Integrand size = 15, antiderivative size = 68 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx=\frac {1}{3 b \left (a+\frac {b}{x^2}\right )^{3/2} x^3}+\frac {1}{b^2 \sqrt {a+\frac {b}{x^2}} x}-\frac {\text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{b^{5/2}} \] Output:
1/3/b/(a+b/x^2)^(3/2)/x^3+1/b^2/(a+b/x^2)^(1/2)/x-arctanh(b^(1/2)/(a+b/x^2 )^(1/2)/x)/b^(5/2)
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx=\frac {\sqrt {b} \left (4 b+3 a x^2\right )-3 \left (b+a x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{3 b^{5/2} \sqrt {a+\frac {b}{x^2}} x \left (b+a x^2\right )} \] Input:
Integrate[1/((a + b/x^2)^(5/2)*x^6),x]
Output:
(Sqrt[b]*(4*b + 3*a*x^2) - 3*(b + a*x^2)^(3/2)*ArcTanh[Sqrt[b + a*x^2]/Sqr t[b]])/(3*b^(5/2)*Sqrt[a + b/x^2]*x*(b + a*x^2))
Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {858, 252, 252, 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.
\(\displaystyle \int \frac {1}{x^6 \left (a+\frac {b}{x^2}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^4}d\frac {1}{x}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{3 b x^3 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {\int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^2}d\frac {1}{x}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{3 b x^3 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {\frac {\int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x}}{b}-\frac {1}{b x \sqrt {a+\frac {b}{x^2}}}}{b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{3 b x^3 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {\frac {\int \frac {1}{1-\frac {b}{x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x}}{b}-\frac {1}{b x \sqrt {a+\frac {b}{x^2}}}}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3 b x^3 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {\frac {\text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{b^{3/2}}-\frac {1}{b x \sqrt {a+\frac {b}{x^2}}}}{b}\) |
Input:
Int[1/((a + b/x^2)^(5/2)*x^6),x]
Output:
1/(3*b*(a + b/x^2)^(3/2)*x^3) - (-(1/(b*Sqrt[a + b/x^2]*x)) + ArcTanh[Sqrt [b]/(Sqrt[a + b/x^2]*x)]/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[(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 = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.13
method | result | size |
default | \(-\frac {\left (a \,x^{2}+b \right ) \left (-3 x^{2} a \,b^{\frac {3}{2}}+3 \left (a \,x^{2}+b \right )^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) b -4 b^{\frac {5}{2}}\right )}{3 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} x^{5} b^{\frac {7}{2}}}\) | \(77\) |
Input:
int(1/(a+b/x^2)^(5/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/3*(a*x^2+b)*(-3*x^2*a*b^(3/2)+3*(a*x^2+b)^(3/2)*ln(2*(b^(1/2)*(a*x^2+b) ^(1/2)+b)/x)*b-4*b^(5/2))/((a*x^2+b)/x^2)^(5/2)/x^5/b^(7/2)
Time = 0.09 (sec) , antiderivative size = 220, normalized size of antiderivative = 3.24 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx=\left [\frac {3 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {b} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \, {\left (3 \, a b x^{3} + 4 \, b^{2} x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{6 \, {\left (a^{2} b^{3} x^{4} + 2 \, a b^{4} x^{2} + b^{5}\right )}}, \frac {3 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{b}\right ) + {\left (3 \, a b x^{3} + 4 \, b^{2} x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a^{2} b^{3} x^{4} + 2 \, a b^{4} x^{2} + b^{5}\right )}}\right ] \] Input:
integrate(1/(a+b/x^2)^(5/2)/x^6,x, algorithm="fricas")
Output:
[1/6*(3*(a^2*x^4 + 2*a*b*x^2 + b^2)*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^3 + 4*b^2*x)*sqrt((a*x^2 + b)/x ^2))/(a^2*b^3*x^4 + 2*a*b^4*x^2 + b^5), 1/3*(3*(a^2*x^4 + 2*a*b*x^2 + b^2) *sqrt(-b)*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/b) + (3*a*b*x^3 + 4*b^2* x)*sqrt((a*x^2 + b)/x^2))/(a^2*b^3*x^4 + 2*a*b^4*x^2 + b^5)]
Leaf count of result is larger than twice the leaf count of optimal. 740 vs. \(2 (58) = 116\).
