Integrand size = 25, antiderivative size = 295 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {3 a \sqrt {1-a^2 x^2}}{4 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {23 a \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}+\frac {7 a \sqrt {1-a^2 x^2} \log (1+a x)}{16 c^2 \sqrt {c-a^2 c x^2}} \]
-(-a^2*x^2+1)^(1/2)/c^2/x/(-a^2*c*x^2+c)^(1/2)+1/8*a*(-a^2*x^2+1)^(1/2)/c^ 2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)+3/4*a*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)/(- a^2*c*x^2+c)^(1/2)-1/8*a*(-a^2*x^2+1)^(1/2)/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/ 2)+a*ln(x)*(-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2+c)^(1/2)-23/16*a*ln(-a*x+1)* (-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2+c)^(1/2)+7/16*a*ln(a*x+1)*(-a^2*x^2+1)^ (1/2)/c^2/(-a^2*c*x^2+c)^(1/2)
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.33 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-\frac {16}{x}+\frac {12 a}{1-a x}+\frac {2 a}{(-1+a x)^2}-\frac {2 a}{1+a x}+16 a \log (x)-23 a \log (1-a x)+7 a \log (1+a x)\right )}{16 c^2 \sqrt {c-a^2 c x^2}} \]
(Sqrt[1 - a^2*x^2]*(-16/x + (12*a)/(1 - a*x) + (2*a)/(-1 + a*x)^2 - (2*a)/ (1 + a*x) + 16*a*Log[x] - 23*a*Log[1 - a*x] + 7*a*Log[1 + a*x]))/(16*c^2*S qrt[c - a^2*c*x^2])
Time = 0.51 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.35, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{x^2 (1-a x)^3 (a x+1)^2}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {23 a^2}{16 (a x-1)}+\frac {7 a^2}{16 (a x+1)}+\frac {3 a^2}{4 (a x-1)^2}+\frac {a^2}{8 (a x+1)^2}-\frac {a^2}{4 (a x-1)^3}+\frac {a}{x}+\frac {1}{x^2}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {3 a}{4 (1-a x)}-\frac {a}{8 (a x+1)}+\frac {a}{8 (1-a x)^2}+a \log (x)-\frac {23}{16} a \log (1-a x)+\frac {7}{16} a \log (a x+1)-\frac {1}{x}\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(-x^(-1) + a/(8*(1 - a*x)^2) + (3*a)/(4*(1 - a*x)) - a/ (8*(1 + a*x)) + a*Log[x] - (23*a*Log[1 - a*x])/16 + (7*a*Log[1 + a*x])/16) )/(c^2*Sqrt[c - a^2*c*x^2])
3.10.84.3.1 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_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar t[p]) Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && !I ntegerQ[n/2]
Time = 0.16 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.75
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (7 \ln \left (a x +1\right ) x^{4} a^{4}+16 \ln \left (x \right ) x^{4} a^{4}-23 \ln \left (a x -1\right ) x^{4} a^{4}-7 a^{3} \ln \left (a x +1\right ) x^{3}-16 a^{3} \ln \left (x \right ) x^{3}+23 a^{3} \ln \left (a x -1\right ) x^{3}-30 a^{3} x^{3}-7 a^{2} \ln \left (a x +1\right ) x^{2}-16 a^{2} \ln \left (x \right ) x^{2}+23 a^{2} \ln \left (a x -1\right ) x^{2}+22 a^{2} x^{2}+7 a \ln \left (a x +1\right ) x +16 a \ln \left (x \right ) x -23 a \ln \left (a x -1\right ) x +28 a x -16\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} \left (a x -1\right )^{2} x \left (a x +1\right )}\) | \(222\) |
-1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(7*ln(a*x+1)*x^4*a^4+16*ln (x)*x^4*a^4-23*ln(a*x-1)*x^4*a^4-7*a^3*ln(a*x+1)*x^3-16*a^3*ln(x)*x^3+23*a ^3*ln(a*x-1)*x^3-30*a^3*x^3-7*a^2*ln(a*x+1)*x^2-16*a^2*ln(x)*x^2+23*a^2*ln (a*x-1)*x^2+22*a^2*x^2+7*a*ln(a*x+1)*x+16*a*ln(x)*x-23*a*ln(a*x-1)*x+28*a* x-16)/(a^2*x^2-1)/c^3/(a*x-1)^2/x/(a*x+1)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x^{2}} \,d x } \]
integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^7*c^3*x^9 - a^6*c^3*x^ 8 - 3*a^5*c^3*x^7 + 3*a^4*c^3*x^6 + 3*a^3*c^3*x^5 - 3*a^2*c^3*x^4 - a*c^3* x^3 + c^3*x^2), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {a x + 1}{x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x^{2}} \,d x } \]
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x^{2}} \,d x } \]
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {a\,x+1}{x^2\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \]