Integrand size = 22, antiderivative size = 361 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {\left (1-a^2 x^2\right )^{7/2}}{a^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^6}+\frac {\left (1-a^2 x^2\right )^{7/2}}{24 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)^3}-\frac {11 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)^2}+\frac {3 \left (1-a^2 x^2\right )^{7/2}}{2 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)^2}-\frac {5 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)}+\frac {51 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7}-\frac {19 \left (1-a^2 x^2\right )^{7/2} \log (1+a x)}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7} \] Output:
(-a^2*x^2+1)^(7/2)/a^7/(c-c/a^2/x^2)^(7/2)/x^6+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)+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)+51/32*(-a^2*x^ 2+1)^(7/2)*ln(-a*x+1)/a^8/(c-c/a^2/x^2)^(7/2)/x^7-19/32*(-a^2*x^2+1)^(7/2) *ln(a*x+1)/a^8/(c-c/a^2/x^2)^(7/2)/x^7
Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.41 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (88+122 a x-338 a^2 x^2-222 a^3 x^3+366 a^4 x^4+96 a^5 x^5-96 a^6 x^6-153 (-1+a x)^3 (1+a x)^2 \log (1-a x)+57 (-1+a x)^3 (1+a x)^2 \log (1+a x)\right )}{96 a^2 c^3 \sqrt {c-\frac {c}{a^2 x^2}} x (-1+a x)^3 (1+a x)^2} \] Input:
Integrate[E^ArcTanh[a*x]/(c - c/(a^2*x^2))^(7/2),x]
Output:
(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 - 153*(-1 + a*x)^3*(1 + a*x)^2*Log[1 - a*x] + 5 7*(-1 + a*x)^3*(1 + a*x)^2*Log[1 + a*x]))/(96*a^2*c^3*Sqrt[c - c/(a^2*x^2) ]*x*(-1 + a*x)^3*(1 + a*x)^2)
Time = 0.52 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, 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 |
\(\displaystyle \frac {\left (1-a^2 x^2\right )^{7/2} \int \frac {x^7}{(1-a x)^4 (a x+1)^3}dx}{x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\left (1-a^2 x^2\right )^{7/2} \int \left (-\frac {19}{32 a^7 (a x+1)}+\frac {5}{16 a^7 (a x+1)^2}-\frac {1}{16 a^7 (a x+1)^3}+\frac {1}{a^7}+\frac {51}{32 a^7 (a x-1)}+\frac {3}{2 a^7 (a x-1)^2}+\frac {11}{16 a^7 (a x-1)^3}+\frac {1}{8 a^7 (a x-1)^4}\right )dx}{x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (1-a^2 x^2\right )^{7/2} \left (\frac {3}{2 a^8 (1-a x)}-\frac {5}{16 a^8 (a x+1)}-\frac {11}{32 a^8 (1-a x)^2}+\frac {1}{32 a^8 (a x+1)^2}+\frac {1}{24 a^8 (1-a x)^3}+\frac {51 \log (1-a x)}{32 a^8}-\frac {19 \log (a x+1)}{32 a^8}+\frac {x}{a^7}\right )}{x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}\) |
Input:
Int[E^ArcTanh[a*x]/(c - c/(a^2*x^2))^(7/2),x]
Output:
((1 - a^2*x^2)^(7/2)*(x/a^7 + 1/(24*a^8*(1 - a*x)^3) - 11/(32*a^8*(1 - a*x )^2) + 3/(2*a^8*(1 - a*x)) + 1/(32*a^8*(1 + a*x)^2) - 5/(16*a^8*(1 + a*x)) + (51*Log[1 - a*x])/(32*a^8) - (19*Log[1 + a*x])/(32*a^8)))/((c - c/(a^2* x^2))^(7/2)*x^7)
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 = 0.19 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.69
method | result | size |
default | \(\frac {\left (-96 x^{6} a^{6}+57 \ln \left (a x +1\right ) x^{5} a^{5}-153 \ln \left (a x -1\right ) x^{5} a^{5}+96 a^{5} x^{5}-57 \ln \left (a x +1\right ) x^{4} a^{4}+153 \ln \left (a x -1\right ) x^{4} a^{4}+366 a^{4} x^{4}-114 \ln \left (a x +1\right ) x^{3} a^{3}+306 a^{3} \ln \left (a x -1\right ) x^{3}-222 a^{3} x^{3}+114 \ln \left (a x +1\right ) x^{2} a^{2}-306 a^{2} \ln \left (a x -1\right ) x^{2}-338 a^{2} x^{2}+57 \ln \left (a x +1\right ) x a -153 a \ln \left (a x -1\right ) x +122 a x -57 \ln \left (a x +1\right )+153 \ln \left (a x -1\right )+88\right ) \left (a x +1\right ) \left (a^{2} x^{2}-1\right )^{2}}{96 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} a^{8} x^{7} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}}}\) | \(250\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(7/2),x,method=_RETURNVERBOSE )
