Integrand size = 22, antiderivative size = 123 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx=-\frac {b}{4 d \sqrt {c+\frac {d}{x^2}} x^5}-\frac {5 b c-4 a d}{4 d^2 \sqrt {c+\frac {d}{x^2}} x^3}+\frac {3 (5 b c-4 a d) \sqrt {c+\frac {d}{x^2}}}{8 d^3 x}-\frac {3 c (5 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{8 d^{7/2}} \]
-3/8*c*(-4*a*d+5*b*c)*arctanh(d^(1/2)/x/(c+d/x^2)^(1/2))/d^(7/2)-1/4*b/d/x ^5/(c+d/x^2)^(1/2)+1/4*(4*a*d-5*b*c)/d^2/x^3/(c+d/x^2)^(1/2)+3/8*(-4*a*d+5 *b*c)*(c+d/x^2)^(1/2)/d^3/x
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.92 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx=\frac {\sqrt {d} \left (-4 a d x^2 \left (d+3 c x^2\right )+b \left (-2 d^2+5 c d x^2+15 c^2 x^4\right )\right )-3 c (5 b c-4 a d) x^4 \sqrt {d+c x^2} \text {arctanh}\left (\frac {\sqrt {d+c x^2}}{\sqrt {d}}\right )}{8 d^{7/2} \sqrt {c+\frac {d}{x^2}} x^5} \]
(Sqrt[d]*(-4*a*d*x^2*(d + 3*c*x^2) + b*(-2*d^2 + 5*c*d*x^2 + 15*c^2*x^4)) - 3*c*(5*b*c - 4*a*d)*x^4*Sqrt[d + c*x^2]*ArcTanh[Sqrt[d + c*x^2]/Sqrt[d]] )/(8*d^(7/2)*Sqrt[c + d/x^2]*x^5)
Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {959, 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.
| \(\displaystyle \int \frac {a+\frac {b}{x^2}}{x^6 \left (c+\frac {d}{x^2}\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle -\frac {(5 b c-4 a d) \int \frac {1}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6}dx}{4 d}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}}\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle \frac {(5 b c-4 a d) \int \frac {1}{\left (c+\frac {d}{x^2}\right )^{3/2} x^4}d\frac {1}{x}}{4 d}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {(5 b c-4 a d) \left (\frac {3 \int \frac {1}{\sqrt {c+\frac {d}{x^2}} x^2}d\frac {1}{x}}{d}-\frac {1}{d x^3 \sqrt {c+\frac {d}{x^2}}}\right )}{4 d}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(5 b c-4 a d) \left (\frac {3 \left (\frac {\sqrt {c+\frac {d}{x^2}}}{2 d x}-\frac {c \int \frac {1}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x}}{2 d}\right )}{d}-\frac {1}{d x^3 \sqrt {c+\frac {d}{x^2}}}\right )}{4 d}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {(5 b c-4 a d) \left (\frac {3 \left (\frac {\sqrt {c+\frac {d}{x^2}}}{2 d x}-\frac {c \int \frac {1}{1-\frac {d}{x^2}}d\frac {1}{\sqrt {c+\frac {d}{x^2}} x}}{2 d}\right )}{d}-\frac {1}{d x^3 \sqrt {c+\frac {d}{x^2}}}\right )}{4 d}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(5 b c-4 a d) \left (\frac {3 \left (\frac {\sqrt {c+\frac {d}{x^2}}}{2 d x}-\frac {c \text {arctanh}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 d^{3/2}}\right )}{d}-\frac {1}{d x^3 \sqrt {c+\frac {d}{x^2}}}\right )}{4 d}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}}\) |
-1/4*b/(d*Sqrt[c + d/x^2]*x^5) + ((5*b*c - 4*a*d)*(-(1/(d*Sqrt[c + d/x^2]* x^3)) + (3*(Sqrt[c + d/x^2]/(2*d*x) - (c*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]* x)])/(2*d^(3/2))))/d))/(4*d)
3.10.85.3.1 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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.26
| method | result | size |
| risch | \(-\frac {\left (c \,x^{2}+d \right ) \left (4 a d \,x^{2}-7 c b \,x^{2}+2 b d \right )}{8 d^{3} x^{5} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}-\frac {c \left (-\frac {4 a d -7 b c}{\sqrt {c \,x^{2}+d}}+3 d \left (4 a d -5 b c \right ) \left (\frac {1}{d \sqrt {c \,x^{2}+d}}-\frac {\ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right )}{d^{\frac {3}{2}}}\right )\right ) \sqrt {c \,x^{2}+d}}{8 d^{3} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x}\) | \(155\) |
| default | \(-\frac {\left (c \,x^{2}+d \right ) \left (12 d^{\frac {5}{2}} a c \,x^{4}-15 d^{\frac {3}{2}} b \,c^{2} x^{4}-12 \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) \sqrt {c \,x^{2}+d}\, a c \,d^{2} x^{4}+15 \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) \sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{4}+4 d^{\frac {7}{2}} a \,x^{2}-5 d^{\frac {5}{2}} b c \,x^{2}+2 d^{\frac {7}{2}} b \right )}{8 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{7} d^{\frac {9}{2}}}\) | \(157\) |
-1/8*(c*x^2+d)*(4*a*d*x^2-7*b*c*x^2+2*b*d)/d^3/x^5/((c*x^2+d)/x^2)^(1/2)-1 /8*c/d^3*(-(4*a*d-7*b*c)/(c*x^2+d)^(1/2)+3*d*(4*a*d-5*b*c)*(1/d/(c*x^2+d)^ (1/2)-1/d^(3/2)*ln((2*d+2*d^(1/2)*(c*x^2+d)^(1/2))/x)))/((c*x^2+d)/x^2)^(1 /2)/x*(c*x^2+d)^(1/2)
Time = 0.28 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.55 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx=\left [-\frac {3 \, {\left ({\left (5 \, b c^{3} - 4 \, a c^{2} d\right )} x^{5} + {\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3}\right )} \sqrt {d} \log \left (-\frac {c x^{2} + 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{4} - 2 \, b d^{3} + {\left (5 \, b c d^{2} - 4 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, {\left (c d^{4} x^{5} + d^{5} x^{3}\right )}}, \frac {3 \, {\left ({\left (5 \, b c^{3} - 4 \, a c^{2} d\right )} x^{5} + {\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (3 \, {\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{4} - 2 \, b d^{3} + {\left (5 \, b c d^{2} - 4 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, {\left (c d^{4} x^{5} + d^{5} x^{3}\right )}}\right ] \]
[-1/16*(3*((5*b*c^3 - 4*a*c^2*d)*x^5 + (5*b*c^2*d - 4*a*c*d^2)*x^3)*sqrt(d )*log(-(c*x^2 + 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) - 2*(3*(5*b* c^2*d - 4*a*c*d^2)*x^4 - 2*b*d^3 + (5*b*c*d^2 - 4*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(c*d^4*x^5 + d^5*x^3), 1/8*(3*((5*b*c^3 - 4*a*c^2*d)*x^5 + (5*b *c^2*d - 4*a*c*d^2)*x^3)*sqrt(-d)*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/ (c*x^2 + d)) + (3*(5*b*c^2*d - 4*a*c*d^2)*x^4 - 2*b*d^3 + (5*b*c*d^2 - 4*a *d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(c*d^4*x^5 + d^5*x^3)]
Time = 11.33 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.46 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx=a \left (- \frac {3 \sqrt {c}}{2 d^{2} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {3 c \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2 d^{\frac {5}{2}}} - \frac {1}{2 \sqrt {c} d x^{3} \sqrt {1 + \frac {d}{c x^{2}}}}\right ) + b \left (\frac {15 c^{\frac {3}{2}}}{8 d^{3} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {5 \sqrt {c}}{8 d^{2} x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {15 c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{8 d^{\frac {7}{2}}} - \frac {1}{4 \sqrt {c} d x^{5} \sqrt {1 + \frac {d}{c x^{2}}}}\right ) \]
a*(-3*sqrt(c)/(2*d**2*x*sqrt(1 + d/(c*x**2))) + 3*c*asinh(sqrt(d)/(sqrt(c) *x))/(2*d**(5/2)) - 1/(2*sqrt(c)*d*x**3*sqrt(1 + d/(c*x**2)))) + b*(15*c** (3/2)/(8*d**3*x*sqrt(1 + d/(c*x**2))) + 5*sqrt(c)/(8*d**2*x**3*sqrt(1 + d/ (c*x**2))) - 15*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*d**(7/2)) - 1/(4*sqrt(c )*d*x**5*sqrt(1 + d/(c*x**2))))
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (103) = 206\).
Time = 0.27 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.98 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx=\frac {1}{16} \, b {\left (\frac {2 \, {\left (15 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} c^{2} x^{4} - 25 \, {\left (c + \frac {d}{x^{2}}\right )} c^{2} d x^{2} + 8 \, c^{2} d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{3} x^{5} - 2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{4} x^{3} + \sqrt {c + \frac {d}{x^{2}}} d^{5} x} + \frac {15 \, c^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {7}{2}}}\right )} - \frac {1}{4} \, a {\left (\frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )} c x^{2} - 2 \, c d\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} x^{3} - \sqrt {c + \frac {d}{x^{2}}} d^{3} x} + \frac {3 \, c \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {5}{2}}}\right )} \]
1/16*b*(2*(15*(c + d/x^2)^2*c^2*x^4 - 25*(c + d/x^2)*c^2*d*x^2 + 8*c^2*d^2 )/((c + d/x^2)^(5/2)*d^3*x^5 - 2*(c + d/x^2)^(3/2)*d^4*x^3 + sqrt(c + d/x^ 2)*d^5*x) + 15*c^2*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(7/2)) - 1/4*a*(2*(3*(c + d/x^2)*c*x^2 - 2*c*d)/((c + d/x^2)^( 3/2)*d^2*x^3 - sqrt(c + d/x^2)*d^3*x) + 3*c*log((sqrt(c + d/x^2)*x - sqrt( d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(5/2))
Time = 0.31 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.20 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx=\frac {3 \, {\left (5 \, b c^{2} - 4 \, a c d\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{8 \, \sqrt {-d} d^{3} \mathrm {sgn}\left (x\right )} + \frac {b c^{2} - a c d}{\sqrt {c x^{2} + d} d^{3} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{2} - 4 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a c d - 9 \, \sqrt {c x^{2} + d} b c^{2} d + 4 \, \sqrt {c x^{2} + d} a c d^{2}}{8 \, c^{2} d^{3} x^{4} \mathrm {sgn}\left (x\right )} \]
3/8*(5*b*c^2 - 4*a*c*d)*arctan(sqrt(c*x^2 + d)/sqrt(-d))/(sqrt(-d)*d^3*sgn (x)) + (b*c^2 - a*c*d)/(sqrt(c*x^2 + d)*d^3*sgn(x)) + 1/8*(7*(c*x^2 + d)^( 3/2)*b*c^2 - 4*(c*x^2 + d)^(3/2)*a*c*d - 9*sqrt(c*x^2 + d)*b*c^2*d + 4*sqr t(c*x^2 + d)*a*c*d^2)/(c^2*d^3*x^4*sgn(x))
Timed out. \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx=\int \frac {a+\frac {b}{x^2}}{x^6\,{\left (c+\frac {d}{x^2}\right )}^{3/2}} \,d x \]