Integrand size = 25, antiderivative size = 255 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{2 c x^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c x \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {2 a^2 \sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {7 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}-\frac {a^2 \sqrt {1-a^2 x^2} \log (1+a x)}{4 c \sqrt {c-a^2 c x^2}} \]
-1/2*(-a^2*x^2+1)^(1/2)/c/x^2/(-a^2*c*x^2+c)^(1/2)-a*(-a^2*x^2+1)^(1/2)/c/ x/(-a^2*c*x^2+c)^(1/2)+1/2*a^2*(-a^2*x^2+1)^(1/2)/c/(-a*x+1)/(-a^2*c*x^2+c )^(1/2)+2*a^2*ln(x)*(-a^2*x^2+1)^(1/2)/c/(-a^2*c*x^2+c)^(1/2)-7/4*a^2*ln(- a*x+1)*(-a^2*x^2+1)^(1/2)/c/(-a^2*c*x^2+c)^(1/2)-1/4*a^2*ln(a*x+1)*(-a^2*x ^2+1)^(1/2)/c/(-a^2*c*x^2+c)^(1/2)
Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.36 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-\frac {2}{x^2}-\frac {4 a}{x}+\frac {2 a^2}{1-a x}+8 a^2 \log (x)-7 a^2 \log (1-a x)-a^2 \log (1+a x)\right )}{4 c \sqrt {c-a^2 c x^2}} \]
(Sqrt[1 - a^2*x^2]*(-2/x^2 - (4*a)/x + (2*a^2)/(1 - a*x) + 8*a^2*Log[x] - 7*a^2*Log[1 - a*x] - a^2*Log[1 + a*x]))/(4*c*Sqrt[c - a^2*c*x^2])
Time = 0.51 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.38, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{x^3 (1-a x)^2 (a x+1)}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {7 a^3}{4 (a x-1)}-\frac {a^3}{4 (a x+1)}+\frac {a^3}{2 (a x-1)^2}+\frac {2 a^2}{x}+\frac {a}{x^2}+\frac {1}{x^3}\right )dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac {7}{4} a^2 \log (1-a x)-\frac {1}{4} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2}\right )}{c \sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(-1/2*1/x^2 - a/x + a^2/(2*(1 - a*x)) + 2*a^2*Log[x] - (7*a^2*Log[1 - a*x])/4 - (a^2*Log[1 + a*x])/4))/(c*Sqrt[c - a^2*c*x^2])
3.10.74.3.1 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 = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.56
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (8 a^{3} \ln \left (x \right ) x^{3}-7 a^{3} \ln \left (a x -1\right ) x^{3}-a^{3} \ln \left (a x +1\right ) x^{3}-8 a^{2} \ln \left (x \right ) x^{2}+7 a^{2} \ln \left (a x -1\right ) x^{2}+a^{2} \ln \left (a x +1\right ) x^{2}-6 a^{2} x^{2}+2 a x +2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} x^{2} \left (a x -1\right )}\) | \(142\) |
-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(8*a^3*ln(x)*x^3-7*a^3*ln(a *x-1)*x^3-a^3*ln(a*x+1)*x^3-8*a^2*ln(x)*x^2+7*a^2*ln(a*x-1)*x^2+a^2*ln(a*x +1)*x^2-6*a^2*x^2+2*a*x+2)/(a^2*x^2-1)/c^2/x^2/(a*x-1)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \]
integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^5*c^2*x^8 - a^4*c^2*x ^7 - 2*a^3*c^2*x^6 + 2*a^2*c^2*x^5 + a*c^2*x^4 - c^2*x^3), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {a x + 1}{x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \]
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \]
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {a\,x+1}{x^3\,{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \]