Integrand size = 19, antiderivative size = 48 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=\frac {2 \sqrt {1-a^2 x^2}}{c (1-a x)}-\frac {\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c} \] Output:
2*(-a^2*x^2+1)^(1/2)/c/(-a*x+1)-arctanh((-a^2*x^2+1)^(1/2))/c
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=\frac {2+2 a x-\sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]/(x*(c - a*c*x)),x]
Output:
(2 + 2*a*x - Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(c*Sqrt[1 - a^2 *x^2])
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6678, 27, 564, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{c^2 x (1-a x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {1-a^2 x^2}}{x (1-a x)^2}dx}{c}\) |
\(\Big \downarrow \) 564 |
\(\displaystyle \frac {\int \frac {1}{x \sqrt {1-a^2 x^2}}dx+\frac {2 \sqrt {1-a^2 x^2}}{1-a x}}{c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2+\frac {2 \sqrt {1-a^2 x^2}}{1-a x}}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{1-a x}-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{1-a x}-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c}\) |
Input:
Int[E^ArcTanh[a*x]/(x*(c - a*c*x)),x]
Output:
((2*Sqrt[1 - a^2*x^2])/(1 - a*x) - ArcTanh[Sqrt[1 - a^2*x^2]])/c
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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)^(p_), x_Symbol ] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b ^(n + 2)*(c + d*x))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(x^m/Sqrt[a + b *x^2])*ExpandToSum[((2^(-n - 1)*(-c)^(m - n - 1))/(d^m*x^m) - (-c + d*x)^(- n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^ 2, 0] && ILtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.27
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}}{c}\) | \(61\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c),x,method=_RETURNVERBOSE)
Output:
-1/c*(arctanh(1/(-a^2*x^2+1)^(1/2))+2/a/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a ))^(1/2))
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=\frac {2 \, a x + {\left (a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} - 2}{a c x - c} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c),x, algorithm="fricas")
Output:
(2*a*x + (a*x - 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - 2*sqrt(-a^2*x^2 + 1) - 2)/(a*c*x - c)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=- \frac {\int \frac {a x}{a x^{2} \sqrt {- a^{2} x^{2} + 1} - x \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a x^{2} \sqrt {- a^{2} x^{2} + 1} - x \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x/(-a*c*x+c),x)
Output:
-(Integral(a*x/(a*x**2*sqrt(-a**2*x**2 + 1) - x*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a*x**2*sqrt(-a**2*x**2 + 1) - x*sqrt(-a**2*x**2 + 1)), x))/c
\[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=\int { -\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )} x} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c),x, algorithm="maxima")
Output:
-integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(a*c*x - c)*x), x)
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.67 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=-\frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c {\left | a \right |}} + \frac {4 \, a}{c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c),x, algorithm="giac")
Output:
-a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a) ) + 4*a/(c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))
Time = 14.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=\frac {2\,a\,\sqrt {1-a^2\,x^2}}{c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{c} \] Input:
int((a*x + 1)/(x*(1 - a^2*x^2)^(1/2)*(c - a*c*x)),x)
Output:
(2*a*(1 - a^2*x^2)^(1/2))/(c*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)*(-a^2)^(1/2 )) - atanh((1 - a^2*x^2)^(1/2))/c
Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)} \, dx=\frac {\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right )-4 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )}{c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c),x)
Output:
(log(tan(asin(a*x)/2))*tan(asin(a*x)/2) - log(tan(asin(a*x)/2)) - 4*tan(as in(a*x)/2))/(c*(tan(asin(a*x)/2) - 1))