Integrand size = 23, antiderivative size = 44 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {1+a x}{c \sqrt {1-a^2 x^2}}-\frac {\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c} \] Output:
(a*x+1)/c/(-a^2*x^2+1)^(1/2)-arctanh((-a^2*x^2+1)^(1/2))/c
Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {\frac {1}{\sqrt {1-a^2 x^2}}+\frac {a x}{\sqrt {1-a^2 x^2}}-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c} \] Input:
Integrate[E^ArcTanh[a*x]/(x*(c - a^2*c*x^2)),x]
Output:
(1/Sqrt[1 - a^2*x^2] + (a*x)/Sqrt[1 - a^2*x^2] - ArcTanh[Sqrt[1 - a^2*x^2] ])/c
Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6698, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \frac {\int \frac {a x+1}{x \left (1-a^2 x^2\right )^{3/2}}dx}{c}\) |
\(\Big \downarrow \) 528 |
\(\displaystyle \frac {\int \frac {1}{x \sqrt {1-a^2 x^2}}dx+\frac {a x+1}{\sqrt {1-a^2 x^2}}}{c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2+\frac {a x+1}{\sqrt {1-a^2 x^2}}}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {a x+1}{\sqrt {1-a^2 x^2}}-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {a x+1}{\sqrt {1-a^2 x^2}}-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c}\) |
Input:
Int[E^ArcTanh[a*x]/(x*(c - a^2*c*x^2)),x]
Output:
((1 + a*x)/Sqrt[1 - a^2*x^2] - ArcTanh[Sqrt[1 - a^2*x^2]])/c
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^p Int[x^m*(1 - a^2*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 + 1)/2, 0] && !IntegerQ[p - n/2]
Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.36
method | result | size |
default | \(-\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}\) | \(60\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c),x,method=_RETURNVERBOSE)
Output:
-1/c*(1/a/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)+arctanh(1/(-a^2*x^2+1 )^(1/2)))
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {a x + {\left (a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} - 1}{a c x - c} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c),x, algorithm="fricas ")
Output:
(a*x + (a*x - 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - sqrt(-a^2*x^2 + 1) - 1) /(a*c*x - c)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {\int \frac {a}{- a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{- a^{2} x^{3} \sqrt {- a^{2} x^{2} + 1} + x \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x/(-a**2*c*x**2+c),x)
Output:
(Integral(a/(-a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(1/(-a**2*x**3*sqrt(-a**2*x**2 + 1) + x*sqrt(-a**2*x**2 + 1)), x) )/c
\[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\int { -\frac {a x + 1}{{\left (a^{2} c x^{2} - c\right )} \sqrt {-a^{2} x^{2} + 1} x} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c),x, algorithm="maxima ")
Output:
-integrate((a*x + 1)/((a^2*c*x^2 - c)*sqrt(-a^2*x^2 + 1)*x), x)
Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.82 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=-\frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c {\left | a \right |}} + \frac {2 \, a}{c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+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) ) + 2*a/(c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))
Time = 14.44 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {a\,\sqrt {1-a^2\,x^2}}{c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{c} \] Input:
int((a*x + 1)/(x*(c - a^2*c*x^2)*(1 - a^2*x^2)^(1/2)),x)
Output:
(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 = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right )-2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )}{c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c),x)
Output:
(log(tan(asin(a*x)/2))*tan(asin(a*x)/2) - log(tan(asin(a*x)/2)) - 2*tan(as in(a*x)/2))/(c*(tan(asin(a*x)/2) - 1))