Integrand size = 15, antiderivative size = 95 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^8} \, dx=\frac {1}{3 b \left (a+\frac {b}{x^2}\right )^{3/2} x^5}+\frac {5}{3 b^2 \sqrt {a+\frac {b}{x^2}} x^3}-\frac {5 \sqrt {a+\frac {b}{x^2}}}{2 b^3 x}+\frac {5 a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{2 b^{7/2}} \]
1/3/b/(a+b/x^2)^(3/2)/x^5+5/2*a*arctanh(b^(1/2)/x/(a+b/x^2)^(1/2))/b^(7/2) +5/3/b^2/x^3/(a+b/x^2)^(1/2)-5/2*(a+b/x^2)^(1/2)/b^3/x
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^8} \, dx=\frac {-\sqrt {b} \left (3 b^2+20 a b x^2+15 a^2 x^4\right )+15 a x^2 \left (b+a x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )}{6 b^{7/2} \sqrt {a+\frac {b}{x^2}} x^3 \left (b+a x^2\right )} \]
(-(Sqrt[b]*(3*b^2 + 20*a*b*x^2 + 15*a^2*x^4)) + 15*a*x^2*(b + a*x^2)^(3/2) *ArcTanh[Sqrt[b + a*x^2]/Sqrt[b]])/(6*b^(7/2)*Sqrt[a + b/x^2]*x^3*(b + a*x ^2))
Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {858, 252, 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.
| \(\displaystyle \int \frac {1}{x^8 \left (a+\frac {b}{x^2}\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^6}d\frac {1}{x}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {1}{3 b x^5 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {5 \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2} x^4}d\frac {1}{x}}{3 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {1}{3 b x^5 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {5 \left (\frac {3 \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^2}d\frac {1}{x}}{b}-\frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}\right )}{3 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {1}{3 b x^5 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x}}{2 b}\right )}{b}-\frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}\right )}{3 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{3 b x^5 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \int \frac {1}{1-\frac {b}{x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x}}{2 b}\right )}{b}-\frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}\right )}{3 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3 b x^5 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\sqrt {a+\frac {b}{x^2}}}{2 b x}-\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 b^{3/2}}\right )}{b}-\frac {1}{b x^3 \sqrt {a+\frac {b}{x^2}}}\right )}{3 b}\) |
1/(3*b*(a + b/x^2)^(3/2)*x^5) - (5*(-(1/(b*Sqrt[a + b/x^2]*x^3)) + (3*(Sqr t[a + b/x^2]/(2*b*x) - (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(2*b^(3/2) )))/b))/(3*b)
3.20.56.3.1 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*(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 = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97
| method | result | size |
| default | \(-\frac {\left (a \,x^{2}+b \right ) \left (15 b^{\frac {3}{2}} a^{2} x^{4}+20 b^{\frac {5}{2}} a \,x^{2}-15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) \left (a \,x^{2}+b \right )^{\frac {3}{2}} a b \,x^{2}+3 b^{\frac {7}{2}}\right )}{6 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} x^{7} b^{\frac {9}{2}}}\) | \(92\) |
| risch | \(-\frac {a \,x^{2}+b}{2 b^{3} x^{3} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}+\frac {\left (\frac {5 a \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right )}{2 b^{\frac {7}{2}}}+\frac {13 a \sqrt {\left (x +\frac {\sqrt {-a b}}{a}\right )^{2} a -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{a}\right )}}{12 b^{3} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{a}\right )}-\frac {13 a \sqrt {\left (x -\frac {\sqrt {-a b}}{a}\right )^{2} a +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{a}\right )}}{12 b^{3} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{a}\right )}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{a}\right )^{2} a +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{a}\right )}}{12 b^{3} \left (x -\frac {\sqrt {-a b}}{a}\right )^{2}}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{a}\right )^{2} a -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{a}\right )}}{12 b^{3} \left (x +\frac {\sqrt {-a b}}{a}\right )^{2}}\right ) \sqrt {a \,x^{2}+b}}{\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\) | \(338\) |
-1/6*(a*x^2+b)*(15*b^(3/2)*a^2*x^4+20*b^(5/2)*a*x^2-15*ln(2*(b^(1/2)*(a*x^ 2+b)^(1/2)+b)/x)*(a*x^2+b)^(3/2)*a*b*x^2+3*b^(7/2))/((a*x^2+b)/x^2)^(5/2)/ x^7/b^(9/2)
Time = 0.32 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.74 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^8} \, dx=\left [\frac {15 \, {\left (a^{3} x^{5} + 2 \, a^{2} b x^{3} + a b^{2} x\right )} \sqrt {b} \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (15 \, a^{2} b x^{4} + 20 \, a b^{2} x^{2} + 3 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{12 \, {\left (a^{2} b^{4} x^{5} + 2 \, a b^{5} x^{3} + b^{6} x\right )}}, -\frac {15 \, {\left (a^{3} x^{5} + 2 \, a^{2} b x^{3} + a b^{2} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (15 \, a^{2} b x^{4} + 20 \, a b^{2} x^{2} + 3 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{6 \, {\left (a^{2} b^{4} x^{5} + 2 \, a b^{5} x^{3} + b^{6} x\right )}}\right ] \]
[1/12*(15*(a^3*x^5 + 2*a^2*b*x^3 + a*b^2*x)*sqrt(b)*log(-(a*x^2 + 2*sqrt(b )*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(15*a^2*b*x^4 + 20*a*b^2*x^2 + 3 *b^3)*sqrt((a*x^2 + b)/x^2))/(a^2*b^4*x^5 + 2*a*b^5*x^3 + b^6*x), -1/6*(15 *(a^3*x^5 + 2*a^2*b*x^3 + a*b^2*x)*sqrt(-b)*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (15*a^2*b*x^4 + 20*a*b^2*x^2 + 3*b^3)*sqrt((a*x^2 + b)/x^2))/(a^2*b^4*x^5 + 2*a*b^5*x^3 + b^6*x)]
Leaf count of result is larger than twice the leaf count of optimal. 864 vs. \(2 (85) = 170\).
