\(\int \sqrt {a+\frac {b}{x^3}} x^2 \, dx\) [1992]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\frac {1}{3} \sqrt {a+\frac {b}{x^3}} x^3+\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{3 \sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 43, 65, 214} \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{3 \sqrt {a}}+\frac {1}{3} x^3 \sqrt {a+\frac {b}{x^3}} \]

[In]

Int[Sqrt[a + b/x^3]*x^2,x]

[Out]

(Sqrt[a + b/x^3]*x^3)/3 + (b*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(3*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}& = -\left (\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,\frac {1}{x^3}\right )\right ) \\ & = \frac {1}{3} \sqrt {a+\frac {b}{x^3}} x^3-\frac {1}{6} b \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right ) \\ & = \frac {1}{3} \sqrt {a+\frac {b}{x^3}} x^3-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^3}}\right ) \\ & = \frac {1}{3} \sqrt {a+\frac {b}{x^3}} x^3+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{3 \sqrt {a}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.45 \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\frac {1}{3} \sqrt {a+\frac {b}{x^3}} x^{3/2} \left (x^{3/2}+\frac {b \log \left (\sqrt {a} x^{3/2}+\sqrt {b+a x^3}\right )}{\sqrt {a} \sqrt {b+a x^3}}\right ) \]

[In]

Integrate[Sqrt[a + b/x^3]*x^2,x]

[Out]

(Sqrt[a + b/x^3]*x^(3/2)*(x^(3/2) + (b*Log[Sqrt[a]*x^(3/2) + Sqrt[b + a*x^3]])/(Sqrt[a]*Sqrt[b + a*x^3])))/3

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.51

method	result	size
default	\(\frac {\sqrt {\frac {a \,x^{3}+b}{x^{3}}}\, x^{2} \left (\sqrt {x \left (a \,x^{3}+b \right )}\, x \sqrt {a}+\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right ) b \right )}{3 \sqrt {x \left (a \,x^{3}+b \right )}\, \sqrt {a}}\)	\(71\)
risch	\(\frac {x^{3} \sqrt {\frac {a \,x^{3}+b}{x^{3}}}}{3}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {x \left (a \,x^{3}+b \right )}}{x^{2} \sqrt {a}}\right ) b x \sqrt {\frac {a \,x^{3}+b}{x^{3}}}\, \sqrt {x \left (a \,x^{3}+b \right )}}{3 \sqrt {a}\, \left (a \,x^{3}+b \right )}\)	\(79\)

[In]

int(x^2*(a+b/x^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 150, normalized size of antiderivative = 3.19 \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\left [\frac {4 \, a x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}} + \sqrt {a} b \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} - 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right )}{12 \, a}, \frac {2 \, a x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}} - \sqrt {-a} b \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right )}{6 \, a}\right ] \]

[In]

integrate(x^2*(a+b/x^3)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(4*a*x^3*sqrt((a*x^3 + b)/x^3) + sqrt(a)*b*log(-8*a^2*x^6 - 8*a*b*x^3 - b^2 - 4*(2*a*x^6 + b*x^3)*sqrt(a
)*sqrt((a*x^3 + b)/x^3)))/a, 1/6*(2*a*x^3*sqrt((a*x^3 + b)/x^3) - sqrt(-a)*b*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3
+ b)/x^3)/(2*a*x^3 + b)))/a]

Sympy [A] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02 \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\frac {\sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a x^{3}}{b} + 1}}{3} + \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{3 \sqrt {a}} \]

[In]

integrate(x**2*(a+b/x**3)**(1/2),x)

[Out]

sqrt(b)*x**(3/2)*sqrt(a*x**3/b + 1)/3 + b*asinh(sqrt(a)*x**(3/2)/sqrt(b))/(3*sqrt(a))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\frac {1}{3} \, \sqrt {a + \frac {b}{x^{3}}} x^{3} - \frac {b \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{6 \, \sqrt {a}} \]

[In]

integrate(x^2*(a+b/x^3)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(a + b/x^3)*x^3 - 1/6*b*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a)))/sqrt(a)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\frac {1}{3} \, \sqrt {a x^{4} + b x} x - \frac {b \arctan \left (\frac {\sqrt {a + \frac {b}{x^{3}}}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a}} \]

[In]

integrate(x^2*(a+b/x^3)^(1/2),x, algorithm="giac")

[Out]

1/3*sqrt(a*x^4 + b*x)*x - 1/3*b*arctan(sqrt(a + b/x^3)/sqrt(-a))/sqrt(-a)

Mupad [B] (verification not implemented)

Time = 6.42 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.19 \[ \int \sqrt {a+\frac {b}{x^3}} x^2 \, dx=\frac {x^3\,\sqrt {a+\frac {b}{x^3}}}{3}+\frac {b\,\ln \left (x^6\,\left (\sqrt {a+\frac {b}{x^3}}-\sqrt {a}\right )\,{\left (\sqrt {a+\frac {b}{x^3}}+\sqrt {a}\right )}^3\right )}{6\,\sqrt {a}} \]

[In]

int(x^2*(a + b/x^3)^(1/2),x)

[Out]

(x^3*(a + b/x^3)^(1/2))/3 + (b*log(x^6*((a + b/x^3)^(1/2) - a^(1/2))*((a + b/x^3)^(1/2) + a^(1/2))^3))/(6*a^(1
/2))