Integrand size = 27, antiderivative size = 52 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\frac {2 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \] Output:
2*(a*x+1)/(-a^2*c*x^2+c)^(1/2)-arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))/c^(1/ 2)
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.27 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\frac {2 \sqrt {c-a^2 c x^2}}{c-a c x}+\frac {\log (x)}{\sqrt {c}}-\frac {\log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right )}{\sqrt {c}} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(x*Sqrt[c - a^2*c*x^2]),x]
Output:
(2*Sqrt[c - a^2*c*x^2])/(c - a*c*x) + Log[x]/Sqrt[c] - Log[c + Sqrt[c]*Sqr t[c - a^2*c*x^2]]/Sqrt[c]
Time = 0.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6701, 528, 243, 73, 221}
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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx\) |
\(\Big \downarrow \) 6701 |
\(\displaystyle c \int \frac {(a x+1)^2}{x \left (c-a^2 c x^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 528 |
\(\displaystyle c \left (\frac {\int \frac {1}{x \sqrt {c-a^2 c x^2}}dx}{c}+\frac {2 (a x+1)}{c \sqrt {c-a^2 c x^2}}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle c \left (\frac {\int \frac {1}{x^2 \sqrt {c-a^2 c x^2}}dx^2}{2 c}+\frac {2 (a x+1)}{c \sqrt {c-a^2 c x^2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle c \left (\frac {2 (a x+1)}{c \sqrt {c-a^2 c x^2}}-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2 c}}d\sqrt {c-a^2 c x^2}}{a^2 c^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle c \left (\frac {2 (a x+1)}{c \sqrt {c-a^2 c x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{c^{3/2}}\right )\) |
Input:
Int[E^(2*ArcTanh[a*x])/(x*Sqrt[c - a^2*c*x^2]),x]
Output:
c*((2*(1 + a*x))/(c*Sqrt[c - a^2*c*x^2]) - ArcTanh[Sqrt[c - a^2*c*x^2]/Sqr t[c]]/c^(3/2))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((x_)^(m_)*((c_) + (d_.)*(x_))^(n_.))/((a_) + (b_.)*(x_)^2)^(3/2), x_Sy mbol] :> Simp[(-2^(n - 1))*c^(m + n - 2)*((c + d*x)/(b*d^(m - 1)*Sqrt[a + b *x^2])), x] + Simp[c^2/a Int[(x^m/Sqrt[a + b*x^2])*ExpandToSum[((c + d*x) ^(n - 1) - (2^(n - 1)*c^(m + n - 1))/(d^m*x^m))/(c - d*x), x], x], x] /; Fr eeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && EqQ[b*c^2 + a*d^2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^(n/2) Int[x^m*(c + d*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ [c, 0]) && IGtQ[n/2, 0]
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.54
method | result | size |
default | \(-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{\sqrt {c}}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{a c \left (x -\frac {1}{a}\right )}\) | \(80\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/c^(1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)-2/a/c/(x-1/a)*(-(x-1 /a)^2*a^2*c-2*(x-1/a)*a*c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.60 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\left [\frac {{\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - 4 \, \sqrt {-a^{2} c x^{2} + c}}{2 \, {\left (a c x - c\right )}}, \frac {{\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c}}{a c x - c}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^(1/2),x, algorithm="fric as")
Output:
[1/2*((a*x - 1)*sqrt(c)*log(-(a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - 4*sqrt(-a^2*c*x^2 + c))/(a*c*x - c), ((a*x - 1)*sqrt(-c)*arct an(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/c) - 2*sqrt(-a^2*c*x^2 + c))/(a*c*x - c)]
\[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=- \int \frac {a x}{a x^{2} \sqrt {- a^{2} c x^{2} + c} - x \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {1}{a x^{2} \sqrt {- a^{2} c x^{2} + c} - x \sqrt {- a^{2} c x^{2} + c}}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/x/(-a**2*c*x**2+c)**(1/2),x)
Output:
-Integral(a*x/(a*x**2*sqrt(-a**2*c*x**2 + c) - x*sqrt(-a**2*c*x**2 + c)), x) - Integral(1/(a*x**2*sqrt(-a**2*c*x**2 + c) - x*sqrt(-a**2*c*x**2 + c)) , x)
\[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\int { -\frac {{\left (a x + 1\right )}^{2}}{\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 1\right )} x} \,d x } \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxi ma")
Output:
-integrate((a*x + 1)^2/(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 - 1)*x), x)
\[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\int { -\frac {{\left (a x + 1\right )}^{2}}{\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 1\right )} x} \,d x } \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac ")
Output:
sage2
Timed out. \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=-\int \frac {{\left (a\,x+1\right )}^2}{x\,\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )} \,d x \] Input:
int(-(a*x + 1)^2/(x*(c - a^2*c*x^2)^(1/2)*(a^2*x^2 - 1)),x)
Output:
-int((a*x + 1)^2/(x*(c - a^2*c*x^2)^(1/2)*(a^2*x^2 - 1)), x)
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\frac {\sqrt {c}\, \left (\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right )-4 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right )}{c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^(1/2),x)
Output:
(sqrt(c)*(log(tan(asin(a*x)/2))*tan(asin(a*x)/2) - log(tan(asin(a*x)/2)) - 4*tan(asin(a*x)/2)))/(c*(tan(asin(a*x)/2) - 1))