Integrand size = 23, antiderivative size = 91 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{2 a^2 c (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2 c \sqrt {c-a^2 c x^2}} \] Output:
1/2*(-a^2*x^2+1)^(1/2)/a^2/c/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/2*(-a^2*x^2+1 )^(1/2)*arctanh(a*x)/a^2/c/(-a^2*c*x^2+c)^(1/2)
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (\frac {1}{2 a^2 (1-a x)}-\frac {\text {arctanh}(a x)}{2 a^2}\right )}{c \sqrt {c-a^2 c x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^(3/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(1/(2*a^2*(1 - a*x)) - ArcTanh[a*x]/(2*a^2)))/(c*Sqrt[c - a^2*c*x^2])
Time = 0.44 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6703, 6700, 86, 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.
\(\displaystyle \int \frac {x e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x}{(1-a x)^2 (a x+1)}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{2 \left (a^2 x^2-1\right ) a}+\frac {1}{2 (a x-1)^2 a}\right )dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {1}{2 a^2 (1-a x)}-\frac {\text {arctanh}(a x)}{2 a^2}\right )}{c \sqrt {c-a^2 c x^2}}\) |
Input:
Int[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^(3/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(1/(2*a^2*(1 - a*x)) - ArcTanh[a*x]/(2*a^2)))/(c*Sqrt[c - a^2*c*x^2])
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {\left (\ln \left (a x +1\right ) x a -a \ln \left (a x -1\right ) x -\ln \left (a x +1\right )+\ln \left (a x -1\right )+2\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{4 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right ) c^{2} a^{2}}\) | \(77\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVERB OSE)
Output:
-1/4*(ln(a*x+1)*x*a-a*ln(a*x-1)*x-ln(a*x+1)+ln(a*x-1)+2)/(-a^2*x^2+1)^(1/2 )/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)/c^2/a^2
Time = 0.10 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.82 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\left [\frac {4 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x + {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} + 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right )}{8 \, {\left (a^{5} c^{2} x^{3} - a^{4} c^{2} x^{2} - a^{3} c^{2} x + a^{2} c^{2}\right )}}, \frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x - {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right )}{4 \, {\left (a^{5} c^{2} x^{3} - a^{4} c^{2} x^{2} - a^{3} c^{2} x + a^{2} c^{2}\right )}}\right ] \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^(3/2),x, algorithm=" fricas")
Output:
[1/8*(4*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*x + (a^3*x^3 - a^2*x^2 - a*x + 1)*sqrt(c)*log(-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a^2*c*x^2 + 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3* a^4*x^4 + 3*a^2*x^2 - 1)))/(a^5*c^2*x^3 - a^4*c^2*x^2 - a^3*c^2*x + a^2*c^ 2), 1/4*(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*x - (a^3*x^3 - a^2*x^ 2 - a*x + 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*s qrt(-c)*x/(a^4*c*x^4 - c)))/(a^5*c^2*x^3 - a^4*c^2*x^2 - a^3*c^2*x + a^2*c ^2)]
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x/(-a**2*c*x**2+c)**(3/2),x)
Output:
Integral(x*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))* *(3/2)), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^(3/2),x, algorithm=" maxima")
Output:
-a*integrate(-x^2/((a^2*c^(3/2)*x^2 - c^(3/2))*(a*x + 1)*(a*x - 1)), x) - 1/2/(a^4*c^(3/2)*x^2 - a^2*c^(3/2))
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.51 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{2} c^{\frac {3}{2}}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{2} c^{\frac {3}{2}}} - \frac {1}{2 \, {\left (a x - 1\right )} a^{2} c^{\frac {3}{2}}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^(3/2),x, algorithm=" giac")
Output:
-1/4*log(abs(a*x + 1))/(a^2*c^(3/2)) + 1/4*log(abs(a*x - 1))/(a^2*c^(3/2)) - 1/2/((a*x - 1)*a^2*c^(3/2))
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x*(a*x + 1))/((c - a^2*c*x^2)^(3/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((x*(a*x + 1))/((c - a^2*c*x^2)^(3/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.60 \[ \int \frac {e^{\text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (\mathrm {log}\left (a x -1\right ) a x -\mathrm {log}\left (a x -1\right )-\mathrm {log}\left (a x +1\right ) a x +\mathrm {log}\left (a x +1\right )-2 a x \right )}{4 a^{2} c^{2} \left (a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*(log(a*x - 1)*a*x - log(a*x - 1) - log(a*x + 1)*a*x + log(a*x + 1 ) - 2*a*x))/(4*a**2*c**2*(a*x - 1))