Time = 2.09 (sec) , antiderivative size = 740, normalized size of antiderivative = 10.88 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b/x**2)**(5/2)/x**6,x)
Output:
3*a**3*b**4*x**6*log(a*x**2/b)/(6*a**3*b**(13/2)*x**6 + 18*a**2*b**(15/2)* x**4 + 18*a*b**(17/2)*x**2 + 6*b**(19/2)) - 6*a**3*b**4*x**6*log(sqrt(a*x* *2/b + 1) + 1)/(6*a**3*b**(13/2)*x**6 + 18*a**2*b**(15/2)*x**4 + 18*a*b**( 17/2)*x**2 + 6*b**(19/2)) + 6*a**2*b**5*x**4*sqrt(a*x**2/b + 1)/(6*a**3*b* *(13/2)*x**6 + 18*a**2*b**(15/2)*x**4 + 18*a*b**(17/2)*x**2 + 6*b**(19/2)) + 9*a**2*b**5*x**4*log(a*x**2/b)/(6*a**3*b**(13/2)*x**6 + 18*a**2*b**(15/ 2)*x**4 + 18*a*b**(17/2)*x**2 + 6*b**(19/2)) - 18*a**2*b**5*x**4*log(sqrt( a*x**2/b + 1) + 1)/(6*a**3*b**(13/2)*x**6 + 18*a**2*b**(15/2)*x**4 + 18*a* b**(17/2)*x**2 + 6*b**(19/2)) + 14*a*b**6*x**2*sqrt(a*x**2/b + 1)/(6*a**3* b**(13/2)*x**6 + 18*a**2*b**(15/2)*x**4 + 18*a*b**(17/2)*x**2 + 6*b**(19/2 )) + 9*a*b**6*x**2*log(a*x**2/b)/(6*a**3*b**(13/2)*x**6 + 18*a**2*b**(15/2 )*x**4 + 18*a*b**(17/2)*x**2 + 6*b**(19/2)) - 18*a*b**6*x**2*log(sqrt(a*x* *2/b + 1) + 1)/(6*a**3*b**(13/2)*x**6 + 18*a**2*b**(15/2)*x**4 + 18*a*b**( 17/2)*x**2 + 6*b**(19/2)) + 8*b**7*sqrt(a*x**2/b + 1)/(6*a**3*b**(13/2)*x* *6 + 18*a**2*b**(15/2)*x**4 + 18*a*b**(17/2)*x**2 + 6*b**(19/2)) + 3*b**7* log(a*x**2/b)/(6*a**3*b**(13/2)*x**6 + 18*a**2*b**(15/2)*x**4 + 18*a*b**(1 7/2)*x**2 + 6*b**(19/2)) - 6*b**7*log(sqrt(a*x**2/b + 1) + 1)/(6*a**3*b**( 13/2)*x**6 + 18*a**2*b**(15/2)*x**4 + 18*a*b**(17/2)*x**2 + 6*b**(19/2))
Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx=\frac {\log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {3 \, {\left (a + \frac {b}{x^{2}}\right )} x^{2} + b}{3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{2} x^{3}} \] Input:
integrate(1/(a+b/x^2)^(5/2)/x^6,x, algorithm="maxima")
Output:
1/2*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/b^(5/ 2) + 1/3*(3*(a + b/x^2)*x^2 + b)/((a + b/x^2)^(3/2)*b^2*x^3)
Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx=-\frac {{\left (3 \, \sqrt {b} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 4 \, \sqrt {-b}\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {-b} b^{\frac {5}{2}}} + \frac {\arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2} \mathrm {sgn}\left (x\right )} + \frac {3 \, a x^{2} + 4 \, b}{3 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} b^{2} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(a+b/x^2)^(5/2)/x^6,x, algorithm="giac")
Output:
-1/3*(3*sqrt(b)*arctan(sqrt(b)/sqrt(-b)) + 4*sqrt(-b))*sgn(x)/(sqrt(-b)*b^ (5/2)) + arctan(sqrt(a*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2*sgn(x)) + 1/3*(3*a *x^2 + 4*b)/((a*x^2 + b)^(3/2)*b^2*sgn(x))
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx=\int \frac {1}{x^6\,{\left (a+\frac {b}{x^2}\right )}^{5/2}} \,d x \] Input:
int(1/(x^6*(a + b/x^2)^(5/2)),x)
Output:
int(1/(x^6*(a + b/x^2)^(5/2)), x)
Time = 0.22 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.50 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6} \, dx=\frac {3 \sqrt {a \,x^{2}+b}\, a b \,x^{2}+4 \sqrt {a \,x^{2}+b}\, b^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x -\sqrt {b}}{\sqrt {b}}\right ) a^{2} x^{4}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x -\sqrt {b}}{\sqrt {b}}\right ) a b \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x -\sqrt {b}}{\sqrt {b}}\right ) b^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x +\sqrt {b}}{\sqrt {b}}\right ) a^{2} x^{4}-6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x +\sqrt {b}}{\sqrt {b}}\right ) a b \,x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x +\sqrt {b}}{\sqrt {b}}\right ) b^{2}}{3 b^{3} \left (a^{2} x^{4}+2 a b \,x^{2}+b^{2}\right )} \] Input:
int(1/(a+b/x^2)^(5/2)/x^6,x)
Output:
(3*sqrt(a*x**2 + b)*a*b*x**2 + 4*sqrt(a*x**2 + b)*b**2 + 3*sqrt(b)*log((sq rt(a*x**2 + b) + sqrt(a)*x - sqrt(b))/sqrt(b))*a**2*x**4 + 6*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))*b**2 - 3*sqrt(b)*log((sq rt(a*x**2 + b) + sqrt(a)*x + sqrt(b))/sqrt(b))*a**2*x**4 - 6*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))*b**2)/(3*b**3*(a**2*x**4 + 2*a*b*x**2 + b**2))