\(\int \frac {(a+\frac {b}{x^2})^{3/2}}{x^4} \, dx\) [1906]

Optimal result

Integrand size = 15, antiderivative size = 95 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}-\frac {a^2 \sqrt {a+\frac {b}{x^2}}}{16 b x}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{16 b^{3/2}} \]

[Out]

-1/6*(a+b/x^2)^(3/2)/x^3+1/16*a^3*arctanh(b^(1/2)/x/(a+b/x^2)^(1/2))/b^(3/2)-1/8*a*(a+b/x^2)^(1/2)/x^3-1/16*a^
2*(a+b/x^2)^(1/2)/b/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, 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 = {342, 285, 327, 223, 212} \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{16 b^{3/2}}-\frac {a^2 \sqrt {a+\frac {b}{x^2}}}{16 b x}-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3} \]

[In]

Int[(a + b/x^2)^(3/2)/x^4,x]

[Out]

-1/8*(a*Sqrt[a + b/x^2])/x^3 - (a + b/x^2)^(3/2)/(6*x^3) - (a^2*Sqrt[a + b/x^2])/(16*b*x) + (a^3*ArcTanh[Sqrt[
b]/(Sqrt[a + b/x^2]*x)])/(16*b^(3/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int x^2 \left (a+b x^2\right )^{3/2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}-\frac {1}{2} a \text {Subst}\left (\int x^2 \sqrt {a+b x^2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {a \sqrt {a+\frac {b}{x^2}}}{8 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}-\frac {1}{8} a^2 \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {a \sqrt {a+\frac {b}{x^2}}}{8 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}-\frac {a^2 \sqrt {a+\frac {b}{x^2}}}{16 b x}+\frac {a^3 \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{16 b} \\ & = -\frac {a \sqrt {a+\frac {b}{x^2}}}{8 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}-\frac {a^2 \sqrt {a+\frac {b}{x^2}}}{16 b x}+\frac {a^3 \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )}{16 b} \\ & = -\frac {a \sqrt {a+\frac {b}{x^2}}}{8 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}-\frac {a^2 \sqrt {a+\frac {b}{x^2}}}{16 b x}+\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{16 b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=\frac {\sqrt {a+\frac {b}{x^2}} \left (-\sqrt {b} \left (8 b^2+14 a b x^2+3 a^2 x^4\right )+\frac {3 a^3 x^6 \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{\sqrt {b+a x^2}}\right )}{48 b^{3/2} x^5} \]

[In]

Integrate[(a + b/x^2)^(3/2)/x^4,x]

[Out]

(Sqrt[a + b/x^2]*(-(Sqrt[b]*(8*b^2 + 14*a*b*x^2 + 3*a^2*x^4)) + (3*a^3*x^6*ArcTanh[Sqrt[b + a*x^2]/Sqrt[b]])/S
qrt[b + a*x^2]))/(48*b^(3/2)*x^5)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.03

method	result	size
risch	\(-\frac {\left (3 a^{2} x^{4}+14 a b \,x^{2}+8 b^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{48 x^{5} b}+\frac {a^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{16 b^{\frac {3}{2}} \sqrt {a \,x^{2}+b}}\)	\(98\)
default	\(\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) a^{3} x^{6}-\left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{3} x^{6}+\left (a \,x^{2}+b \right )^{\frac {5}{2}} a^{2} x^{4}-3 \sqrt {a \,x^{2}+b}\, a^{3} b \,x^{6}+2 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a b \,x^{2}-8 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{2}\right )}{48 x^{3} \left (a \,x^{2}+b \right )^{\frac {3}{2}} b^{3}}\)	\(145\)

[In]

int((a+b/x^2)^(3/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/48*(3*a^2*x^4+14*a*b*x^2+8*b^2)/x^5/b*((a*x^2+b)/x^2)^(1/2)+1/16/b^(3/2)*a^3*ln((2*b+2*b^(1/2)*(a*x^2+b)^(1
/2))/x)*((a*x^2+b)/x^2)^(1/2)*x/(a*x^2+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} x^{5} \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (3 \, a^{2} b x^{4} + 14 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{96 \, b^{2} x^{5}}, -\frac {3 \, a^{3} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (3 \, a^{2} b x^{4} + 14 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{48 \, b^{2} x^{5}}\right ] \]

[In]

integrate((a+b/x^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(3*a^3*sqrt(b)*x^5*log(-(a*x^2 + 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(3*a^2*b*x^4 + 14*a*b
^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b^2*x^5), -1/48*(3*a^3*sqrt(-b)*x^5*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)
/x^2)/(a*x^2 + b)) + (3*a^2*b*x^4 + 14*a*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b^2*x^5)]

Sympy [A] (verification not implemented)

Time = 3.42 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=- \frac {a^{\frac {5}{2}}}{16 b x \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {17 a^{\frac {3}{2}}}{48 x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {11 \sqrt {a} b}{24 x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{16 b^{\frac {3}{2}}} - \frac {b^{2}}{6 \sqrt {a} x^{7} \sqrt {1 + \frac {b}{a x^{2}}}} \]

[In]

integrate((a+b/x**2)**(3/2)/x**4,x)

[Out]

-a**(5/2)/(16*b*x*sqrt(1 + b/(a*x**2))) - 17*a**(3/2)/(48*x**3*sqrt(1 + b/(a*x**2))) - 11*sqrt(a)*b/(24*x**5*s
qrt(1 + b/(a*x**2))) + a**3*asinh(sqrt(b)/(sqrt(a)*x))/(16*b**(3/2)) - b**2/(6*sqrt(a)*x**7*sqrt(1 + b/(a*x**2
)))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (75) = 150\).

Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{32 \, b^{\frac {3}{2}}} - \frac {3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} a^{3} x^{5} + 8 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{3} b x^{3} - 3 \, \sqrt {a + \frac {b}{x^{2}}} a^{3} b^{2} x}{48 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{3} b x^{6} - 3 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} b^{2} x^{4} + 3 \, {\left (a + \frac {b}{x^{2}}\right )} b^{3} x^{2} - b^{4}\right )}} \]

[In]

integrate((a+b/x^2)^(3/2)/x^4,x, algorithm="maxima")

[Out]

-1/32*a^3*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/b^(3/2) - 1/48*(3*(a + b/x^2)^(5/2)
*a^3*x^5 + 8*(a + b/x^2)^(3/2)*a^3*b*x^3 - 3*sqrt(a + b/x^2)*a^3*b^2*x)/((a + b/x^2)^3*b*x^6 - 3*(a + b/x^2)^2
*b^2*x^4 + 3*(a + b/x^2)*b^3*x^2 - b^4)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {\frac {3 \, a^{4} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b} b} + \frac {3 \, {\left (a x^{2} + b\right )}^{\frac {5}{2}} a^{4} \mathrm {sgn}\left (x\right ) + 8 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{4} b \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {a x^{2} + b} a^{4} b^{2} \mathrm {sgn}\left (x\right )}{a^{3} b x^{6}}}{48 \, a} \]

[In]

integrate((a+b/x^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

-1/48*(3*a^4*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b) + (3*(a*x^2 + b)^(5/2)*a^4*sgn(x) + 8*(a*x^2
+ b)^(3/2)*a^4*b*sgn(x) - 3*sqrt(a*x^2 + b)*a^4*b^2*sgn(x))/(a^3*b*x^6))/a

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (a+\frac {b}{x^2}\right )}^{3/2}}{x^4} \,d x \]

[In]

int((a + b/x^2)^(3/2)/x^4,x)

[Out]

int((a + b/x^2)^(3/2)/x^4, x)