Time = 3.14 (sec) , antiderivative size = 864, normalized size of antiderivative = 9.09 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^8} \, dx=- \frac {15 a^{4} b^{13} x^{8} \log {\left (\frac {a x^{2}}{b} \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} + \frac {30 a^{4} b^{13} x^{8} \log {\left (\sqrt {\frac {a x^{2}}{b} + 1} + 1 \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} - \frac {30 a^{3} b^{14} x^{6} \sqrt {\frac {a x^{2}}{b} + 1}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} - \frac {45 a^{3} b^{14} x^{6} \log {\left (\frac {a x^{2}}{b} \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} + \frac {90 a^{3} b^{14} x^{6} \log {\left (\sqrt {\frac {a x^{2}}{b} + 1} + 1 \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} - \frac {70 a^{2} b^{15} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} - \frac {45 a^{2} b^{15} x^{4} \log {\left (\frac {a x^{2}}{b} \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} + \frac {90 a^{2} b^{15} x^{4} \log {\left (\sqrt {\frac {a x^{2}}{b} + 1} + 1 \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} - \frac {46 a b^{16} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} - \frac {15 a b^{16} x^{2} \log {\left (\frac {a x^{2}}{b} \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} + \frac {30 a b^{16} x^{2} \log {\left (\sqrt {\frac {a x^{2}}{b} + 1} + 1 \right )}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} - \frac {6 b^{17} \sqrt {\frac {a x^{2}}{b} + 1}}{12 a^{3} b^{\frac {33}{2}} x^{8} + 36 a^{2} b^{\frac {35}{2}} x^{6} + 36 a b^{\frac {37}{2}} x^{4} + 12 b^{\frac {39}{2}} x^{2}} \]
-15*a**4*b**13*x**8*log(a*x**2/b)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35 /2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) + 30*a**4*b**13*x**8*l og(sqrt(a*x**2/b + 1) + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x** 6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 30*a**3*b**14*x**6*sqrt(a*x **2/b + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/ 2)*x**4 + 12*b**(39/2)*x**2) - 45*a**3*b**14*x**6*log(a*x**2/b)/(12*a**3*b **(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2 )*x**2) + 90*a**3*b**14*x**6*log(sqrt(a*x**2/b + 1) + 1)/(12*a**3*b**(33/2 )*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 70*a**2*b**15*x**4*sqrt(a*x**2/b + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2 *b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 45*a**2*b**15 *x**4*log(a*x**2/b)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36* a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) + 90*a**2*b**15*x**4*log(sqrt(a*x**2 /b + 1) + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(3 7/2)*x**4 + 12*b**(39/2)*x**2) - 46*a*b**16*x**2*sqrt(a*x**2/b + 1)/(12*a* *3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**( 39/2)*x**2) - 15*a*b**16*x**2*log(a*x**2/b)/(12*a**3*b**(33/2)*x**8 + 36*a **2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) + 30*a*b**16 *x**2*log(sqrt(a*x**2/b + 1) + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35 /2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 6*b**17*sqrt(a*x*...
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^8} \, dx=-\frac {15 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} a x^{4} - 10 \, {\left (a + \frac {b}{x^{2}}\right )} a b x^{2} - 2 \, a b^{2}}{6 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} b^{3} x^{5} - {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{4} x^{3}\right )}} - \frac {5 \, a \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{4 \, b^{\frac {7}{2}}} \]
-1/6*(15*(a + b/x^2)^2*a*x^4 - 10*(a + b/x^2)*a*b*x^2 - 2*a*b^2)/((a + b/x ^2)^(5/2)*b^3*x^5 - (a + b/x^2)^(3/2)*b^4*x^3) - 5/4*a*log((sqrt(a + b/x^2 )*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/b^(7/2)
Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^8} \, dx=-\frac {5 \, a \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right )}{2 \, \sqrt {-b} b^{3} \mathrm {sgn}\left (x\right )} - \frac {6 \, {\left (a x^{2} + b\right )} a + a b}{3 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\left (x\right )} - \frac {\sqrt {a x^{2} + b}}{2 \, b^{3} x^{2} \mathrm {sgn}\left (x\right )} \]
-5/2*a*arctan(sqrt(a*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^3*sgn(x)) - 1/3*(6*(a* x^2 + b)*a + a*b)/((a*x^2 + b)^(3/2)*b^3*sgn(x)) - 1/2*sqrt(a*x^2 + b)/(b^ 3*x^2*sgn(x))
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2} x^8} \, dx=\int \frac {1}{x^8\,{\left (a+\frac {b}{x^2}\right )}^{5/2}} \,d x \]