Integrand size = 24, antiderivative size = 218 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}{4 \left (1-a^2 x^2\right )^{5/2}}+\frac {a \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}{\left (1-a^2 x^2\right )^{5/2}}-\frac {a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}{\left (1-a^2 x^2\right )^{5/2}}-\frac {2 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}{\left (1-a^2 x^2\right )^{5/2}}+\frac {a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^6}{\left (1-a^2 x^2\right )^{5/2}}-\frac {3 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (1-a^2 x^2\right )^{5/2}} \]
-1/4*(c-c/a^2/x^2)^(5/2)*x/(-a^2*x^2+1)^(5/2)+a*(c-c/a^2/x^2)^(5/2)*x^2/(- a^2*x^2+1)^(5/2)-a^2*(c-c/a^2/x^2)^(5/2)*x^3/(-a^2*x^2+1)^(5/2)-2*a^3*(c-c /a^2/x^2)^(5/2)*x^4/(-a^2*x^2+1)^(5/2)+a^5*(c-c/a^2/x^2)^(5/2)*x^6/(-a^2*x ^2+1)^(5/2)-3*a^4*(c-c/a^2/x^2)^(5/2)*x^5*ln(x)/(-a^2*x^2+1)^(5/2)
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.42 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 \left (-\frac {5 a^4}{4}-\frac {1}{4 x^4}+\frac {a}{x^3}-\frac {a^2}{x^2}-\frac {2 a^3}{x}+a^5 x-3 a^4 \log (x)\right )}{\sqrt {1-a^2 x^2} \left (-1+a^2 x^2\right )^2} \]
((c - c/(a^2*x^2))^(5/2)*x^5*((-5*a^4)/4 - 1/(4*x^4) + a/x^3 - a^2/x^2 - ( 2*a^3)/x + a^5*x - 3*a^4*Log[x]))/(Sqrt[1 - a^2*x^2]*(-1 + a^2*x^2)^2)
Time = 0.47 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.34, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6710, 6700, 84, 2009}
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.
| \(\displaystyle \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx\) |
| \(\Big \downarrow \) 6710 |
| \(\displaystyle \frac {x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \int \frac {e^{-3 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^{5/2}}{x^5}dx}{\left (1-a^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \int \frac {(1-a x)^4 (a x+1)}{x^5}dx}{\left (1-a^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 84 |
| \(\displaystyle \frac {x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \int \left (a^5-\frac {3 a^4}{x}+\frac {2 a^3}{x^2}+\frac {2 a^2}{x^3}-\frac {3 a}{x^4}+\frac {1}{x^5}\right )dx}{\left (1-a^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \left (a^5 x-3 a^4 \log (x)-\frac {2 a^3}{x}-\frac {a^2}{x^2}+\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{\left (1-a^2 x^2\right )^{5/2}}\) |
((c - c/(a^2*x^2))^(5/2)*x^5*(-1/4*1/x^4 + a/x^3 - a^2/x^2 - (2*a^3)/x + a ^5*x - 3*a^4*Log[x]))/(1 - a^2*x^2)^(5/2)
3.8.34.3.1 Defintions of rubi rules used
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[x^(2*p)*((c + d/x^2)^p/(1 - a^2*x^2)^p) Int[(u/x^(2*p))*(1 - a ^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[p] && !IntegerQ[n/2]
Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.39
| method | result | size |
| default | \(\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}} x \sqrt {-a^{2} x^{2}+1}\, \left (-4 a^{5} x^{5}+12 \ln \left (x \right ) x^{4} a^{4}+8 a^{3} x^{3}+4 a^{2} x^{2}-4 a x +1\right )}{4 \left (a^{2} x^{2}-1\right )^{3}}\) | \(86\) |
1/4*(c*(a^2*x^2-1)/a^2/x^2)^(5/2)*x/(a^2*x^2-1)^3*(-a^2*x^2+1)^(1/2)*(-4*a ^5*x^5+12*ln(x)*x^4*a^4+8*a^3*x^3+4*a^2*x^2-4*a*x+1)
Time = 0.29 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.20 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\left [\frac {6 \, {\left (a^{5} c^{2} x^{5} - a^{3} c^{2} x^{3}\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} - {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - {\left (4 \, a^{5} c^{2} x^{5} - 8 \, a^{3} c^{2} x^{3} - {\left (4 \, a^{5} - 8 \, a^{3} - 4 \, a^{2} + 4 \, a - 1\right )} c^{2} x^{4} - 4 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x - c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{6} x^{5} - a^{4} x^{3}\right )}}, \frac {12 \, {\left (a^{5} c^{2} x^{5} - a^{3} c^{2} x^{3}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} + a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} - {\left (a^{2} + 1\right )} c x^{2} + c}\right ) - {\left (4 \, a^{5} c^{2} x^{5} - 8 \, a^{3} c^{2} x^{3} - {\left (4 \, a^{5} - 8 \, a^{3} - 4 \, a^{2} + 4 \, a - 1\right )} c^{2} x^{4} - 4 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x - c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{6} x^{5} - a^{4} x^{3}\right )}}\right ] \]
[1/4*(6*(a^5*c^2*x^5 - a^3*c^2*x^3)*sqrt(-c)*log((a^2*c*x^6 + a^2*c*x^2 - c*x^4 - (a*x^5 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^ 2*x^2)) - c)/(a^2*x^4 - x^2)) - (4*a^5*c^2*x^5 - 8*a^3*c^2*x^3 - (4*a^5 - 8*a^3 - 4*a^2 + 4*a - 1)*c^2*x^4 - 4*a^2*c^2*x^2 + 4*a*c^2*x - c^2)*sqrt(- a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^6*x^5 - a^4*x^3), 1/4*(12 *(a^5*c^2*x^5 - a^3*c^2*x^3)*sqrt(c)*arctan(sqrt(-a^2*x^2 + 1)*(a*x^3 + a* x)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^4 - (a^2 + 1)*c*x^2 + c)) - (4*a^5*c^2*x^5 - 8*a^3*c^2*x^3 - (4*a^5 - 8*a^3 - 4*a^2 + 4*a - 1)*c ^2*x^4 - 4*a^2*c^2*x^2 + 4*a*c^2*x - c^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x ^2 - c)/(a^2*x^2)))/(a^6*x^5 - a^4*x^3)]
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}}}{\left (a x + 1\right )^{3}}\, dx \]
Integral((-(a*x - 1)*(a*x + 1))**(3/2)*(-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))** (5/2)/(a*x + 1)**3, x)
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}{{\left (a x + 1\right )}^{3}} \,d x } \]
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}{{\left (a x + 1\right )}^{3}} \,d x } \]
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \]