Integrand size = 15, antiderivative size = 71 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=\frac {1}{b \sqrt {a+\frac {b}{x^2}} x^3}-\frac {3 \sqrt {a+\frac {b}{x^2}}}{2 b^2 x}+\frac {3 a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{2 b^{5/2}} \] Output:
1/b/(a+b/x^2)^(1/2)/x^3-3/2*(a+b/x^2)^(1/2)/b^2/x+3/2*a*arctanh(b^(1/2)/(a +b/x^2)^(1/2)/x)/b^(5/2)
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=\frac {-\sqrt {b} \left (b+3 a x^2\right )+3 a x^2 \sqrt {b+a x^2} \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{2 b^{5/2} \sqrt {a+\frac {b}{x^2}} x^3} \] Input:
Integrate[1/((a + b/x^2)^(3/2)*x^6),x]
Output:
(-(Sqrt[b]*(b + 3*a*x^2)) + 3*a*x^2*Sqrt[b + a*x^2]*ArcTanh[Sqrt[b + a*x^2 ]/Sqrt[b]])/(2*b^(5/2)*Sqrt[a + b/x^2]*x^3)
Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {858, 252, 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.
\(\displaystyle \int \frac {1}{x^6 \left (a+\frac {b}{x^2}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^4}d\frac {1}{x}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}-\frac {3 \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^2}d\frac {1}{x}}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}-\frac {3 \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x}}{2 b}\right )}{b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}-\frac {3 \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \int \frac {1}{1-\frac {b}{x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x}}{2 b}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}-\frac {3 \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 b^{3/2}}\right )}{b}\) |
Input:
Int[1/((a + b/x^2)^(3/2)*x^6),x]
Output:
1/(b*Sqrt[a + b/x^2]*x^3) - (3*(Sqrt[a + b/x^2]/(2*b*x) - (a*ArcTanh[Sqrt[ b]/(Sqrt[a + b/x^2]*x)])/(2*b^(3/2))))/b
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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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 = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {\left (a \,x^{2}+b \right ) \left (3 \sqrt {a \,x^{2}+b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) a b \,x^{2}-3 x^{2} a \,b^{\frac {3}{2}}-b^{\frac {5}{2}}\right )}{2 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} x^{5} b^{\frac {7}{2}}}\) | \(81\) |
risch | \(-\frac {a \,x^{2}+b}{2 b^{2} x^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {\left (-\frac {a}{b^{2} \sqrt {a \,x^{2}+b}}+\frac {3 a \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right )}{2 b^{\frac {5}{2}}}\right ) \sqrt {a \,x^{2}+b}}{\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\) | \(101\) |
Input:
int(1/(a+b/x^2)^(3/2)/x^6,x,method=_RETURNVERBOSE)
Output:
1/2*(a*x^2+b)*(3*(a*x^2+b)^(1/2)*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*a*b*x ^2-3*x^2*a*b^(3/2)-b^(5/2))/((a*x^2+b)/x^2)^(3/2)/x^5/b^(7/2)
Time = 0.08 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.59 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=\left [\frac {3 \, {\left (a^{2} x^{3} + a b x\right )} \sqrt {b} \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (3 \, a b x^{2} + b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{4 \, {\left (a b^{3} x^{3} + b^{4} x\right )}}, -\frac {3 \, {\left (a^{2} x^{3} + a b x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{b}\right ) + {\left (3 \, a b x^{2} + b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{2 \, {\left (a b^{3} x^{3} + b^{4} x\right )}}\right ] \] Input:
integrate(1/(a+b/x^2)^(3/2)/x^6,x, algorithm="fricas")
Output:
[1/4*(3*(a^2*x^3 + a*b*x)*sqrt(b)*log(-(a*x^2 + 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(3*a*b*x^2 + b^2)*sqrt((a*x^2 + b)/x^2))/(a*b^3*x^ 3 + b^4*x), -1/2*(3*(a^2*x^3 + a*b*x)*sqrt(-b)*arctan(sqrt(-b)*x*sqrt((a*x ^2 + b)/x^2)/b) + (3*a*b*x^2 + b^2)*sqrt((a*x^2 + b)/x^2))/(a*b^3*x^3 + b^ 4*x)]
Time = 2.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=- \frac {3 \sqrt {a}}{2 b^{2} x \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{2 b^{\frac {5}{2}}} - \frac {1}{2 \sqrt {a} b x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} \] Input:
integrate(1/(a+b/x**2)**(3/2)/x**6,x)
Output:
-3*sqrt(a)/(2*b**2*x*sqrt(1 + b/(a*x**2))) + 3*a*asinh(sqrt(b)/(sqrt(a)*x) )/(2*b**(5/2)) - 1/(2*sqrt(a)*b*x**3*sqrt(1 + b/(a*x**2)))
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=-\frac {3 \, {\left (a + \frac {b}{x^{2}}\right )} a x^{2} - 2 \, a b}{2 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{2} x^{3} - \sqrt {a + \frac {b}{x^{2}}} b^{3} x\right )}} - \frac {3 \, a \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{4 \, b^{\frac {5}{2}}} \] Input:
integrate(1/(a+b/x^2)^(3/2)/x^6,x, algorithm="maxima")
Output:
-1/2*(3*(a + b/x^2)*a*x^2 - 2*a*b)/((a + b/x^2)^(3/2)*b^2*x^3 - sqrt(a + b /x^2)*b^3*x) - 3/4*a*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/b^(5/2)
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=-\frac {3 \, a \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right )}{2 \, \sqrt {-b} b^{2} \mathrm {sgn}\left (x\right )} - \frac {3 \, {\left (a x^{2} + b\right )} a - 2 \, a b}{2 \, {\left ({\left (a x^{2} + b\right )}^{\frac {3}{2}} - \sqrt {a x^{2} + b} b\right )} b^{2} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(a+b/x^2)^(3/2)/x^6,x, algorithm="giac")
Output:
-3/2*a*arctan(sqrt(a*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2*sgn(x)) - 1/2*(3*(a* x^2 + b)*a - 2*a*b)/(((a*x^2 + b)^(3/2) - sqrt(a*x^2 + b)*b)*b^2*sgn(x))
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=\int \frac {1}{x^6\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \,d x \] Input:
int(1/(x^6*(a + b/x^2)^(3/2)),x)
Output:
int(1/(x^6*(a + b/x^2)^(3/2)), x)
Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.42 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^6} \, dx=\frac {-3 \sqrt {a \,x^{2}+b}\, a b \,x^{2}-\sqrt {a \,x^{2}+b}\, b^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x -\sqrt {b}}{\sqrt {b}}\right ) a^{2} x^{4}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x -\sqrt {b}}{\sqrt {b}}\right ) a b \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x +\sqrt {b}}{\sqrt {b}}\right ) a^{2} x^{4}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x +\sqrt {b}}{\sqrt {b}}\right ) a b \,x^{2}}{2 b^{3} x^{2} \left (a \,x^{2}+b \right )} \] Input:
int(1/(a+b/x^2)^(3/2)/x^6,x)
Output:
( - 3*sqrt(a*x**2 + b)*a*b*x**2 - sqrt(a*x**2 + b)*b**2 - 3*sqrt(b)*log((s qrt(a*x**2 + b) + sqrt(a)*x - sqrt(b))/sqrt(b))*a**2*x**4 - 3*sqrt(b)*log( (sqrt(a*x**2 + b) + sqrt(a)*x - sqrt(b))/sqrt(b))*a*b*x**2 + 3*sqrt(b)*log ((sqrt(a*x**2 + b) + sqrt(a)*x + sqrt(b))/sqrt(b))*a**2*x**4 + 3*sqrt(b)*l og((sqrt(a*x**2 + b) + sqrt(a)*x + sqrt(b))/sqrt(b))*a*b*x**2)/(2*b**3*x** 2*(a*x**2 + b))