2 \sqrt{a+\frac{b}{x}}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{a^2}\\ &=-\frac{2 x}{3 a \left (a+\frac{b}{x}\right )^{3/2}}-\frac{10 x}{3 a^2 \sqrt{a+\frac{b}{x}}}+\frac{5 \sqrt{a+\frac{b}{x}} x}{a^3}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 a^3}\\ &=-\frac{2 x}{3 a \left (a+\frac{b}{x}\right )^{3/2}}-\frac{10 x}{3 a^2 \sqrt{a+\frac{b}{x}}}+\frac{5 \sqrt{a+\frac{b}{x}} x}{a^3}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{a^3}\\ &=-\frac{2 x}{3 a \left (a+\frac{b}{x}\right )^{3/2}}-\frac{10 x}{3 a^2 \sqrt{a+\frac{b}{x}}}+\frac{5 \sqrt{a+\frac{b}{x}} x}{a^3}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0157974, size = 38, normalized size = 0.48 \[ \frac{2 b \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{a+\frac{b}{x}}{a}\right )}{3 a^2 \left (a+\frac{b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(-5/2),x]

[Out]

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

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Maple [B]  time = 0.008, size = 271, normalized size = 3.4 \begin{align*} -{\frac{x}{6\, \left ( ax+b \right ) ^{3}}\sqrt{{\frac{ax+b}{x}}} \left ( 15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{3}{a}^{3}b-30\,{a}^{7/2}\sqrt{ \left ( ax+b \right ) x}{x}^{3}+45\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{2}{a}^{2}{b}^{2}+24\,{a}^{5/2} \left ( \left ( ax+b \right ) x \right ) ^{3/2}x-90\,{a}^{5/2}\sqrt{ \left ( ax+b \right ) x}{x}^{2}b+45\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) xa{b}^{3}+20\,b{a}^{3/2} \left ( \left ( ax+b \right ) x \right ) ^{3/2}-90\,{a}^{3/2}\sqrt{ \left ( ax+b \right ) x}x{b}^{2}+15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){b}^{4}-30\,\sqrt{a}\sqrt{ \left ( ax+b \right ) x}{b}^{3} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(5/2),x)

[Out]

-1/6*((a*x+b)/x)^(1/2)*x/a^(7/2)*(15*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^3*a^3*b-30*a^(7/2
)*((a*x+b)*x)^(1/2)*x^3+45*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^2*b^2+24*a^(5/2)*((a*x+
b)*x)^(3/2)*x-90*a^(5/2)*((a*x+b)*x)^(1/2)*x^2*b+45*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a*
b^3+20*b*a^(3/2)*((a*x+b)*x)^(3/2)-90*a^(3/2)*((a*x+b)*x)^(1/2)*x*b^2+15*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2
*a*x+b)/a^(1/2))*b^4-30*a^(1/2)*((a*x+b)*x)^(1/2)*b^3)/((a*x+b)*x)^(1/2)/(a*x+b)^3

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27383, size = 501, normalized size = 6.34 \begin{align*} \left [\frac{15 \,{\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt{a} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (3 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt{\frac{a x + b}{x}}}{6 \,{\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}, \frac{15 \,{\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (3 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt{\frac{a x + b}{x}}}{3 \,{\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(15*(a^2*b*x^2 + 2*a*b^2*x + b^3)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(3*a^3*x^3 +
20*a^2*b*x^2 + 15*a*b^2*x)*sqrt((a*x + b)/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2), 1/3*(15*(a^2*b*x^2 + 2*a*b^2*x
+ b^3)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3*a^3*x^3 + 20*a^2*b*x^2 + 15*a*b^2*x)*sqrt((a*x + b)
/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2)]

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Sympy [B]  time = 4.93673, size = 774, normalized size = 9.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(5/2),x)

[Out]

6*a**17*x**4*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**
3) + 46*a**16*b*x**3*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(3
3/2)*b**3) + 15*a**16*b*x**3*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a*
*(33/2)*b**3) - 30*a**16*b*x**3*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35
/2)*b**2*x + 6*a**(33/2)*b**3) + 70*a**15*b**2*x**2*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2
+ 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**15*b**2*x**2*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b
*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**15*b**2*x**2*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x
**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 30*a**14*b**3*x*sqrt(1 + b/(a*x))/(6*a**
(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**14*b**3*x*log(b/(a*x))/(6*
a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**14*b**3*x*log(sqrt(1 +
b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 15*a**13*b**
4*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 30*a**13*b*
*4*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3
)

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Giac [A]  time = 1.22499, size = 132, normalized size = 1.67 \begin{align*} \frac{1}{3} \, b{\left (\frac{2 \,{\left (a + \frac{6 \,{\left (a x + b\right )}}{x}\right )} x}{{\left (a x + b\right )} a^{3} \sqrt{\frac{a x + b}{x}}} + \frac{15 \, \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} - \frac{3 \, \sqrt{\frac{a x + b}{x}}}{{\left (a - \frac{a x + b}{x}\right )} a^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(5/2),x, algorithm="giac")

[Out]

1/3*b*(2*(a + 6*(a*x + b)/x)*x/((a*x + b)*a^3*sqrt((a*x + b)/x)) + 15*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt
(-a)*a^3) - 3*sqrt((a*x + b)/x)/((a - (a*x + b)/x)*a^3))