\(\int (a+\frac {b}{x})^{3/2} x^2 \, dx\) [1702]

Optimal result

Integrand size = 15, antiderivative size = 90 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {b^2 \sqrt {a+\frac {b}{x}} x}{8 a}+\frac {1}{4} b \sqrt {a+\frac {b}{x}} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} x^3-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{3/2}} \]

[Out]

1/3*(a+b/x)^(3/2)*x^3-1/8*b^3*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(3/2)+1/8*b^2*x*(a+b/x)^(1/2)/a+1/4*b*x^2*(a+b/
x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 44, 65, 214} \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {b^2 x \sqrt {a+\frac {b}{x}}}{8 a}+\frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{3/2}+\frac {1}{4} b x^2 \sqrt {a+\frac {b}{x}} \]

[In]

Int[(a + b/x)^(3/2)*x^2,x]

[Out]

(b^2*Sqrt[a + b/x]*x)/(8*a) + (b*Sqrt[a + b/x]*x^2)/4 + ((a + b/x)^(3/2)*x^3)/3 - (b^3*ArcTanh[Sqrt[a + b/x]/S
qrt[a]])/(8*a^(3/2))

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 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[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^4} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} x^3-\frac {1}{2} b \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{4} b \sqrt {a+\frac {b}{x}} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} x^3-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {b^2 \sqrt {a+\frac {b}{x}} x}{8 a}+\frac {1}{4} b \sqrt {a+\frac {b}{x}} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} x^3+\frac {b^3 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{16 a} \\ & = \frac {b^2 \sqrt {a+\frac {b}{x}} x}{8 a}+\frac {1}{4} b \sqrt {a+\frac {b}{x}} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} x^3+\frac {b^2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{8 a} \\ & = \frac {b^2 \sqrt {a+\frac {b}{x}} x}{8 a}+\frac {1}{4} b \sqrt {a+\frac {b}{x}} x^2+\frac {1}{3} \left (a+\frac {b}{x}\right )^{3/2} x^3-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {\sqrt {a} \sqrt {a+\frac {b}{x}} x \left (3 b^2+14 a b x+8 a^2 x^2\right )-3 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{24 a^{3/2}} \]

[In]

Integrate[(a + b/x)^(3/2)*x^2,x]

[Out]

(Sqrt[a]*Sqrt[a + b/x]*x*(3*b^2 + 14*a*b*x + 8*a^2*x^2) - 3*b^3*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(24*a^(3/2))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {\left (8 a^{2} x^{2}+14 a b x +3 b^{2}\right ) x \sqrt {\frac {a x +b}{x}}}{24 a}-\frac {b^{3} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{16 a^{\frac {3}{2}} \left (a x +b \right )}\)	\(97\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}}+12 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b x +6 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{2}-3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{3}\right )}{48 a^{\frac {5}{2}} \sqrt {x \left (a x +b \right )}}\)	\(115\)

[In]

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

[Out]

1/24*(8*a^2*x^2+14*a*b*x+3*b^2)*x/a*((a*x+b)/x)^(1/2)-1/16*b^3/a^(3/2)*ln((1/2*b+a*x)/a^(1/2)+(a*x^2+b*x)^(1/2
))*((a*x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)/(a*x+b)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.68 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\left [\frac {3 \, \sqrt {a} b^{3} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (8 \, a^{3} x^{3} + 14 \, a^{2} b x^{2} + 3 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{48 \, a^{2}}, \frac {3 \, \sqrt {-a} b^{3} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (8 \, a^{3} x^{3} + 14 \, a^{2} b x^{2} + 3 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{24 \, a^{2}}\right ] \]

[In]

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

[Out]

[1/48*(3*sqrt(a)*b^3*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(8*a^3*x^3 + 14*a^2*b*x^2 + 3*a*b^2*x)
*sqrt((a*x + b)/x))/a^2, 1/24*(3*sqrt(-a)*b^3*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (8*a^3*x^3 + 14*a^2*b*x^2
+ 3*a*b^2*x)*sqrt((a*x + b)/x))/a^2]

Sympy [A] (verification not implemented)

Time = 3.47 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.38 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {a^{2} x^{\frac {7}{2}}}{3 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {11 a \sqrt {b} x^{\frac {5}{2}}}{12 \sqrt {\frac {a x}{b} + 1}} + \frac {17 b^{\frac {3}{2}} x^{\frac {3}{2}}}{24 \sqrt {\frac {a x}{b} + 1}} + \frac {b^{\frac {5}{2}} \sqrt {x}}{8 a \sqrt {\frac {a x}{b} + 1}} - \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{8 a^{\frac {3}{2}}} \]

[In]

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

[Out]

a**2*x**(7/2)/(3*sqrt(b)*sqrt(a*x/b + 1)) + 11*a*sqrt(b)*x**(5/2)/(12*sqrt(a*x/b + 1)) + 17*b**(3/2)*x**(3/2)/
(24*sqrt(a*x/b + 1)) + b**(5/2)*sqrt(x)/(8*a*sqrt(a*x/b + 1)) - b**3*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(8*a**(3/2
))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.50 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{16 \, a^{\frac {3}{2}}} + \frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{3} + 8 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {a + \frac {b}{x}} a^{2} b^{3}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{3} a - 3 \, {\left (a + \frac {b}{x}\right )}^{2} a^{2} + 3 \, {\left (a + \frac {b}{x}\right )} a^{3} - a^{4}\right )}} \]

[In]

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

[Out]

1/16*b^3*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(3/2) + 1/24*(3*(a + b/x)^(5/2)*b^3 + 8*(a
+ b/x)^(3/2)*a*b^3 - 3*sqrt(a + b/x)*a^2*b^3)/((a + b/x)^3*a - 3*(a + b/x)^2*a^2 + 3*(a + b/x)*a^3 - a^4)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {b^{3} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {3}{2}}} - \frac {b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {3}{2}}} + \frac {1}{24} \, \sqrt {a x^{2} + b x} {\left (2 \, {\left (4 \, a x \mathrm {sgn}\left (x\right ) + 7 \, b \mathrm {sgn}\left (x\right )\right )} x + \frac {3 \, b^{2} \mathrm {sgn}\left (x\right )}{a}\right )} \]

[In]

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

[Out]

1/16*b^3*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))*sgn(x)/a^(3/2) - 1/16*b^3*log(abs(b))*sgn(x)/
a^(3/2) + 1/24*sqrt(a*x^2 + b*x)*(2*(4*a*x*sgn(x) + 7*b*sgn(x))*x + 3*b^2*sgn(x)/a)

Mupad [B] (verification not implemented)

Time = 6.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.80 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {x^3\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3}-\frac {a\,x^3\,\sqrt {a+\frac {b}{x}}}{8}+\frac {x^3\,{\left (a+\frac {b}{x}\right )}^{5/2}}{8\,a}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{3/2}} \]

[In]

int(x^2*(a + b/x)^(3/2),x)

[Out]

(x^3*(a + b/x)^(3/2))/3 + (b^3*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*1i)/(8*a^(3/2)) - (a*x^3*(a + b/x)^(1/2))/8
+ (x^3*(a + b/x)^(5/2))/(8*a)