Integrand size = 24, antiderivative size = 299 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{9/2} x}{8 \left (1-a^2 x^2\right )^{9/2}}+\frac {3 a \left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^2}{7 \left (1-a^2 x^2\right )^{9/2}}-\frac {8 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^4}{5 \left (1-a^2 x^2\right )^{9/2}}+\frac {3 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^5}{2 \left (1-a^2 x^2\right )^{9/2}}+\frac {2 a^5 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^6}{\left (1-a^2 x^2\right )^{9/2}}-\frac {4 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^7}{\left (1-a^2 x^2\right )^{9/2}}+\frac {a^9 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^{10}}{\left (1-a^2 x^2\right )^{9/2}}-\frac {3 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^9 \log (x)}{\left (1-a^2 x^2\right )^{9/2}} \]
-1/8*(c-c/a^2/x^2)^(9/2)*x/(-a^2*x^2+1)^(9/2)+3/7*a*(c-c/a^2/x^2)^(9/2)*x^ 2/(-a^2*x^2+1)^(9/2)-8/5*a^3*(c-c/a^2/x^2)^(9/2)*x^4/(-a^2*x^2+1)^(9/2)+3/ 2*a^4*(c-c/a^2/x^2)^(9/2)*x^5/(-a^2*x^2+1)^(9/2)+2*a^5*(c-c/a^2/x^2)^(9/2) *x^6/(-a^2*x^2+1)^(9/2)-4*a^6*(c-c/a^2/x^2)^(9/2)*x^7/(-a^2*x^2+1)^(9/2)+a ^9*(c-c/a^2/x^2)^(9/2)*x^10/(-a^2*x^2+1)^(9/2)-3*a^8*(c-c/a^2/x^2)^(9/2)*x ^9*ln(x)/(-a^2*x^2+1)^(9/2)
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.32 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{9/2} x^9 \left (-\frac {1}{8 x^8}+\frac {3 a}{7 x^7}-\frac {8 a^3}{5 x^5}+\frac {3 a^4}{2 x^4}+\frac {2 a^5}{x^3}-\frac {4 a^6}{x^2}+a^9 x-3 a^8 \log (x)\right )}{\left (1-a^2 x^2\right )^{9/2}} \]
((c - c/(a^2*x^2))^(9/2)*x^9*(-1/8*1/x^8 + (3*a)/(7*x^7) - (8*a^3)/(5*x^5) + (3*a^4)/(2*x^4) + (2*a^5)/x^3 - (4*a^6)/x^2 + a^9*x - 3*a^8*Log[x]))/(1 - a^2*x^2)^(9/2)
Time = 0.47 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.32, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6710, 6700, 99, 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.
| \(\displaystyle \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx\) |
| \(\Big \downarrow \) 6710 |
| \(\displaystyle \frac {x^9 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \int \frac {e^{-3 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^{9/2}}{x^9}dx}{\left (1-a^2 x^2\right )^{9/2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {x^9 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \int \frac {(1-a x)^6 (a x+1)^3}{x^9}dx}{\left (1-a^2 x^2\right )^{9/2}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {x^9 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \int \left (a^9-\frac {3 a^8}{x}+\frac {8 a^6}{x^3}-\frac {6 a^5}{x^4}-\frac {6 a^4}{x^5}+\frac {8 a^3}{x^6}-\frac {3 a}{x^8}+\frac {1}{x^9}\right )dx}{\left (1-a^2 x^2\right )^{9/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^9 \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \left (a^9 x-3 a^8 \log (x)-\frac {4 a^6}{x^2}+\frac {2 a^5}{x^3}+\frac {3 a^4}{2 x^4}-\frac {8 a^3}{5 x^5}+\frac {3 a}{7 x^7}-\frac {1}{8 x^8}\right )}{\left (1-a^2 x^2\right )^{9/2}}\) |
((c - c/(a^2*x^2))^(9/2)*x^9*(-1/8*1/x^8 + (3*a)/(7*x^7) - (8*a^3)/(5*x^5) + (3*a^4)/(2*x^4) + (2*a^5)/x^3 - (4*a^6)/x^2 + a^9*x - 3*a^8*Log[x]))/(1 - a^2*x^2)^(9/2)
