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

Optimal. Leaf size=71 \[ 5 a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+x \left (a+\frac{b}{x}\right )^{5/2}-\frac{5}{3} b \left (a+\frac{b}{x}\right )^{3/2}-5 a b \sqrt{a+\frac{b}{x}} \]

[Out]

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

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Rubi [A]  time = 0.0343127, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {242, 47, 50, 63, 208} \[ 5 a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+x \left (a+\frac{b}{x}\right )^{5/2}-\frac{5}{3} b \left (a+\frac{b}{x}\right )^{3/2}-5 a b \sqrt{a+\frac{b}{x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0442537, size = 64, normalized size = 0.9 \[ \frac{\sqrt{a+\frac{b}{x}} \left (3 a^2 x^2-14 a b x-2 b^2\right )}{3 x}+5 a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2),x]

[Out]

(Sqrt[a + b/x]*(-2*b^2 - 14*a*b*x + 3*a^2*x^2))/(3*x) + 5*a^(3/2)*b*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

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Maple [B]  time = 0.007, size = 120, normalized size = 1.7 \begin{align*} -{\frac{1}{6\,{x}^{2}}\sqrt{{\frac{ax+b}{x}}} \left ( -30\,\sqrt{a{x}^{2}+bx}{a}^{5/2}{x}^{3}-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{3}{a}^{2}b+24\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}x+4\,b \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a} \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((a*x+b)/x)^(1/2)/x^2*(-30*(a*x^2+b*x)^(1/2)*a^(5/2)*x^3-15*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/
a^(1/2))*x^3*a^2*b+24*(a*x^2+b*x)^(3/2)*a^(3/2)*x+4*b*(a*x^2+b*x)^(3/2)*a^(1/2))/((a*x+b)*x)^(1/2)/a^(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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.42284, size = 329, normalized size = 4.63 \begin{align*} \left [\frac{15 \, a^{\frac{3}{2}} b x \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (3 \, a^{2} x^{2} - 14 \, a b x - 2 \, b^{2}\right )} \sqrt{\frac{a x + b}{x}}}{6 \, x}, -\frac{15 \, \sqrt{-a} a b x \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (3 \, a^{2} x^{2} - 14 \, a b x - 2 \, b^{2}\right )} \sqrt{\frac{a x + b}{x}}}{3 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a^(3/2)*b*x*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(3*a^2*x^2 - 14*a*b*x - 2*b^2)*sqrt((a
*x + b)/x))/x, -1/3*(15*sqrt(-a)*a*b*x*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (3*a^2*x^2 - 14*a*b*x - 2*b^2)*s
qrt((a*x + b)/x))/x]

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Sympy [A]  time = 3.37097, size = 99, normalized size = 1.39 \begin{align*} a^{\frac{5}{2}} x \sqrt{1 + \frac{b}{a x}} - \frac{14 a^{\frac{3}{2}} b \sqrt{1 + \frac{b}{a x}}}{3} - \frac{5 a^{\frac{3}{2}} b \log{\left (\frac{b}{a x} \right )}}{2} + 5 a^{\frac{3}{2}} b \log{\left (\sqrt{1 + \frac{b}{a x}} + 1 \right )} - \frac{2 \sqrt{a} b^{2} \sqrt{1 + \frac{b}{a x}}}{3 x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/2)*x*sqrt(1 + b/(a*x)) - 14*a**(3/2)*b*sqrt(1 + b/(a*x))/3 - 5*a**(3/2)*b*log(b/(a*x))/2 + 5*a**(3/2)*b*
log(sqrt(1 + b/(a*x)) + 1) - 2*sqrt(a)*b**2*sqrt(1 + b/(a*x))/(3*x)

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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, algorithm="giac")

[Out]

Exception raised: TypeError