Integrand size = 24, antiderivative size = 361 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=-\frac {\left (1-a^2 x^2\right )^{7/2}}{a^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^6}+\frac {\left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)^2}-\frac {5 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{7/2}}{24 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)^3}-\frac {11 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)^2}+\frac {3 \left (1-a^2 x^2\right )^{7/2}}{2 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)}-\frac {19 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7}+\frac {51 \left (1-a^2 x^2\right )^{7/2} \log (1+a x)}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7} \]
-(-a^2*x^2+1)^(7/2)/a^7/(c-c/a^2/x^2)^(7/2)/x^6+1/32*(-a^2*x^2+1)^(7/2)/a^ 8/(c-c/a^2/x^2)^(7/2)/x^7/(-a*x+1)^2-5/16*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/ x^2)^(7/2)/x^7/(-a*x+1)+1/24*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/x^2)^(7/2)/x^ 7/(a*x+1)^3-11/32*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/x^2)^(7/2)/x^7/(a*x+1)^2 +3/2*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/x^2)^(7/2)/x^7/(a*x+1)-19/32*(-a^2*x^ 2+1)^(7/2)*ln(-a*x+1)/a^8/(c-c/a^2/x^2)^(7/2)/x^7+51/32*(-a^2*x^2+1)^(7/2) *ln(a*x+1)/a^8/(c-c/a^2/x^2)^(7/2)/x^7
Time = 0.09 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.40 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-88+122 a x+338 a^2 x^2-222 a^3 x^3-366 a^4 x^4+96 a^5 x^5+96 a^6 x^6+57 (-1+a x)^2 (1+a x)^3 \log (1-a x)-153 (-1+a x)^2 (1+a x)^3 \log (1+a x)\right )}{96 a^2 \sqrt {c-\frac {c}{a^2 x^2}} x (-1+a x)^2 (c+a c x)^3} \]
(Sqrt[1 - a^2*x^2]*(-88 + 122*a*x + 338*a^2*x^2 - 222*a^3*x^3 - 366*a^4*x^ 4 + 96*a^5*x^5 + 96*a^6*x^6 + 57*(-1 + a*x)^2*(1 + a*x)^3*Log[1 - a*x] - 1 53*(-1 + a*x)^2*(1 + a*x)^3*Log[1 + a*x]))/(96*a^2*Sqrt[c - c/(a^2*x^2)]*x *(-1 + a*x)^2*(c + a*c*x)^3)
Time = 0.52 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.39, 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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 6710 |
\(\displaystyle \frac {\left (1-a^2 x^2\right )^{7/2} \int \frac {e^{-\text {arctanh}(a x)} x^7}{\left (1-a^2 x^2\right )^{7/2}}dx}{x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\left (1-a^2 x^2\right )^{7/2} \int \frac {x^7}{(1-a x)^3 (a x+1)^4}dx}{x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\left (1-a^2 x^2\right )^{7/2} \int \left (\frac {51}{32 a^7 (a x+1)}-\frac {3}{2 a^7 (a x+1)^2}+\frac {11}{16 a^7 (a x+1)^3}-\frac {1}{8 a^7 (a x+1)^4}-\frac {1}{a^7}-\frac {19}{32 a^7 (a x-1)}-\frac {5}{16 a^7 (a x-1)^2}-\frac {1}{16 a^7 (a x-1)^3}\right )dx}{x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (1-a^2 x^2\right )^{7/2} \left (-\frac {5}{16 a^8 (1-a x)}+\frac {3}{2 a^8 (a x+1)}+\frac {1}{32 a^8 (1-a x)^2}-\frac {11}{32 a^8 (a x+1)^2}+\frac {1}{24 a^8 (a x+1)^3}-\frac {19 \log (1-a x)}{32 a^8}+\frac {51 \log (a x+1)}{32 a^8}-\frac {x}{a^7}\right )}{x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}\) |
((1 - a^2*x^2)^(7/2)*(-(x/a^7) + 1/(32*a^8*(1 - a*x)^2) - 5/(16*a^8*(1 - a *x)) + 1/(24*a^8*(1 + a*x)^3) - 11/(32*a^8*(1 + a*x)^2) + 3/(2*a^8*(1 + a* x)) - (19*Log[1 - a*x])/(32*a^8) + (51*Log[1 + a*x])/(32*a^8)))/((c - c/(a ^2*x^2))^(7/2)*x^7)
3.8.22.3.1 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 = 239, normalized size of antiderivative = 0.66
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right ) \left (-96 a^{6} x^{6}+153 \ln \left (a x +1\right ) x^{5} a^{5}-57 \ln \left (a x -1\right ) x^{5} a^{5}-96 a^{5} x^{5}+153 \ln \left (a x +1\right ) x^{4} a^{4}-57 \ln \left (a x -1\right ) x^{4} a^{4}+366 a^{4} x^{4}-306 a^{3} \ln \left (a x +1\right ) x^{3}+114 a^{3} \ln \left (a x -1\right ) x^{3}+222 a^{3} x^{3}-306 a^{2} \ln \left (a x +1\right ) x^{2}+114 a^{2} \ln \left (a x -1\right ) x^{2}-338 a^{2} x^{2}+153 a \ln \left (a x +1\right ) x -57 a \ln \left (a x -1\right ) x -122 a x +153 \ln \left (a x +1\right )-57 \ln \left (a x -1\right )+88\right )}{96 a^{8} x^{7} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}}}\) | \(239\) |
-1/96*(-a^2*x^2+1)^(1/2)*(a*x-1)*(-96*a^6*x^6+153*ln(a*x+1)*x^5*a^5-57*ln( a*x-1)*x^5*a^5-96*a^5*x^5+153*ln(a*x+1)*x^4*a^4-57*ln(a*x-1)*x^4*a^4+366*a ^4*x^4-306*a^3*ln(a*x+1)*x^3+114*a^3*ln(a*x-1)*x^3+222*a^3*x^3-306*a^2*ln( a*x+1)*x^2+114*a^2*ln(a*x-1)*x^2-338*a^2*x^2+153*a*ln(a*x+1)*x-57*a*ln(a*x -1)*x-122*a*x+153*ln(a*x+1)-57*ln(a*x-1)+88)/a^8/x^7/(c*(a^2*x^2-1)/a^2/x^ 2)^(7/2)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}}} \,d x } \]
integral(sqrt(-a^2*x^2 + 1)*a^8*x^8*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^9*c ^4*x^9 + a^8*c^4*x^8 - 4*a^7*c^4*x^7 - 4*a^6*c^4*x^6 + 6*a^5*c^4*x^5 + 6*a ^4*c^4*x^4 - 4*a^3*c^4*x^3 - 4*a^2*c^4*x^2 + a*c^4*x + c^4), x)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {7}{2}} \left (a x + 1\right )}\, dx \]
Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))**(7 /2)*(a*x + 1)), x)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}}} \,d x } \]
Exception generated. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{7/2}\,\left (a\,x+1\right )} \,d x \]