Integrand size = 27, antiderivative size = 149 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{2 x^2 \sqrt {1-a^2 x^2}}-\frac {3 a \sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}+\frac {4 a^2 \sqrt {c-a^2 c x^2} \log (x)}{\sqrt {1-a^2 x^2}}-\frac {4 a^2 \sqrt {c-a^2 c x^2} \log (1-a x)}{\sqrt {1-a^2 x^2}} \]
-1/2*(-a^2*c*x^2+c)^(1/2)/x^2/(-a^2*x^2+1)^(1/2)-3*a*(-a^2*c*x^2+c)^(1/2)/ x/(-a^2*x^2+1)^(1/2)+4*a^2*ln(x)*(-a^2*c*x^2+c)^(1/2)/(-a^2*x^2+1)^(1/2)-4 *a^2*ln(-a*x+1)*(-a^2*c*x^2+c)^(1/2)/(-a^2*x^2+1)^(1/2)
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.42 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (-\frac {1}{2 x^2}-\frac {3 a}{x}+4 a^2 \log (x)-4 a^2 \log (1-a x)\right )}{\sqrt {1-a^2 x^2}} \]
(Sqrt[c - a^2*c*x^2]*(-1/2*1/x^2 - (3*a)/x + 4*a^2*Log[x] - 4*a^2*Log[1 - a*x]))/Sqrt[1 - a^2*x^2]
Time = 0.45 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.
\(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {1-a^2 x^2}}{x^3}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {(a x+1)^2}{x^3 (1-a x)}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \left (-\frac {4 a^3}{a x-1}+\frac {4 a^2}{x}+\frac {3 a}{x^2}+\frac {1}{x^3}\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (4 a^2 \log (x)-4 a^2 \log (1-a x)-\frac {3 a}{x}-\frac {1}{2 x^2}\right )}{\sqrt {1-a^2 x^2}}\) |
(Sqrt[c - a^2*c*x^2]*(-1/2*1/x^2 - (3*a)/x + 4*a^2*Log[x] - 4*a^2*Log[1 - a*x]))/Sqrt[1 - a^2*x^2]
3.12.62.3.1 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 = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.49
method | result | size |
default | \(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (8 a^{2} \ln \left (x \right ) x^{2}-8 a^{2} \ln \left (a x -1\right ) x^{2}-6 a x -1\right )}{2 \left (a^{2} x^{2}-1\right ) x^{2}}\) | \(73\) |
-1/2*(-c*(a^2*x^2-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(8*a^2*ln(x)*x^2-8*a^2*ln(a *x-1)*x^2-6*a*x-1)/(a^2*x^2-1)/x^2
Time = 0.29 (sec) , antiderivative size = 450, normalized size of antiderivative = 3.02 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\left [\frac {4 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {c} \log \left (-\frac {4 \, a^{5} c x^{5} - {\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} - {\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x + {\left (4 \, a^{3} x^{3} - {\left (4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} x^{4} - 6 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) - \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (6 \, a + 1\right )} x^{2} - 6 \, a x - 1\right )}}{2 \, {\left (a^{2} x^{4} - x^{2}\right )}}, -\frac {8 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {-c} \arctan \left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a^{2} - 2 \, a + 1\right )} x^{2} - 2 \, a x + 1\right )} \sqrt {-c}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - a^{2}\right )} c x^{4} - {\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right ) + \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (6 \, a + 1\right )} x^{2} - 6 \, a x - 1\right )}}{2 \, {\left (a^{2} x^{4} - x^{2}\right )}}\right ] \]
[1/2*(4*(a^4*x^4 - a^2*x^2)*sqrt(c)*log(-(4*a^5*c*x^5 - (2*a^6 - 4*a^5 + 6 *a^4 - 4*a^3 + a^2)*c*x^6 - (4*a^4 + 4*a^3 - 6*a^2 + 4*a - 1)*c*x^4 + 5*a^ 2*c*x^2 - 4*a*c*x + (4*a^3*x^3 - (4*a^3 - 6*a^2 + 4*a - 1)*x^4 - 6*a^2*x^2 + 4*a*x - 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) + c)/(a^4*x^ 6 - 2*a^3*x^5 + 2*a*x^3 - x^2)) - sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)* ((6*a + 1)*x^2 - 6*a*x - 1))/(a^2*x^4 - x^2), -1/2*(8*(a^4*x^4 - a^2*x^2)* sqrt(-c)*arctan(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((2*a^2 - 2*a + 1 )*x^2 - 2*a*x + 1)*sqrt(-c)/(2*a^3*c*x^3 - (2*a^3 - a^2)*c*x^4 - (a^2 - 2* a + 1)*c*x^2 - 2*a*c*x + c)) + sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((6 *a + 1)*x^2 - 6*a*x - 1))/(a^2*x^4 - x^2)]
\[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int \frac {\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )^{3}}{x^{3} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.07 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-2 \, \left (-1\right )^{-2 \, a^{2} c x^{2} + 2 \, c} a^{2} \sqrt {c} \log \left (-2 \, a^{2} c + \frac {2 \, c}{x^{2}}\right ) + \frac {1}{2} \, a^{3} {\left (\frac {\sqrt {c} \log \left (a x + 1\right )}{a} - \frac {\sqrt {c} \log \left (a x - 1\right )}{a}\right )} + \frac {a^{2} c}{2 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {3}{2} \, {\left (a \sqrt {c} \log \left (a x + 1\right ) - a \sqrt {c} \log \left (a x - 1\right ) - \frac {2 \, \sqrt {c}}{x}\right )} a - \frac {c}{2 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} x^{2}} \]
-2*(-1)^(-2*a^2*c*x^2 + 2*c)*a^2*sqrt(c)*log(-2*a^2*c + 2*c/x^2) + 1/2*a^3 *(sqrt(c)*log(a*x + 1)/a - sqrt(c)*log(a*x - 1)/a) + 1/2*a^2*c/sqrt(a^4*c* x^4 - 2*a^2*c*x^2 + c) + 3/2*(a*sqrt(c)*log(a*x + 1) - a*sqrt(c)*log(a*x - 1) - 2*sqrt(c)/x)*a - 1/2*c/(sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*x^2)
\[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
Timed out. \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int \frac {\sqrt {c-a^2\,c\,x^2}\,{\left (a\,x+1\right )}^3}{x^3\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]