3.8.32.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 102, normalized size of antiderivative = 0.34
| method | result | size |
| default | \(\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {9}{2}} x \sqrt {-a^{2} x^{2}+1}\, \left (-280 a^{9} x^{9}+840 a^{8} \ln \left (x \right ) x^{8}+1120 a^{6} x^{6}-560 a^{5} x^{5}-420 a^{4} x^{4}+448 a^{3} x^{3}-120 a x +35\right )}{280 \left (a^{2} x^{2}-1\right )^{5}}\) | \(102\) |
1/280*(c*(a^2*x^2-1)/a^2/x^2)^(9/2)*x/(a^2*x^2-1)^5*(-a^2*x^2+1)^(1/2)*(-2 80*a^9*x^9+840*a^8*ln(x)*x^8+1120*a^6*x^6-560*a^5*x^5-420*a^4*x^4+448*a^3* x^3-120*a*x+35)
Time = 0.29 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.82 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\left [\frac {420 \, {\left (a^{9} c^{4} x^{9} - a^{7} c^{4} x^{7}\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} - {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - {\left (280 \, a^{9} c^{4} x^{9} - 1120 \, a^{6} c^{4} x^{6} + 560 \, a^{5} c^{4} x^{5} - {\left (280 \, a^{9} - 1120 \, a^{6} + 560 \, a^{5} + 420 \, a^{4} - 448 \, a^{3} + 120 \, a - 35\right )} c^{4} x^{8} + 420 \, a^{4} c^{4} x^{4} - 448 \, a^{3} c^{4} x^{3} + 120 \, a c^{4} x - 35 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{280 \, {\left (a^{10} x^{9} - a^{8} x^{7}\right )}}, \frac {840 \, {\left (a^{9} c^{4} x^{9} - a^{7} c^{4} x^{7}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} + a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} - {\left (a^{2} + 1\right )} c x^{2} + c}\right ) - {\left (280 \, a^{9} c^{4} x^{9} - 1120 \, a^{6} c^{4} x^{6} + 560 \, a^{5} c^{4} x^{5} - {\left (280 \, a^{9} - 1120 \, a^{6} + 560 \, a^{5} + 420 \, a^{4} - 448 \, a^{3} + 120 \, a - 35\right )} c^{4} x^{8} + 420 \, a^{4} c^{4} x^{4} - 448 \, a^{3} c^{4} x^{3} + 120 \, a c^{4} x - 35 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{280 \, {\left (a^{10} x^{9} - a^{8} x^{7}\right )}}\right ] \]
[1/280*(420*(a^9*c^4*x^9 - a^7*c^4*x^7)*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)) - (280*a^9*c^4*x^9 - 1120*a^6*c^4*x^6 + 560*a^5*c^4*x^5 - (280*a^9 - 1120*a^6 + 560*a^5 + 420*a^4 - 448*a^3 + 120* a - 35)*c^4*x^8 + 420*a^4*c^4*x^4 - 448*a^3*c^4*x^3 + 120*a*c^4*x - 35*c^4 )*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^10*x^9 - a^8*x^7) , 1/280*(840*(a^9*c^4*x^9 - a^7*c^4*x^7)*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)) - (280*a^9*c^4*x^9 - 1120*a^6*c^4*x^6 + 560*a^5*c^4*x^5 - (280*a^9 - 1120*a^6 + 560*a^5 + 420*a^4 - 448*a^3 + 120*a - 35)*c^4*x^8 + 420*a^4*c^4*x^4 - 448*a^3*c^4*x^3 + 120*a*c^4*x - 35*c^4)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^10*x^9 - a^8*x^7)]
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\text {Timed out} \]
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {9}{2}}}{{\left (a x + 1\right )}^{3}} \,d x } \]
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {9}{2}}}{{\left (a x + 1\right )}^{3}} \,d x } \]
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{9/2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \]