3.3.41 \(\int (a+\frac {b}{x})^{5/2} \, dx\) [241]

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

[Out]

-5/3*b*(a+b/x)^(3/2)+(a+b/x)^(5/2)*x+5*a^(3/2)*b*arctanh((a+b/x)^(1/2)/a^(1/2))-5*a*b*(a+b/x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 = {248, 43, 52, 65, 214} \begin {gather*} 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}} \end {gather*}

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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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 65

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 214

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

Rule 248

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]

Rubi steps

\begin {align*} \int \left (a+\frac {b}{x}\right )^{5/2} \, dx &=-\text {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) \text {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) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.04, size = 64, normalized size = 0.90 \begin {gather*} \frac {\sqrt {a+\frac {b}{x}} \left (-2 b^2-14 a b x+3 a^2 x^2\right )}{3 x}+5 a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(119\) vs. \(2(57)=114\).
time = 0.00, size = 120, normalized size = 1.69

method	result	size
risch	\(\frac {\left (3 a^{2} x^{2}-14 a b x -2 b^{2}\right ) \sqrt {\frac {a x +b}{x}}}{3 x}+\frac {5 a^{\frac {3}{2}} b \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{2 \left (a x +b \right )}\)	\(94\)
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-30 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} x^{3}-15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b \,x^{3}+24 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} x +4 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b \sqrt {a}\right )}{6 x^{2} \sqrt {x \left (a x +b \right )}\, \sqrt {a}}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+1/x*b)^(5/2),x,method=_RETURNVERBOSE)

[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))*a^2*b*x^3+24*(a*x^2+b*x)^(3/2)*a^(3/2)*x+4*(a*x^2+b*x)^(3/2)*b*a^(1/2))/(x*(a*x+b))^(1/2)/a^(1/2)

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Maxima [A]
time = 0.49, size = 78, normalized size = 1.10 \begin {gather*} \sqrt {a + \frac {b}{x}} a^{2} x - \frac {5}{2} \, a^{\frac {3}{2}} b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - \frac {2}{3} \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b - 4 \, \sqrt {a + \frac {b}{x}} a b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a + b/x)*a^2*x - 5/2*a^(3/2)*b*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) - 2/3*(a + b/x)^(
3/2)*b - 4*sqrt(a + b/x)*a*b

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Fricas [A]
time = 3.22, size = 139, normalized size = 1.96 \begin {gather*} \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 {gather*}

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 = 2.14, size = 99, normalized size = 1.39 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 1.63, size = 34, normalized size = 0.48 \begin {gather*} -\frac {2\,x\,{\left (a+\frac {b}{x}\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {3}{2};\ -\frac {1}{2};\ -\frac {a\,x}{b}\right )}{3\,{\left (\frac {a\,x}{b}+1\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(5/2),x)

[Out]

-(2*x*(a + b/x)^(5/2)*hypergeom([-5/2, -3/2], -1/2, -(a*x)/b))/(3*((a*x)/b + 1)^(5/2))

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