Integrand size = 25, antiderivative size = 107 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{x \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} \log (x)}{\sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2} \log (1-a x)}{\sqrt {c-a^2 c x^2}} \] Output:
-(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c)^(1/2)+a*(-a^2*x^2+1)^(1/2)*ln(x)/(-a^ 2*c*x^2+c)^(1/2)-a*(-a^2*x^2+1)^(1/2)*ln(-a*x+1)/(-a^2*c*x^2+c)^(1/2)
Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.47 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-\frac {1}{x}+a \log (x)-a \log (1-a x)\right )}{\sqrt {c-a^2 c x^2}} \] Input:
Integrate[E^ArcTanh[a*x]/(x^2*Sqrt[c - a^2*c*x^2]),x]
Output:
(Sqrt[1 - a^2*x^2]*(-x^(-1) + a*Log[x] - a*Log[1 - a*x]))/Sqrt[c - a^2*c*x ^2]
Time = 0.44 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.47, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {1-a^2 x^2}}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{x^2 (1-a x)}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {a^2}{a x-1}+\frac {a}{x}+\frac {1}{x^2}\right )dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (a \log (x)-a \log (1-a x)-\frac {1}{x}\right )}{\sqrt {c-a^2 c x^2}}\) |
Input:
Int[E^ArcTanh[a*x]/(x^2*Sqrt[c - a^2*c*x^2]),x]
Output:
(Sqrt[1 - a^2*x^2]*(-x^(-1) + a*Log[x] - a*Log[1 - a*x]))/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_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar t[p]) Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && !I ntegerQ[n/2]
Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {\left (-a \ln \left (a x -1\right ) x +a \ln \left (x \right ) x -1\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{\sqrt {-a^{2} x^{2}+1}\, c x}\) | \(51\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^2/(-a^2*c*x^2+c)^(1/2),x,method=_RETURNVE RBOSE)
Output:
(-a*ln(a*x-1)*x+a*ln(x)*x-1)/(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)/c/x
Time = 0.11 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.89 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\left [\frac {{\left (a^{3} x^{3} - a x\right )} \sqrt {c} \log \left (-\frac {4 \, a^{5} c x^{5} - {\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} - {\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x + {\left (4 \, a^{3} x^{3} - {\left (4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} x^{4} - 6 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (x - 1\right )}}{2 \, {\left (a^{2} c x^{3} - c x\right )}}, -\frac {{\left (a^{3} x^{3} - a x\right )} \sqrt {-c} \arctan \left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a^{2} - 2 \, a + 1\right )} x^{2} - 2 \, a x + 1\right )} \sqrt {-c}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - a^{2}\right )} c x^{4} - {\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right ) + \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (x - 1\right )}}{a^{2} c x^{3} - c x}\right ] \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^2/(-a^2*c*x^2+c)^(1/2),x, algorithm ="fricas")
Output:
[1/2*((a^3*x^3 - a*x)*sqrt(c)*log(-(4*a^5*c*x^5 - (2*a^6 - 4*a^5 + 6*a^4 - 4*a^3 + a^2)*c*x^6 - (4*a^4 + 4*a^3 - 6*a^2 + 4*a - 1)*c*x^4 + 5*a^2*c*x^ 2 - 4*a*c*x + (4*a^3*x^3 - (4*a^3 - 6*a^2 + 4*a - 1)*x^4 - 6*a^2*x^2 + 4*a *x - 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) + c)/(a^4*x^6 - 2* a^3*x^5 + 2*a*x^3 - x^2)) - 2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(x - 1))/(a^2*c*x^3 - c*x), -((a^3*x^3 - a*x)*sqrt(-c)*arctan(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*((2*a^2 - 2*a + 1)*x^2 - 2*a*x + 1)*sqrt(-c)/(2*a ^3*c*x^3 - (2*a^3 - a^2)*c*x^4 - (a^2 - 2*a + 1)*c*x^2 - 2*a*c*x + c)) + s qrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(x - 1))/(a^2*c*x^3 - c*x)]
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int \frac {a x + 1}{x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x**2/(-a**2*c*x**2+c)**(1/2),x)
Output:
Integral((a*x + 1)/(x**2*sqrt(-(a*x - 1)*(a*x + 1))*sqrt(-c*(a*x - 1)*(a*x + 1))), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int { \frac {a x + 1}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{2}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^2/(-a^2*c*x^2+c)^(1/2),x, algorithm ="maxima")
Output:
integrate((a*x + 1)/(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^2), x)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.29 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=-\frac {a \log \left ({\left | -a x + 1 \right |}\right )}{\sqrt {c}} + \frac {a \log \left ({\left | x \right |}\right )}{\sqrt {c}} - \frac {1}{\sqrt {c} x} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^2/(-a^2*c*x^2+c)^(1/2),x, algorithm ="giac")
Output:
-a*log(abs(-a*x + 1))/sqrt(c) + a*log(abs(x))/sqrt(c) - 1/(sqrt(c)*x)
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int \frac {a\,x+1}{x^2\,\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((a*x + 1)/(x^2*(c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((a*x + 1)/(x^2*(c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.24 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\frac {\sqrt {c}\, \left (-\mathrm {log}\left (a x -1\right ) a x +\mathrm {log}\left (x \right ) a x -1\right )}{c x} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^2/(-a^2*c*x^2+c)^(1/2),x)
Output:
(sqrt(c)*( - log(a*x - 1)*a*x + log(x)*a*x - 1))/(c*x)