3.1775 \(\int \frac{(a+\frac{b}{x})^{5/2}}{x^{3/2}} \, dx\)

Optimal. Leaf size=100 \[ -\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{8 \sqrt{x}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{8 \sqrt{b}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{12 \sqrt{x}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{3 \sqrt{x}} \]

[Out]

(-5*a^2*Sqrt[a + b/x])/(8*Sqrt[x]) - (5*a*(a + b/x)^(3/2))/(12*Sqrt[x]) - (a + b/x)^(5/2)/(3*Sqrt[x]) - (5*a^3
*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(8*Sqrt[b])

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Rubi [A]  time = 0.045123, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {337, 195, 217, 206} \[ -\frac{5 a^2 \sqrt{a+\frac{b}{x}}}{8 \sqrt{x}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{x} \sqrt{a+\frac{b}{x}}}\right )}{8 \sqrt{b}}-\frac{5 a \left (a+\frac{b}{x}\right )^{3/2}}{12 \sqrt{x}}-\frac{\left (a+\frac{b}{x}\right )^{5/2}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/x^(3/2),x]

[Out]

(-5*a^2*Sqrt[a + b/x])/(8*Sqrt[x]) - (5*a*(a + b/x)^(3/2))/(12*Sqrt[x]) - (a + b/x)^(5/2)/(3*Sqrt[x]) - (5*a^3
*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/(8*Sqrt[b])

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.159139, size = 85, normalized size = 0.85 \[ \frac{1}{24} \sqrt{a+\frac{b}{x}} \left (-\frac{33 a^2 x^2+26 a b x+8 b^2}{x^{5/2}}-\frac{15 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}}\right )}{\sqrt{b} \sqrt{\frac{b}{a x}+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)/x^(3/2),x]

[Out]

(Sqrt[a + b/x]*(-((8*b^2 + 26*a*b*x + 33*a^2*x^2)/x^(5/2)) - (15*a^(5/2)*ArcSinh[Sqrt[b]/(Sqrt[a]*Sqrt[x])])/(
Sqrt[b]*Sqrt[1 + b/(a*x)])))/24

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Maple [A]  time = 0.015, size = 92, normalized size = 0.9 \begin{align*} -{\frac{1}{24}\sqrt{{\frac{ax+b}{x}}} \left ( 15\,{\it Artanh} \left ({\frac{\sqrt{ax+b}}{\sqrt{b}}} \right ){a}^{3}{x}^{3}+33\,{x}^{2}{a}^{2}\sqrt{b}\sqrt{ax+b}+26\,xa{b}^{3/2}\sqrt{ax+b}+8\,{b}^{5/2}\sqrt{ax+b} \right ){x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{ax+b}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)/x^(3/2),x)

[Out]

-1/24*((a*x+b)/x)^(1/2)/x^(5/2)*(15*arctanh((a*x+b)^(1/2)/b^(1/2))*a^3*x^3+33*x^2*a^2*b^(1/2)*(a*x+b)^(1/2)+26
*x*a*b^(3/2)*(a*x+b)^(1/2)+8*b^(5/2)*(a*x+b)^(1/2))/(a*x+b)^(1/2)/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^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55714, size = 414, normalized size = 4.14 \begin{align*} \left [\frac{15 \, a^{3} \sqrt{b} x^{3} \log \left (\frac{a x - 2 \, \sqrt{b} \sqrt{x} \sqrt{\frac{a x + b}{x}} + 2 \, b}{x}\right ) - 2 \,{\left (33 \, a^{2} b x^{2} + 26 \, a b^{2} x + 8 \, b^{3}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{48 \, b x^{3}}, \frac{15 \, a^{3} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{-b} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{b}\right ) -{\left (33 \, a^{2} b x^{2} + 26 \, a b^{2} x + 8 \, b^{3}\right )} \sqrt{x} \sqrt{\frac{a x + b}{x}}}{24 \, b x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/x^(3/2),x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*x^3*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) - 2*(33*a^2*b*x^2 + 26*a*b^
2*x + 8*b^3)*sqrt(x)*sqrt((a*x + b)/x))/(b*x^3), 1/24*(15*a^3*sqrt(-b)*x^3*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x +
b)/x)/b) - (33*a^2*b*x^2 + 26*a*b^2*x + 8*b^3)*sqrt(x)*sqrt((a*x + b)/x))/(b*x^3)]

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Sympy [A]  time = 70.646, size = 104, normalized size = 1.04 \begin{align*} - \frac{11 a^{\frac{5}{2}} \sqrt{1 + \frac{b}{a x}}}{8 \sqrt{x}} - \frac{13 a^{\frac{3}{2}} b \sqrt{1 + \frac{b}{a x}}}{12 x^{\frac{3}{2}}} - \frac{\sqrt{a} b^{2} \sqrt{1 + \frac{b}{a x}}}{3 x^{\frac{5}{2}}} - \frac{5 a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} \sqrt{x}} \right )}}{8 \sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)/x**(3/2),x)

[Out]

-11*a**(5/2)*sqrt(1 + b/(a*x))/(8*sqrt(x)) - 13*a**(3/2)*b*sqrt(1 + b/(a*x))/(12*x**(3/2)) - sqrt(a)*b**2*sqrt
(1 + b/(a*x))/(3*x**(5/2)) - 5*a**3*asinh(sqrt(b)/(sqrt(a)*sqrt(x)))/(8*sqrt(b))

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Giac [A]  time = 1.31902, size = 90, normalized size = 0.9 \begin{align*} \frac{1}{24} \, a^{3}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{a x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{33 \,{\left (a x + b\right )}^{\frac{5}{2}} - 40 \,{\left (a x + b\right )}^{\frac{3}{2}} b + 15 \, \sqrt{a x + b} b^{2}}{a^{3} x^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/x^(3/2),x, algorithm="giac")

[Out]

1/24*a^3*(15*arctan(sqrt(a*x + b)/sqrt(-b))/sqrt(-b) - (33*(a*x + b)^(5/2) - 40*(a*x + b)^(3/2)*b + 15*sqrt(a*
x + b)*b^2)/(a^3*x^3))