Integrand size = 24, antiderivative size = 90 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{2 (1+a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \text {arctanh}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}} \] Output:
1/2*a^2*(1-1/a^2/x^2)^(3/2)*x^3/(a*x+1)/(-a^2*c*x^2+c)^(3/2)-1/2*a^2*(1-1/ a^2/x^2)^(3/2)*x^3*arctanh(a*x)/(-a^2*c*x^2+c)^(3/2)
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.60 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (-1+(1+a x) \text {arctanh}(a x))}{2 (c+a c x) \sqrt {c-a^2 c x^2}} \] Input:
Integrate[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(3/2)),x]
Output:
(Sqrt[1 - 1/(a^2*x^2)]*x*(-1 + (1 + a*x)*ArcTanh[a*x]))/(2*(c + a*c*x)*Sqr t[c - a^2*c*x^2])
Time = 0.73 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6746, 6747, 25, 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.
\(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int -\frac {1}{(1-a x) (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \frac {1}{(1-a x) (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \left (\frac {1}{2 (a x+1)^2}-\frac {1}{2 \left (a^2 x^2-1\right )}\right )dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (\frac {\text {arctanh}(a x)}{2 a}-\frac {1}{2 a (a x+1)}\right )}{\left (c-a^2 c x^2\right )^{3/2}}\) |
Input:
Int[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(3/2)),x]
Output:
-((a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*(-1/2*1/(a*(1 + a*x)) + ArcTanh[a*x]/(2 *a)))/(c - a^2*c*x^2)^(3/2))
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^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && !IntegerQ[p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[c^p/a^(2*p) Int[(u/x^(2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x +1\right ) x a -a \ln \left (a x -1\right ) x +\ln \left (a x +1\right )-\ln \left (a x -1\right )-2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} a}\) | \(84\) |
Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/4*((a*x-1)/(a*x+1))^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(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)/c^2/a
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {{\left (a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, \sqrt {-a^{2} c}}{4 \, {\left (a^{3} c^{2} x + a^{2} c^{2}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="frica s")
Output:
-1/4*((a^2*x + a)*sqrt(-c)*log((a^2*c*x^2 - 2*sqrt(-a^2*c)*sqrt(-c)*x + c) /(a^2*x^2 - 1)) - 2*sqrt(-a^2*c))/(a^3*c^2*x + a^2*c^2)
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(((a*x-1)/(a*x+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), x)
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxim a")
Output:
integrate(sqrt((a*x - 1)/(a*x + 1))/(-a^2*c*x^2 + c)^(3/2), x)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {{\left (\frac {\log \left ({\left | a x + 1 \right |}\right )}{a c} - \frac {\log \left ({\left | a x - 1 \right |}\right )}{a c} - \frac {2}{{\left (a x + 1\right )} a c}\right )} \mathrm {sgn}\left (a x + 1\right )}{4 \, \sqrt {-c}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac" )
Output:
1/4*(log(abs(a*x + 1))/(a*c) - log(abs(a*x - 1))/(a*c) - 2/((a*x + 1)*a*c) )*sgn(a*x + 1)/sqrt(-c)
Timed out. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}} \,d x \] Input:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(3/2),x)
Output:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, i \left (\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )+\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-2 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x -2 \,\mathrm {log}\left (\sqrt {-a x +1}\right )+a x -1\right )}{4 a \,c^{2} \left (a x +1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*i*(log(sqrt( - a*x + 1) - sqrt(2))*a*x + log(sqrt( - a*x + 1) - s qrt(2)) + log(sqrt( - a*x + 1) + sqrt(2))*a*x + log(sqrt( - a*x + 1) + sqr t(2)) - 2*log(sqrt( - a*x + 1))*a*x - 2*log(sqrt( - a*x + 1)) + a*x - 1))/ (4*a*c**2*(a*x + 1))