3.1776 \(\int \frac{(a+\frac{b}{x})^{5/2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac{5 a^4 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{64 b^{3/2}}-\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{32 x^{3/2}}-\frac{5 a^3 \sqrt{a+\frac{b}{x}}}{64 b \sqrt{x}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}} \]

[Out]

(-5*a^2*Sqrt[a + b/x])/(32*x^(3/2)) - (5*a*(a + b/x)^(3/2))/(24*x^(3/2)) - (a + b/x)^(5/2)/(4*x^(3/2)) - (5*a^
3*Sqrt[a + b/x])/(64*b*Sqrt[x]) + (5*a^4*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(64*b^(3/2))

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Rubi [A]  time = 0.0682302, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {337, 279, 321, 217, 206} \[ \frac{5 a^4 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{64 b^{3/2}}-\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{32 x^{3/2}}-\frac{5 a^3 \sqrt{a+\frac{b}{x}}}{64 b \sqrt{x}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/x^(5/2),x]

[Out]

(-5*a^2*Sqrt[a + b/x])/(32*x^(3/2)) - (5*a*(a + b/x)^(3/2))/(24*x^(3/2)) - (a + b/x)^(5/2)/(4*x^(3/2)) - (5*a^
3*Sqrt[a + b/x])/(64*b*Sqrt[x]) + (5*a^4*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(64*b^(3/2))

Rule 337

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, -Dist[k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{x}\right )^{5/2}}{x^{5/2}} \, dx &=-\left (2 \operatorname{Subst}\left (\int x^2 \left (a+b x^2\right )^{5/2} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac{1}{4} (5 a) \operatorname{Subst}\left (\int x^2 \left (a+b x^2\right )^{3/2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac{1}{8} \left (5 a^2\right ) \operatorname{Subst}\left (\int x^2 \sqrt{a+b x^2} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{32 x^{3/2}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac{1}{32} \left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )\\ &=-\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{32 x^{3/2}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac{5 a^3 \sqrt{a+\frac{b}{x}}}{64 b \sqrt{x}}+\frac{\left (5 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{64 b}\\ &=-\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{32 x^{3/2}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac{5 a^3 \sqrt{a+\frac{b}{x}}}{64 b \sqrt{x}}+\frac{\left (5 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )}{64 b}\\ &=-\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{32 x^{3/2}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{24 x^{3/2}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{4 x^{3/2}}-\frac{5 a^3 \sqrt{a+\frac{b}{x}}}{64 b \sqrt{x}}+\frac{5 a^4 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x}} \sqrt{x}}\right )}{64 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.146404, size = 101, normalized size = 0.8 \[ \frac{\sqrt{a+\frac{b}{x}} \left (\frac{15 a^{7/2} \sinh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}}\right )}{\sqrt{\frac{b}{a x}+1}}-\frac{\sqrt{b} \left (118 a^2 b x^2+15 a^3 x^3+136 a b^2 x+48 b^3\right )}{x^{7/2}}\right )}{192 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)/x^(5/2),x]

[Out]

(Sqrt[a + b/x]*(-((Sqrt[b]*(48*b^3 + 136*a*b^2*x + 118*a^2*b*x^2 + 15*a^3*x^3))/x^(7/2)) + (15*a^(7/2)*ArcSinh
[Sqrt[b]/(Sqrt[a]*Sqrt[x])])/Sqrt[1 + b/(a*x)]))/(192*b^(3/2))

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Maple [A]  time = 0.017, size = 110, normalized size = 0.9 \begin{align*} -{\frac{1}{192}\sqrt{{\frac{ax+b}{x}}} \left ( -15\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ){a}^{4}{x}^{4}+48\,{b}^{7/2}\sqrt{ax+b}+136\,xa{b}^{5/2}\sqrt{ax+b}+118\,{x}^{2}{a}^{2}{b}^{3/2}\sqrt{ax+b}+15\,{x}^{3}{a}^{3}\sqrt{ax+b}\sqrt{b} \right ){x}^{-{\frac{7}{2}}}{b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ax+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*((a*x+b)/x)^(1/2)*(-15*arctanh((a*x+b)^(1/2)/b^(1/2))*a^4*x^4+48*b^(7/2)*(a*x+b)^(1/2)+136*x*a*b^(5/2)*
(a*x+b)^(1/2)+118*x^2*a^2*b^(3/2)*(a*x+b)^(1/2)+15*x^3*a^3*(a*x+b)^(1/2)*b^(1/2))/x^(7/2)/b^(3/2)/(a*x+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((a+b/x)^(5/2)/x^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53178, size = 478, normalized size = 3.79 \begin{align*} \left [\frac{15 \, a^{4} \sqrt{b} x^{4} \log \left (\frac{a x + 2 \, \sqrt{b} \sqrt{x} \sqrt{\frac{a x + b}{x}} + 2 \, b}{x}\right ) - 2 \,{\left (15 \, a^{3} b x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a b^{3} x + 48 \, b^{4}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{384 \, b^{2} x^{4}}, -\frac{15 \, a^{4} \sqrt{-b} x^{4} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) +{\left (15 \, a^{3} b x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a b^{3} x + 48 \, b^{4}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{192 \, b^{2} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*a^4*sqrt(b)*x^4*log((a*x + 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) - 2*(15*a^3*b*x^3 + 118*a^
2*b^2*x^2 + 136*a*b^3*x + 48*b^4)*sqrt(x)*sqrt((a*x + b)/x))/(b^2*x^4), -1/192*(15*a^4*sqrt(-b)*x^4*arctan(sqr
t(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (15*a^3*b*x^3 + 118*a^2*b^2*x^2 + 136*a*b^3*x + 48*b^4)*sqrt(x)*sqrt((a*x
+ b)/x))/(b^2*x^4)]

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Sympy [A]  time = 74.2673, size = 155, normalized size = 1.23 \begin{align*} - \frac{5 a^{\frac{7}{2}}}{64 b \sqrt{x} \sqrt{1 + \frac{b}{a x}}} - \frac{133 a^{\frac{5}{2}}}{192 x^{\frac{3}{2}} \sqrt{1 + \frac{b}{a x}}} - \frac{127 a^{\frac{3}{2}} b}{96 x^{\frac{5}{2}} \sqrt{1 + \frac{b}{a x}}} - \frac{23 \sqrt{a} b^{2}}{24 x^{\frac{7}{2}} \sqrt{1 + \frac{b}{a x}}} + \frac{5 a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}} \right )}}{64 b^{\frac{3}{2}}} - \frac{b^{3}}{4 \sqrt{a} x^{\frac{9}{2}} \sqrt{1 + \frac{b}{a x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*a**(7/2)/(64*b*sqrt(x)*sqrt(1 + b/(a*x))) - 133*a**(5/2)/(192*x**(3/2)*sqrt(1 + b/(a*x))) - 127*a**(3/2)*b/
(96*x**(5/2)*sqrt(1 + b/(a*x))) - 23*sqrt(a)*b**2/(24*x**(7/2)*sqrt(1 + b/(a*x))) + 5*a**4*asinh(sqrt(b)/(sqrt
(a)*sqrt(x)))/(64*b**(3/2)) - b**3/(4*sqrt(a)*x**(9/2)*sqrt(1 + b/(a*x)))

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Giac [A]  time = 1.354, size = 113, normalized size = 0.9 \begin{align*} -\frac{1}{192} \, a^{4}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{15 \,{\left (a x + b\right )}^{\frac{7}{2}} + 73 \,{\left (a x + b\right )}^{\frac{5}{2}} b - 55 \,{\left (a x + b\right )}^{\frac{3}{2}} b^{2} + 15 \, \sqrt{a x + b} b^{3}}{a^{4} b x^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*a^4*(15*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b) + (15*(a*x + b)^(7/2) + 73*(a*x + b)^(5/2)*b - 55*(
a*x + b)^(3/2)*b^2 + 15*sqrt(a*x + b)*b^3)/(a^4*b*x^4))