Output:
1/96*(-96*x^6*a^6+57*ln(a*x+1)*x^5*a^5-153*ln(a*x-1)*x^5*a^5+96*a^5*x^5-57 *ln(a*x+1)*x^4*a^4+153*ln(a*x-1)*x^4*a^4+366*a^4*x^4-114*ln(a*x+1)*x^3*a^3 +306*a^3*ln(a*x-1)*x^3-222*a^3*x^3+114*ln(a*x+1)*x^2*a^2-306*a^2*ln(a*x-1) *x^2-338*a^2*x^2+57*ln(a*x+1)*x*a-153*a*ln(a*x-1)*x+122*a*x-57*ln(a*x+1)+1 53*ln(a*x-1)+88)*(a*x+1)*(a^2*x^2-1)^2/(-a^2*x^2+1)^(3/2)/a^8/x^7/(c*(a^2* x^2-1)/a^2/x^2)^(7/2)
\[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int { \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(7/2),x, algorithm="fri cas")
Output:
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)
\[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int \frac {a x + 1}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/(c-c/a**2/x**2)**(7/2),x)
Output:
Integral((a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(-1 + 1/(a*x))*(1 + 1/( a*x)))**(7/2)), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int { \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(7/2),x, algorithm="max ima")
Output:
integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(c - c/(a^2*x^2))^(7/2)), x)
Exception generated. \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(7/2),x, algorithm="gia c")
Output:
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. \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\int \frac {a\,x+1}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{7/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((a*x + 1)/((c - c/(a^2*x^2))^(7/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((a*x + 1)/((c - c/(a^2*x^2))^(7/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.14 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx=\frac {\sqrt {c}\, i \left (153 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}-153 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-306 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+306 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+153 \,\mathrm {log}\left (a x -1\right ) a x -153 \,\mathrm {log}\left (a x -1\right )-57 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}+57 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+114 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-114 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-57 \,\mathrm {log}\left (a x +1\right ) a x +57 \,\mathrm {log}\left (a x +1\right )+96 a^{6} x^{6}-462 a^{5} x^{5}+954 a^{3} x^{3}-394 a^{2} x^{2}-488 a x +278\right )}{96 a \,c^{4} \left (a^{5} x^{5}-a^{4} x^{4}-2 a^{3} x^{3}+2 a^{2} x^{2}+a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(7/2),x)
Output:
(sqrt(c)*i*(153*log(a*x - 1)*a**5*x**5 - 153*log(a*x - 1)*a**4*x**4 - 306* log(a*x - 1)*a**3*x**3 + 306*log(a*x - 1)*a**2*x**2 + 153*log(a*x - 1)*a*x - 153*log(a*x - 1) - 57*log(a*x + 1)*a**5*x**5 + 57*log(a*x + 1)*a**4*x** 4 + 114*log(a*x + 1)*a**3*x**3 - 114*log(a*x + 1)*a**2*x**2 - 57*log(a*x + 1)*a*x + 57*log(a*x + 1) + 96*a**6*x**6 - 462*a**5*x**5 + 954*a**3*x**3 - 394*a**2*x**2 - 488*a*x + 278))/(96*a*c**4*(a**5*x**5 - a**4*x**4 - 2*a** 3*x**3 + 2*a**2*x**2 + a*x - 1))