\(\int \sqrt {a+\frac {b}{x^2}} x^3 \, dx\) [1891]

Optimal result

Integrand size = 15, antiderivative size = 71 \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=\frac {b \sqrt {a+\frac {b}{x^2}} x^2}{8 a}+\frac {1}{4} \sqrt {a+\frac {b}{x^2}} x^4-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 44, 65, 214} \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {b x^2 \sqrt {a+\frac {b}{x^2}}}{8 a}+\frac {1}{4} x^4 \sqrt {a+\frac {b}{x^2}} \]

[In]

Int[Sqrt[a + b/x^2]*x^3,x]

[Out]

(b*Sqrt[a + b/x^2]*x^2)/(8*a) + (Sqrt[a + b/x^2]*x^4)/4 - (b^2*ArcTanh[Sqrt[a + b/x^2]/Sqrt[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}& = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = \frac {1}{4} \sqrt {a+\frac {b}{x^2}} x^4-\frac {1}{8} b \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x^2}\right ) \\ & = \frac {b \sqrt {a+\frac {b}{x^2}} x^2}{8 a}+\frac {1}{4} \sqrt {a+\frac {b}{x^2}} x^4+\frac {b^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^2}\right )}{16 a} \\ & = \frac {b \sqrt {a+\frac {b}{x^2}} x^2}{8 a}+\frac {1}{4} \sqrt {a+\frac {b}{x^2}} x^4+\frac {b \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^2}}\right )}{8 a} \\ & = \frac {b \sqrt {a+\frac {b}{x^2}} x^2}{8 a}+\frac {1}{4} \sqrt {a+\frac {b}{x^2}} x^4-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=\frac {\sqrt {a+\frac {b}{x^2}} x^2 \left (b+2 a x^2\right )}{8 a}-\frac {b^2 \sqrt {a+\frac {b}{x^2}} x \text {arctanh}\left (\frac {\sqrt {a} x}{-\sqrt {b}+\sqrt {b+a x^2}}\right )}{4 a^{3/2} \sqrt {b+a x^2}} \]

[In]

Integrate[Sqrt[a + b/x^2]*x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10

method	result	size
risch	\(\frac {x^{2} \left (2 a \,x^{2}+b \right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{8 a}-\frac {b^{2} \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{8 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b}}\)	\(78\)
default	\(-\frac {\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x \left (-2 x \left (a \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {a}+\sqrt {a}\, \sqrt {a \,x^{2}+b}\, b x +\ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) b^{2}\right )}{8 \sqrt {a \,x^{2}+b}\, a^{\frac {3}{2}}}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.14 \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=\left [\frac {\sqrt {a} b^{2} \log \left (-2 \, a x^{2} + 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) + 2 \, {\left (2 \, a^{2} x^{4} + a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, a^{2}}, \frac {\sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (2 \, a^{2} x^{4} + a b x^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, a^{2}}\right ] \]

[In]

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

[Out]

[1/16*(sqrt(a)*b^2*log(-2*a*x^2 + 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 2*(2*a^2*x^4 + a*b*x^2)*sqrt((a*x
^2 + b)/x^2))/a^2, 1/8*(sqrt(-a)*b^2*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (2*a^2*x^4 + a*b
*x^2)*sqrt((a*x^2 + b)/x^2))/a^2]

Sympy [A] (verification not implemented)

Time = 2.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30 \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=\frac {a x^{5}}{4 \sqrt {b} \sqrt {\frac {a x^{2}}{b} + 1}} + \frac {3 \sqrt {b} x^{3}}{8 \sqrt {\frac {a x^{2}}{b} + 1}} + \frac {b^{\frac {3}{2}} x}{8 a \sqrt {\frac {a x^{2}}{b} + 1}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x}{\sqrt {b}} \right )}}{8 a^{\frac {3}{2}}} \]

[In]

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

[Out]

a*x**5/(4*sqrt(b)*sqrt(a*x**2/b + 1)) + 3*sqrt(b)*x**3/(8*sqrt(a*x**2/b + 1)) + b**(3/2)*x/(8*a*sqrt(a*x**2/b
+ 1)) - b**2*asinh(sqrt(a)*x/sqrt(b))/(8*a**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=\frac {b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right )}{16 \, a^{\frac {3}{2}}} + \frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{2} + \sqrt {a + \frac {b}{x^{2}}} a b^{2}}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{2} a - 2 \, {\left (a + \frac {b}{x^{2}}\right )} a^{2} + a^{3}\right )}} \]

[In]

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

[Out]

1/16*b^2*log((sqrt(a + b/x^2) - sqrt(a))/(sqrt(a + b/x^2) + sqrt(a)))/a^(3/2) + 1/8*((a + b/x^2)^(3/2)*b^2 + s
qrt(a + b/x^2)*a*b^2)/((a + b/x^2)^2*a - 2*(a + b/x^2)*a^2 + a^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97 \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=\frac {1}{8} \, \sqrt {a x^{2} + b} {\left (2 \, x^{2} \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{a}\right )} x + \frac {b^{2} \log \left ({\left | -\sqrt {a} x + \sqrt {a x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, a^{\frac {3}{2}}} - \frac {b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {3}{2}}} \]

[In]

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

[Out]

1/8*sqrt(a*x^2 + b)*(2*x^2*sgn(x) + b*sgn(x)/a)*x + 1/8*b^2*log(abs(-sqrt(a)*x + sqrt(a*x^2 + b)))*sgn(x)/a^(3
/2) - 1/16*b^2*log(abs(b))*sgn(x)/a^(3/2)

Mupad [B] (verification not implemented)

Time = 6.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76 \[ \int \sqrt {a+\frac {b}{x^2}} x^3 \, dx=\frac {x^4\,\sqrt {a+\frac {b}{x^2}}}{8}-\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )}{8\,a^{3/2}}+\frac {x^4\,{\left (a+\frac {b}{x^2}\right )}^{3/2}}{8\,a} \]

[In]

int(x^3*(a + b/x^2)^(1/2),x)

[Out]

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