Integrand size = 24, antiderivative size = 90 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{2 a c (1+a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a c \sqrt {c-a^2 c x^2}} \] Output:
-1/2*(-a^2*x^2+1)^(1/2)/a/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/c/(-a^2*c*x^2+c)^(1/2)
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-\frac {1}{2 a (1+a x)}+\frac {\text {arctanh}(a x)}{2 a}\right )}{c \sqrt {c-a^2 c x^2}} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2)),x]
Output:
(Sqrt[1 - a^2*x^2]*(-1/2*1/(a*(1 + a*x)) + ArcTanh[a*x]/(2*a)))/(c*Sqrt[c - a^2*c*x^2])
Time = 0.38 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6693, 6690, 54, 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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{-\text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{(1-a x) (a x+1)^2}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{2 (a x+1)^2}-\frac {1}{2 \left (a^2 x^2-1\right )}\right )dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {\text {arctanh}(a x)}{2 a}-\frac {1}{2 a (a x+1)}\right )}{c \sqrt {c-a^2 c x^2}}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2)),x]
Output:
(Sqrt[1 - a^2*x^2]*(-1/2*1/(a*(1 + a*x)) + ArcTanh[a*x]/(2*a)))/(c*Sqrt[c - a^2*c*x^2])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p Int[(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a , c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPart[p]) Int [(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {\left (\ln \left (a x +1\right ) x a -a \ln \left (a x -1\right ) x +\ln \left (a x +1\right )-\ln \left (a x -1\right )-2\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{4 \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) c^{2} a}\) | \(77\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-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
Time = 0.10 (sec) , antiderivative size = 343, normalized size of antiderivative = 3.81 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\left [-\frac {4 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x - {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right )}{8 \, {\left (a^{4} c^{2} x^{3} + a^{3} c^{2} x^{2} - a^{2} c^{2} x - a c^{2}\right )}}, -\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x - {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right )}{4 \, {\left (a^{4} c^{2} x^{3} + a^{3} c^{2} x^{2} - a^{2} c^{2} x - a c^{2}\right )}}\right ] \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-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^4*c^2*x^3 + a^3*c^2*x^2 - a^2*c^2*x - a*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^4*c^2*x^3 + a^3*c^2*x^2 - a^2*c^2*x - a*c^2 )]
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (a x + 1\right )}\, dx \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a**2*c*x**2+c)**(3/2),x)
Output:
Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(a*x - 1)*(a*x + 1))**(3/2)*(a*x + 1)), x)
Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.56 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c}}{2 \, {\left (a^{2} c^{2} x + a c^{2}\right )}} + \frac {\log \left (a x + 1\right )}{4 \, a c^{\frac {3}{2}}} - \frac {\log \left (a x - 1\right )}{4 \, a c^{\frac {3}{2}}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm=" maxima")
Output:
-1/2*sqrt(c)/(a^2*c^2*x + a*c^2) + 1/4*log(a*x + 1)/(a*c^(3/2)) - 1/4*log( a*x - 1)/(a*c^(3/2))
Time = 0.14 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a c^{\frac {3}{2}}} - \frac {\log \left ({\left | -a x + 1 \right |}\right )}{4 \, a c^{\frac {3}{2}}} - \frac {1}{2 \, {\left (a x + 1\right )} a c^{\frac {3}{2}}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm=" giac")
Output:
1/4*log(abs(a*x + 1))/(a*c^(3/2)) - 1/4*log(abs(-a*x + 1))/(a*c^(3/2)) - 1 /2/((a*x + 1)*a*c^(3/2))
Timed out. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\left (a\,x+1\right )} \,d x \] Input:
int((1 - a^2*x^2)^(1/2)/((c - a^2*c*x^2)^(3/2)*(a*x + 1)),x)
Output:
int((1 - a^2*x^2)^(1/2)/((c - a^2*c*x^2)^(3/2)*(a*x + 1)), x)
Time = 0.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-\mathrm {log}\left (a x -1\right ) a x -\mathrm {log}\left (a x -1\right )+\mathrm {log}\left (a x +1\right ) a x +\mathrm {log}\left (a x +1\right )+2 a x \right )}{4 a \,c^{2} \left (a x +1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-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*c**2*(a*x + 1))