Integrand size = 27, antiderivative size = 148 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{2 x^2 \sqrt {1-a^2 x^2}}+\frac {3 a \sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}+\frac {4 a^2 \sqrt {c-a^2 c x^2} \log (x)}{\sqrt {1-a^2 x^2}}-\frac {4 a^2 \sqrt {c-a^2 c x^2} \log (1+a x)}{\sqrt {1-a^2 x^2}} \] Output:
-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*(-a^2*c*x^2+c)^(1/2)*ln(x)/(-a^2*x^2+1)^(1/2)-4 *a^2*(-a^2*c*x^2+c)^(1/2)*ln(a*x+1)/(-a^2*x^2+1)^(1/2)
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.42 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (-\frac {1}{2 x^2}+\frac {3 a}{x}+4 a^2 \log (x)-4 a^2 \log (1+a x)\right )}{\sqrt {1-a^2 x^2}} \] Input:
Integrate[Sqrt[c - a^2*c*x^2]/(E^(3*ArcTanh[a*x])*x^3),x]
Output:
(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 = 0.48 (sec) , antiderivative size = 62, 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 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {(1-a x)^2}{x^3 (a x+1)}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (4 a^2 \log (x)-4 a^2 \log (a x+1)+\frac {3 a}{x}-\frac {1}{2 x^2}\right )}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[Sqrt[c - a^2*c*x^2]/(E^(3*ArcTanh[a*x])*x^3),x]
Output:
(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]
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 = 0.34 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.42
method | result | size |
default | \(-\frac {\left (8 \ln \left (a x +1\right ) x^{2} a^{2}-8 a^{2} \ln \left (x \right ) x^{2}-6 a x +1\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{2 \sqrt {-a^{2} x^{2}+1}\, x^{2}}\) | \(62\) |
Input:
int((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x,method=_RETURN VERBOSE)
Output:
-1/2*(8*ln(a*x+1)*x^2*a^2-8*a^2*ln(x)*x^2-6*a*x+1)/(-a^2*x^2+1)^(1/2)*(-c* (a^2*x^2-1))^(1/2)/x^2
Time = 0.15 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.02 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\left [\frac {4 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {c} \log \left (\frac {4 \, a^{5} c x^{5} + {\left (2 \, a^{6} + 4 \, a^{5} + 6 \, a^{4} + 4 \, a^{3} + a^{2}\right )} c x^{6} + {\left (4 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} - 4 \, a - 1\right )} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, a c x + {\left (4 \, a^{3} x^{3} - {\left (4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} x^{4} + 6 \, a^{2} x^{2} + 4 \, a x + 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{4} x^{6} + 2 \, a^{3} x^{5} - 2 \, a x^{3} - x^{2}}\right ) + \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (6 \, a - 1\right )} x^{2} - 6 \, a x + 1\right )}}{2 \, {\left (a^{2} x^{4} - x^{2}\right )}}, \frac {8 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {-c} \arctan \left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a^{2} + 2 \, a + 1\right )} x^{2} + 2 \, a x + 1\right )} \sqrt {-c}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + a^{2}\right )} c x^{4} + {\left (a^{2} + 2 \, a + 1\right )} c x^{2} - 2 \, a c x - c}\right ) + \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (6 \, a - 1\right )} x^{2} - 6 \, a x + 1\right )}}{2 \, {\left (a^{2} x^{4} - x^{2}\right )}}\right ] \] Input:
integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorit hm="fricas")
Output:
[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)*sq rt(-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)]
\[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{x^{3} \left (a x + 1\right )^{3}}\, dx \] Input:
integrate((-a**2*c*x**2+c)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/x**3,x)
Output:
Integral((-(a*x - 1)*(a*x + 1))**(3/2)*sqrt(-c*(a*x - 1)*(a*x + 1))/(x**3* (a*x + 1)**3), x)
\[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x^{3}} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorit hm="maxima")
Output:
integrate(sqrt(-a^2*c*x^2 + c)*(-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*x^3), x)
Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.25 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-\frac {1}{2} \, {\left (8 \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) - 8 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {6 \, a x - 1}{x^{2}}\right )} \sqrt {c} \] Input:
integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorit hm="giac")
Output:
-1/2*(8*a^2*log(abs(a*x + 1)) - 8*a^2*log(abs(x)) - (6*a*x - 1)/x^2)*sqrt( c)
Timed out. \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int \frac {\sqrt {c-a^2\,c\,x^2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^3\,{\left (a\,x+1\right )}^3} \,d x \] Input:
int(((c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(3/2))/(x^3*(a*x + 1)^3),x)
Output:
int(((c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(3/2))/(x^3*(a*x + 1)^3), x)
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.25 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\frac {\sqrt {c}\, \left (-8 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+8 \,\mathrm {log}\left (x \right ) a^{2} x^{2}+6 a x -1\right )}{2 x^{2}} \] Input:
int((-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x)
Output:
(sqrt(c)*( - 8*log(a*x + 1)*a**2*x**2 + 8*log(x)*a**2*x**2 + 6*a*x - 1))/( 2*x**2)