Integrand size = 15, antiderivative size = 92 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{16 x}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{24 x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{16 \sqrt {b}} \] Output:
-5/16*a^2*(a+b/x^2)^(1/2)/x-5/24*a*(a+b/x^2)^(3/2)/x-1/6*(a+b/x^2)^(5/2)/x -5/16*a^3*arctanh(b^(1/2)/(a+b/x^2)^(1/2)/x)/b^(1/2)
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {a+\frac {b}{x^2}} \left (-8 b^2-26 a b x^2-33 a^2 x^4-\frac {15 a^3 x^6 \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {b+a x^2}}\right )}{48 x^5} \] Input:
Integrate[(a + b/x^2)^(5/2)/x^2,x]
Output:
(Sqrt[a + b/x^2]*(-8*b^2 - 26*a*b*x^2 - 33*a^2*x^4 - (15*a^3*x^6*ArcTanh[S qrt[b + a*x^2]/Sqrt[b]])/(Sqrt[b]*Sqrt[b + a*x^2])))/(48*x^5)
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {858, 211, 211, 211, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \left (a+\frac {b}{x^2}\right )^{5/2}d\frac {1}{x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -\frac {5}{6} a \int \left (a+\frac {b}{x^2}\right )^{3/2}d\frac {1}{x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {a+\frac {b}{x^2}}d\frac {1}{x}+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b}{x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}\) |
Input:
Int[(a + b/x^2)^(5/2)/x^2,x]
Output:
-1/6*(a + b/x^2)^(5/2)/x - (5*a*((a + b/x^2)^(3/2)/(4*x) + (3*a*(Sqrt[a + b/x^2]/(2*x) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(2*Sqrt[b])))/4))/ 6
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
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] && Int egerQ[m]
Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {\left (33 a^{2} x^{4}+26 a b \,x^{2}+8 b^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{48 x^{5}}-\frac {5 a^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{16 \sqrt {b}\, \sqrt {a \,x^{2}+b}}\) | \(95\) |
default | \(-\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} \left (-3 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a^{3} x^{6}+15 b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) a^{3} x^{6}+3 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a^{2} x^{4}-5 \left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{3} b \,x^{6}-15 \sqrt {a \,x^{2}+b}\, a^{3} b^{2} x^{6}+2 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a b \,x^{2}+8 \left (a \,x^{2}+b \right )^{\frac {7}{2}} b^{2}\right )}{48 x \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{3}}\) | \(166\) |
Input:
int((a+b/x^2)^(5/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-1/48*(33*a^2*x^4+26*a*b*x^2+8*b^2)/x^5*((a*x^2+b)/x^2)^(1/2)-5/16*a^3/b^( 1/2)*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)
Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.95 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} x^{5} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (33 \, a^{2} b x^{4} + 26 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{96 \, b x^{5}}, \frac {15 \, a^{3} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{b}\right ) - {\left (33 \, a^{2} b x^{4} + 26 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{48 \, b x^{5}}\right ] \] Input:
integrate((a+b/x^2)^(5/2)/x^2,x, algorithm="fricas")
Output:
[1/96*(15*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*(33*a^2*b*x^4 + 26*a*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2 ))/(b*x^5), 1/48*(15*a^3*sqrt(-b)*x^5*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x ^2)/b) - (33*a^2*b*x^4 + 26*a*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b*x ^5)]
Time = 2.87 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=- \frac {11 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x^{2}}}}{16 x} - \frac {13 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{2}}}}{24 x^{3}} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{2}}}}{6 x^{5}} - \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{16 \sqrt {b}} \] Input:
integrate((a+b/x**2)**(5/2)/x**2,x)
Output:
-11*a**(5/2)*sqrt(1 + b/(a*x**2))/(16*x) - 13*a**(3/2)*b*sqrt(1 + b/(a*x** 2))/(24*x**3) - sqrt(a)*b**2*sqrt(1 + b/(a*x**2))/(6*x**5) - 5*a**3*asinh( sqrt(b)/(sqrt(a)*x))/(16*sqrt(b))
Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (72) = 144\).
Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.65 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=\frac {5 \, a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{32 \, \sqrt {b}} - \frac {33 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} a^{3} x^{5} - 40 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{3} b x^{3} + 15 \, \sqrt {a + \frac {b}{x^{2}}} a^{3} b^{2} x}{48 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{3} x^{6} - 3 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} b x^{4} + 3 \, {\left (a + \frac {b}{x^{2}}\right )} b^{2} x^{2} - b^{3}\right )}} \] Input:
integrate((a+b/x^2)^(5/2)/x^2,x, algorithm="maxima")
Output:
5/32*a^3*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/ sqrt(b) - 1/48*(33*(a + b/x^2)^(5/2)*a^3*x^5 - 40*(a + b/x^2)^(3/2)*a^3*b* x^3 + 15*sqrt(a + b/x^2)*a^3*b^2*x)/((a + b/x^2)^3*x^6 - 3*(a + b/x^2)^2*b *x^4 + 3*(a + b/x^2)*b^2*x^2 - b^3)
Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=\frac {1}{48} \, {\left (\frac {15 \, \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} - \frac {33 \, {\left (a x^{2} + b\right )}^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 40 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} b \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {a x^{2} + b} b^{2} \mathrm {sgn}\left (x\right )}{a^{3} x^{6}}\right )} a^{3} \] Input:
integrate((a+b/x^2)^(5/2)/x^2,x, algorithm="giac")
Output:
1/48*(15*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) - (33*(a*x^2 + b )^(5/2)*sgn(x) - 40*(a*x^2 + b)^(3/2)*b*sgn(x) + 15*sqrt(a*x^2 + b)*b^2*sg n(x))/(a^3*x^6))*a^3
Time = 1.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=-\frac {{\left (a\,x^2+b\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b}{a\,x^2}\right )}{x\,{\left (\frac {b}{a}+x^2\right )}^{5/2}} \] Input:
int((a + b/x^2)^(5/2)/x^2,x)
Output:
-((b + a*x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, -b/(a*x^2)))/(x*(b/a + x^2 )^(5/2))
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx=\frac {-33 \sqrt {a \,x^{2}+b}\, a^{2} b \,x^{4}-26 \sqrt {a \,x^{2}+b}\, a \,b^{2} x^{2}-8 \sqrt {a \,x^{2}+b}\, b^{3}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x -\sqrt {b}}{\sqrt {b}}\right ) a^{3} x^{6}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x +\sqrt {b}}{\sqrt {b}}\right ) a^{3} x^{6}}{48 b \,x^{6}} \] Input:
int((a+b/x^2)^(5/2)/x^2,x)
Output:
( - 33*sqrt(a*x**2 + b)*a**2*b*x**4 - 26*sqrt(a*x**2 + b)*a*b**2*x**2 - 8* sqrt(a*x**2 + b)*b**3 + 15*sqrt(b)*log((sqrt(a*x**2 + b) + sqrt(a)*x - sqr t(b))/sqrt(b))*a**3*x**6 - 15*sqrt(b)*log((sqrt(a*x**2 + b) + sqrt(a)*x + sqrt(b))/sqrt(b))*a**3*x**6)/(48*b*x**6)