Integrand size = 15, antiderivative size = 95 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {a \sqrt {a+\frac {b}{x^2}}}{8 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}-\frac {a^2 \sqrt {a+\frac {b}{x^2}}}{16 b x}+\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{16 b^{3/2}} \]
-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
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=\frac {\sqrt {a+\frac {b}{x^2}} \left (-\sqrt {b} \left (8 b^2+14 a b x^2+3 a^2 x^4\right )+\frac {3 a^3 x^6 \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{\sqrt {b+a x^2}}\right )}{48 b^{3/2} x^5} \]
(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]])/Sqrt[b + a*x^2]))/(48*b^(3/2)*x^5)
Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {858, 248, 248, 262, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle -\frac {1}{2} a \int \frac {\sqrt {a+\frac {b}{x^2}}}{x^2}d\frac {1}{x}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle -\frac {1}{2} a \left (\frac {1}{4} a \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^2}d\frac {1}{x}+\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x}}{2 b}\right )+\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \int \frac {1}{1-\frac {b}{x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x}}{2 b}\right )+\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 b^{3/2}}\right )+\frac {\sqrt {a+\frac {b}{x^2}}}{4 x^3}\right )-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{6 x^3}\) |
-1/6*(a + b/x^2)^(3/2)/x^3 - (a*(Sqrt[a + b/x^2]/(4*x^3) + (a*(Sqrt[a + b/ x^2]/(2*b*x) - (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(2*b^(3/2))))/4))/ 2
3.20.6.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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 = 0.04 (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 | \(\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) a^{3} x^{6}-\left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{3} x^{6}+\left (a \,x^{2}+b \right )^{\frac {5}{2}} a^{2} x^{4}-3 \sqrt {a \,x^{2}+b}\, a^{3} b \,x^{6}+2 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a b \,x^{2}-8 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{2}\right )}{48 x^{3} \left (a \,x^{2}+b \right )^{\frac {3}{2}} b^{3}}\) | \(145\) |
-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)
Time = 0.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} x^{5} \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (3 \, a^{2} b x^{4} + 14 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{96 \, b^{2} x^{5}}, -\frac {3 \, a^{3} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (3 \, a^{2} b x^{4} + 14 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{48 \, b^{2} x^{5}}\right ] \]
[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)]
Time = 3.42 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=- \frac {a^{\frac {5}{2}}}{16 b x \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {17 a^{\frac {3}{2}}}{48 x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {11 \sqrt {a} b}{24 x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{16 b^{\frac {3}{2}}} - \frac {b^{2}}{6 \sqrt {a} x^{7} \sqrt {1 + \frac {b}{a x^{2}}}} \]
-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*sqrt(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)) )
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 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{32 \, b^{\frac {3}{2}}} - \frac {3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} a^{3} x^{5} + 8 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{3} b x^{3} - 3 \, \sqrt {a + \frac {b}{x^{2}}} a^{3} b^{2} x}{48 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{3} b x^{6} - 3 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} b^{2} x^{4} + 3 \, {\left (a + \frac {b}{x^{2}}\right )} b^{3} x^{2} - b^{4}\right )}} \]
-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)
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {\frac {3 \, a^{4} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b} b} + \frac {3 \, {\left (a x^{2} + b\right )}^{\frac {5}{2}} a^{4} \mathrm {sgn}\left (x\right ) + 8 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{4} b \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {a x^{2} + b} a^{4} b^{2} \mathrm {sgn}\left (x\right )}{a^{3} b x^{6}}}{48 \, a} \]
-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
Timed out. \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (a+\frac {b}{x^2}\right )}^{3/2}}{x^4} \,d x \]