\(\int (a+\frac {b}{x})^{3/2} x \, dx\) [1703]

Optimal result

Integrand size = 13, antiderivative size = 66 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=\frac {3}{4} b \sqrt {a+\frac {b}{x}} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{3/2} x^2+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {272, 43, 65, 214} \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{3/2}+\frac {3}{4} b x \sqrt {a+\frac {b}{x}} \]

[In]

Int[(a + b/x)^(3/2)*x,x]

[Out]

(3*b*Sqrt[a + b/x]*x)/4 + ((a + b/x)^(3/2)*x^2)/2 + (3*b^2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(4*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 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 272

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]]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^3} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{2} \left (a+\frac {b}{x}\right )^{3/2} x^2-\frac {1}{4} (3 b) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {3}{4} b \sqrt {a+\frac {b}{x}} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{3/2} x^2-\frac {1}{8} \left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {3}{4} b \sqrt {a+\frac {b}{x}} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{3/2} x^2-\frac {1}{4} (3 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right ) \\ & = \frac {3}{4} b \sqrt {a+\frac {b}{x}} x+\frac {1}{2} \left (a+\frac {b}{x}\right )^{3/2} x^2+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 \sqrt {a}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=\frac {1}{4} \left (\sqrt {a+\frac {b}{x}} x (5 b+2 a x)+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \]

[In]

Integrate[(a + b/x)^(3/2)*x,x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26

method	result	size
risch	\(\frac {\left (2 a x +5 b \right ) x \sqrt {\frac {a x +b}{x}}}{4}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{8 \sqrt {a}\, \left (a x +b \right )}\)	\(83\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (4 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x}\, x +10 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, b +3 b^{2} \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \right )}{8 \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}}}\)	\(96\)

[In]

int((a+b/x)^(3/2)*x,x,method=_RETURNVERBOSE)

[Out]

1/4*(2*a*x+5*b)*x*((a*x+b)/x)^(1/2)+3/8*b^2*ln((1/2*b+a*x)/a^(1/2)+(a*x^2+b*x)^(1/2))/a^(1/2)*((a*x+b)/x)^(1/2
)*(x*(a*x+b))^(1/2)/(a*x+b)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.97 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (2 \, a^{2} x^{2} + 5 \, a b x\right )} \sqrt {\frac {a x + b}{x}}}{8 \, a}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (2 \, a^{2} x^{2} + 5 \, a b x\right )} \sqrt {\frac {a x + b}{x}}}{4 \, a}\right ] \]

[In]

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

[Out]

[1/8*(3*sqrt(a)*b^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(2*a^2*x^2 + 5*a*b*x)*sqrt((a*x + b)/x)
)/a, -1/4*(3*sqrt(-a)*b^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (2*a^2*x^2 + 5*a*b*x)*sqrt((a*x + b)/x))/a]

Sympy [A] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.14 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=\frac {a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a x}{b} + 1}}{2} + \frac {5 b^{\frac {3}{2}} \sqrt {x} \sqrt {\frac {a x}{b} + 1}}{4} + \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4 \sqrt {a}} \]

[In]

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

[Out]

a*sqrt(b)*x**(3/2)*sqrt(a*x/b + 1)/2 + 5*b**(3/2)*sqrt(x)*sqrt(a*x/b + 1)/4 + 3*b**2*asinh(sqrt(a)*sqrt(x)/sqr
t(b))/(4*sqrt(a))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.48 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=-\frac {3 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{8 \, \sqrt {a}} + \frac {5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {a + \frac {b}{x}} a b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} - 2 \, {\left (a + \frac {b}{x}\right )} a + a^{2}\right )}} \]

[In]

integrate((a+b/x)^(3/2)*x,x, algorithm="maxima")

[Out]

-3/8*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a) + 1/4*(5*(a + b/x)^(3/2)*b^2 - 3*sqr
t(a + b/x)*a*b^2)/((a + b/x)^2 - 2*(a + b/x)*a + a^2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.17 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=-\frac {3 \, b^{2} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, \sqrt {a}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, \sqrt {a}} + \frac {1}{4} \, \sqrt {a x^{2} + b x} {\left (2 \, a x \mathrm {sgn}\left (x\right ) + 5 \, b \mathrm {sgn}\left (x\right )\right )} \]

[In]

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

[Out]

-3/8*b^2*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))*sgn(x)/sqrt(a) + 3/8*b^2*log(abs(b))*sgn(x)/s
qrt(a) + 1/4*sqrt(a*x^2 + b*x)*(2*a*x*sgn(x) + 5*b*sgn(x))

Mupad [B] (verification not implemented)

Time = 5.82 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x \, dx=\frac {5\,x^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}{4}+\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4\,\sqrt {a}}-\frac {3\,a\,x^2\,\sqrt {a+\frac {b}{x}}}{4} \]

[In]

int(x*(a + b/x)^(3/2),x)

[Out]

(5*x^2*(a + b/x)^(3/2))/4 + (3*b^2*atanh((a + b/x)^(1/2)/a^(1/2)))/(4*a^(1/2)) - (3*a*x^2*(a + b/x)^(1/2))/4