3.1714 \(\int (a+\frac{b}{x})^{5/2} x \, dx\)

Optimal. Leaf size=84 \[ -\frac{15}{4} b^2 \sqrt{a+\frac{b}{x}}+\frac{15}{4} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+\frac{1}{2} x^2 \left (a+\frac{b}{x}\right )^{5/2}+\frac{5}{4} b x \left (a+\frac{b}{x}\right )^{3/2} \]

[Out]

(-15*b^2*Sqrt[a + b/x])/4 + (5*b*(a + b/x)^(3/2)*x)/4 + ((a + b/x)^(5/2)*x^2)/2 + (15*Sqrt[a]*b^2*ArcTanh[Sqrt
[a + b/x]/Sqrt[a]])/4

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Rubi [A]  time = 0.034874, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {266, 47, 50, 63, 208} \[ -\frac{15}{4} b^2 \sqrt{a+\frac{b}{x}}+\frac{15}{4} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+\frac{1}{2} x^2 \left (a+\frac{b}{x}\right )^{5/2}+\frac{5}{4} b x \left (a+\frac{b}{x}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*x,x]

[Out]

(-15*b^2*Sqrt[a + b/x])/4 + (5*b*(a + b/x)^(3/2)*x)/4 + ((a + b/x)^(5/2)*x^2)/2 + (15*Sqrt[a]*b^2*ArcTanh[Sqrt
[a + b/x]/Sqrt[a]])/4

Rule 266

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \left (a+\frac{b}{x}\right )^{5/2} x \, dx &=-\operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} \left (a+\frac{b}{x}\right )^{5/2} x^2-\frac{1}{4} (5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{5}{4} b \left (a+\frac{b}{x}\right )^{3/2} x+\frac{1}{2} \left (a+\frac{b}{x}\right )^{5/2} x^2-\frac{1}{8} \left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{15}{4} b^2 \sqrt{a+\frac{b}{x}}+\frac{5}{4} b \left (a+\frac{b}{x}\right )^{3/2} x+\frac{1}{2} \left (a+\frac{b}{x}\right )^{5/2} x^2-\frac{1}{8} \left (15 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{15}{4} b^2 \sqrt{a+\frac{b}{x}}+\frac{5}{4} b \left (a+\frac{b}{x}\right )^{3/2} x+\frac{1}{2} \left (a+\frac{b}{x}\right )^{5/2} x^2-\frac{1}{4} (15 a b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )\\ &=-\frac{15}{4} b^2 \sqrt{a+\frac{b}{x}}+\frac{5}{4} b \left (a+\frac{b}{x}\right )^{3/2} x+\frac{1}{2} \left (a+\frac{b}{x}\right )^{5/2} x^2+\frac{15}{4} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0151367, size = 39, normalized size = 0.46 \[ \frac{2 b^2 \left (a+\frac{b}{x}\right )^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{b}{a x}+1\right )}{7 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)*x,x]

[Out]

(2*b^2*(a + b/x)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, 1 + b/(a*x)])/(7*a^3)

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Maple [A]  time = 0.009, size = 125, normalized size = 1.5 \begin{align*} -{\frac{1}{8\,x}\sqrt{{\frac{ax+b}{x}}} \left ( -4\,\sqrt{a{x}^{2}+bx}{a}^{7/2}{x}^{3}-34\,\sqrt{a{x}^{2}+bx}{a}^{5/2}{x}^{2}b-15\,{a}^{2}\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{2}{b}^{2}+16\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}b \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((a*x+b)/x)^(1/2)/x*(-4*(a*x^2+b*x)^(1/2)*a^(7/2)*x^3-34*(a*x^2+b*x)^(1/2)*a^(5/2)*x^2*b-15*a^2*ln(1/2*(2
*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*b^2+16*(a*x^2+b*x)^(3/2)*a^(3/2)*b)/((a*x+b)*x)^(1/2)/a^(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((a+b/x)^(5/2)*x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85353, size = 316, normalized size = 3.76 \begin{align*} \left [\frac{15}{8} \, \sqrt{a} b^{2} \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + \frac{1}{4} \,{\left (2 \, a^{2} x^{2} + 9 \, a b x - 8 \, b^{2}\right )} \sqrt{\frac{a x + b}{x}}, -\frac{15}{4} \, \sqrt{-a} b^{2} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) + \frac{1}{4} \,{\left (2 \, a^{2} x^{2} + 9 \, a b x - 8 \, b^{2}\right )} \sqrt{\frac{a x + b}{x}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[15/8*sqrt(a)*b^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 1/4*(2*a^2*x^2 + 9*a*b*x - 8*b^2)*sqrt((a*x
+ b)/x), -15/4*sqrt(-a)*b^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + 1/4*(2*a^2*x^2 + 9*a*b*x - 8*b^2)*sqrt((a*
x + b)/x)]

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Sympy [A]  time = 5.75697, size = 126, normalized size = 1.5 \begin{align*} \frac{15 \sqrt{a} b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )}}{4} + \frac{a^{3} x^{\frac{5}{2}}}{2 \sqrt{b} \sqrt{\frac{a x}{b} + 1}} + \frac{11 a^{2} \sqrt{b} x^{\frac{3}{2}}}{4 \sqrt{\frac{a x}{b} + 1}} + \frac{a b^{\frac{3}{2}} \sqrt{x}}{4 \sqrt{\frac{a x}{b} + 1}} - \frac{2 b^{\frac{5}{2}}}{\sqrt{x} \sqrt{\frac{a x}{b} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(a)*b**2*asinh(sqrt(a)*sqrt(x)/sqrt(b))/4 + a**3*x**(5/2)/(2*sqrt(b)*sqrt(a*x/b + 1)) + 11*a**2*sqrt(b)
*x**(3/2)/(4*sqrt(a*x/b + 1)) + a*b**(3/2)*sqrt(x)/(4*sqrt(a*x/b + 1)) - 2*b**(5/2)/(sqrt(x)*sqrt(a*x/b + 1))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError