\(\int \frac {x^2}{(a+\frac {b}{x})^{3/2}} \, dx\) [1732]

Optimal result

Integrand size = 15, antiderivative size = 117 \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {35 b^3}{8 a^4 \sqrt {a+\frac {b}{x}}}+\frac {35 b^2 x}{24 a^3 \sqrt {a+\frac {b}{x}}}-\frac {7 b x^2}{12 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x}}}-\frac {35 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{9/2}} \]

[Out]

-35/8*b^3*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(9/2)+35/8*b^3/a^4/(a+b/x)^(1/2)+35/24*b^2*x/a^3/(a+b/x)^(1/2)-7/12
*b*x^2/a^2/(a+b/x)^(1/2)+1/3*x^3/a/(a+b/x)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 44, 53, 65, 214} \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=-\frac {35 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {35 b^3}{8 a^4 \sqrt {a+\frac {b}{x}}}+\frac {35 b^2 x}{24 a^3 \sqrt {a+\frac {b}{x}}}-\frac {7 b x^2}{12 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x}}} \]

[In]

Int[x^2/(a + b/x)^(3/2),x]

[Out]

(35*b^3)/(8*a^4*Sqrt[a + b/x]) + (35*b^2*x)/(24*a^3*Sqrt[a + b/x]) - (7*b*x^2)/(12*a^2*Sqrt[a + b/x]) + x^3/(3
*a*Sqrt[a + b/x]) - (35*b^3*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(8*a^(9/2))

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 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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 {1}{x^4 (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {x^3}{3 a \sqrt {a+\frac {b}{x}}}+\frac {(7 b) \text {Subst}\left (\int \frac {1}{x^3 (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{6 a} \\ & = -\frac {7 b x^2}{12 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x}}}-\frac {\left (35 b^2\right ) \text {Subst}\left (\int \frac {1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{24 a^2} \\ & = \frac {35 b^2 x}{24 a^3 \sqrt {a+\frac {b}{x}}}-\frac {7 b x^2}{12 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x}}}+\frac {\left (35 b^3\right ) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{16 a^3} \\ & = \frac {35 b^3}{8 a^4 \sqrt {a+\frac {b}{x}}}+\frac {35 b^2 x}{24 a^3 \sqrt {a+\frac {b}{x}}}-\frac {7 b x^2}{12 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x}}}+\frac {\left (35 b^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{16 a^4} \\ & = \frac {35 b^3}{8 a^4 \sqrt {a+\frac {b}{x}}}+\frac {35 b^2 x}{24 a^3 \sqrt {a+\frac {b}{x}}}-\frac {7 b x^2}{12 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x}}}+\frac {\left (35 b^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{8 a^4} \\ & = \frac {35 b^3}{8 a^4 \sqrt {a+\frac {b}{x}}}+\frac {35 b^2 x}{24 a^3 \sqrt {a+\frac {b}{x}}}-\frac {7 b x^2}{12 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x}}}-\frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {\sqrt {a+\frac {b}{x}} x \left (105 b^3+35 a b^2 x-14 a^2 b x^2+8 a^3 x^3\right )}{24 a^4 (b+a x)}-\frac {35 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{9/2}} \]

[In]

Integrate[x^2/(a + b/x)^(3/2),x]

[Out]

(Sqrt[a + b/x]*x*(105*b^3 + 35*a*b^2*x - 14*a^2*b*x^2 + 8*a^3*x^3))/(24*a^4*(b + a*x)) - (35*b^3*ArcTanh[Sqrt[
a + b/x]/Sqrt[a]])/(8*a^(9/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {\left (8 a^{2} x^{2}-22 a b x +57 b^{2}\right ) \left (a x +b \right )}{24 a^{4} \sqrt {\frac {a x +b}{x}}}+\frac {\left (-\frac {35 b^{3} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right )}{16 a^{\frac {9}{2}}}+\frac {2 b^{3} \sqrt {a \left (x +\frac {b}{a}\right )^{2}-b \left (x +\frac {b}{a}\right )}}{a^{5} \left (x +\frac {b}{a}\right )}\right ) \sqrt {x \left (a x +b \right )}}{x \sqrt {\frac {a x +b}{x}}}\)	\(140\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {9}{2}} x^{2}-60 \sqrt {a \,x^{2}+b x}\, a^{\frac {9}{2}} b \,x^{3}+32 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}} b x -150 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} b^{2} x^{2}+240 a^{\frac {7}{2}} \sqrt {x \left (a x +b \right )}\, b^{2} x^{2}-120 a^{3} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{3} x^{2}+16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2}-120 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b^{3} x -96 a^{\frac {5}{2}} \left (x \left (a x +b \right )\right )^{\frac {3}{2}} b^{2}+480 a^{\frac {5}{2}} \sqrt {x \left (a x +b \right )}\, b^{3} x -240 a^{2} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{4} x +15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{3} x^{2}-30 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{4}+240 a^{\frac {3}{2}} \sqrt {x \left (a x +b \right )}\, b^{4}-120 a \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{5}+30 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b^{4} x +15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{5}\right )}{48 a^{\frac {11}{2}} \sqrt {x \left (a x +b \right )}\, \left (a x +b \right )^{2}}\)	\(458\)

[In]

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

[Out]

1/24*(8*a^2*x^2-22*a*b*x+57*b^2)*(a*x+b)/a^4/((a*x+b)/x)^(1/2)+(-35/16*b^3/a^(9/2)*ln((1/2*b+a*x)/a^(1/2)+(a*x
^2+b*x)^(1/2))+2*b^3/a^5/(x+b/a)*(a*(x+b/a)^2-b*(x+b/a))^(1/2))/x/((a*x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.77 \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\left [\frac {105 \, {\left (a b^{3} x + b^{4}\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (8 \, a^{4} x^{4} - 14 \, a^{3} b x^{3} + 35 \, a^{2} b^{2} x^{2} + 105 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{48 \, {\left (a^{6} x + a^{5} b\right )}}, \frac {105 \, {\left (a b^{3} x + b^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (8 \, a^{4} x^{4} - 14 \, a^{3} b x^{3} + 35 \, a^{2} b^{2} x^{2} + 105 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{24 \, {\left (a^{6} x + a^{5} b\right )}}\right ] \]

[In]

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

[Out]

[1/48*(105*(a*b^3*x + b^4)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(8*a^4*x^4 - 14*a^3*b*x^
3 + 35*a^2*b^2*x^2 + 105*a*b^3*x)*sqrt((a*x + b)/x))/(a^6*x + a^5*b), 1/24*(105*(a*b^3*x + b^4)*sqrt(-a)*arcta
n(sqrt(-a)*sqrt((a*x + b)/x)/a) + (8*a^4*x^4 - 14*a^3*b*x^3 + 35*a^2*b^2*x^2 + 105*a*b^3*x)*sqrt((a*x + b)/x))
/(a^6*x + a^5*b)]

Sympy [A] (verification not implemented)

Time = 12.63 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.14 \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {x^{\frac {7}{2}}}{3 a \sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {7 \sqrt {b} x^{\frac {5}{2}}}{12 a^{2} \sqrt {\frac {a x}{b} + 1}} + \frac {35 b^{\frac {3}{2}} x^{\frac {3}{2}}}{24 a^{3} \sqrt {\frac {a x}{b} + 1}} + \frac {35 b^{\frac {5}{2}} \sqrt {x}}{8 a^{4} \sqrt {\frac {a x}{b} + 1}} - \frac {35 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{8 a^{\frac {9}{2}}} \]

[In]

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

[Out]

x**(7/2)/(3*a*sqrt(b)*sqrt(a*x/b + 1)) - 7*sqrt(b)*x**(5/2)/(12*a**2*sqrt(a*x/b + 1)) + 35*b**(3/2)*x**(3/2)/(
24*a**3*sqrt(a*x/b + 1)) + 35*b**(5/2)*sqrt(x)/(8*a**4*sqrt(a*x/b + 1)) - 35*b**3*asinh(sqrt(a)*sqrt(x)/sqrt(b
))/(8*a**(9/2))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.32 \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {105 \, {\left (a + \frac {b}{x}\right )}^{3} b^{3} - 280 \, {\left (a + \frac {b}{x}\right )}^{2} a b^{3} + 231 \, {\left (a + \frac {b}{x}\right )} a^{2} b^{3} - 48 \, a^{3} b^{3}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} a^{4} - 3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{5} + 3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{6} - \sqrt {a + \frac {b}{x}} a^{7}\right )}} + \frac {35 \, b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{16 \, a^{\frac {9}{2}}} \]

[In]

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

[Out]

1/24*(105*(a + b/x)^3*b^3 - 280*(a + b/x)^2*a*b^3 + 231*(a + b/x)*a^2*b^3 - 48*a^3*b^3)/((a + b/x)^(7/2)*a^4 -
3*(a + b/x)^(5/2)*a^5 + 3*(a + b/x)^(3/2)*a^6 - sqrt(a + b/x)*a^7) + 35/16*b^3*log((sqrt(a + b/x) - sqrt(a))/
(sqrt(a + b/x) + sqrt(a)))/a^(9/2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.29 \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {1}{24} \, \sqrt {a x^{2} + b x} {\left (2 \, x {\left (\frac {4 \, x}{a^{2} \mathrm {sgn}\left (x\right )} - \frac {11 \, b}{a^{3} \mathrm {sgn}\left (x\right )}\right )} + \frac {57 \, b^{2}}{a^{4} \mathrm {sgn}\left (x\right )}\right )} + \frac {35 \, b^{3} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right )}{16 \, a^{\frac {9}{2}} \mathrm {sgn}\left (x\right )} + \frac {2 \, b^{4}}{{\left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b\right )} a^{\frac {9}{2}} \mathrm {sgn}\left (x\right )} - \frac {{\left (35 \, b^{3} \log \left ({\left | b \right |}\right ) + 32 \, b^{3}\right )} \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {9}{2}}} \]

[In]

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

[Out]

1/24*sqrt(a*x^2 + b*x)*(2*x*(4*x/(a^2*sgn(x)) - 11*b/(a^3*sgn(x))) + 57*b^2/(a^4*sgn(x))) + 35/16*b^3*log(abs(
2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))/(a^(9/2)*sgn(x)) + 2*b^4/(((sqrt(a)*x - sqrt(a*x^2 + b*x))*sqr
t(a) + b)*a^(9/2)*sgn(x)) - 1/16*(35*b^3*log(abs(b)) + 32*b^3)*sgn(x)/a^(9/2)

Mupad [B] (verification not implemented)

Time = 6.10 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {35\,b^3}{8\,a^4\,\sqrt {a+\frac {b}{x}}}-\frac {35\,b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8\,a^{9/2}}+\frac {x^3}{3\,a\,\sqrt {a+\frac {b}{x}}}-\frac {7\,b\,x^2}{12\,a^2\,\sqrt {a+\frac {b}{x}}}+\frac {35\,b^2\,x}{24\,a^3\,\sqrt {a+\frac {b}{x}}} \]

[In]

int(x^2/(a + b/x)^(3/2),x)

[Out]

(35*b^3)/(8*a^4*(a + b/x)^(1/2)) - (35*b^3*atanh((a + b/x)^(1/2)/a^(1/2)))/(8*a^(9/2)) + x^3/(3*a*(a + b/x)^(1
/2)) - (7*b*x^2)/(12*a^2*(a + b/x)^(1/2)) + (35*b^2*x)/(24*a^3*(a + b/x)^(1/2))