Integrand size = 27, antiderivative size = 167 \[ \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=-\frac {(1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1+n}{2}} \sqrt {1-a^2 x^2}}{x \sqrt {c-a^2 c x^2}}-\frac {2 a n (1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1-n}{2},\frac {3-n}{2},\frac {1-a x}{1+a x}\right )}{(1-n) \sqrt {c-a^2 c x^2}} \]
-(-a*x+1)^(1/2-1/2*n)*(a*x+1)^(1/2+1/2*n)*(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2 +c)^(1/2)-2*a*n*(-a*x+1)^(1/2-1/2*n)*(a*x+1)^(-1/2+1/2*n)*hypergeom([1, 1/ 2-1/2*n],[3/2-1/2*n],(-a*x+1)/(a*x+1))*(-a^2*x^2+1)^(1/2)/(1-n)/(-a^2*c*x^ 2+c)^(1/2)
Time = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.70 \[ \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\frac {(1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \left (-((-1+n) (1+a x))+2 a n x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},\frac {1-a x}{1+a x}\right )\right )}{(-1+n) x \sqrt {c-a^2 c x^2}} \]
((1 - a*x)^(1/2 - n/2)*(1 + a*x)^((-1 + n)/2)*Sqrt[1 - a^2*x^2]*(-((-1 + n )*(1 + a*x)) + 2*a*n*x*Hypergeometric2F1[1, 1/2 - n/2, 3/2 - n/2, (1 - a*x )/(1 + a*x)]))/((-1 + n)*x*Sqrt[c - a^2*c*x^2])
Time = 0.50 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6703, 6700, 107, 141}
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.
\(\displaystyle \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {1-a^2 x^2}}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {(1-a x)^{\frac {1}{2} (-n-1)} (a x+1)^{\frac {n-1}{2}}}{x^2}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 107 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (a n \int \frac {(1-a x)^{\frac {1}{2} (-n-1)} (a x+1)^{\frac {n-1}{2}}}{x}dx-\frac {(1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n+1}{2}}}{x}\right )}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (-\frac {2 a n (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1-n}{2}} \operatorname {Hypergeometric2F1}\left (1,\frac {1-n}{2},\frac {3-n}{2},\frac {1-a x}{a x+1}\right )}{1-n}-\frac {(a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{x}\right )}{\sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(-(((1 - a*x)^((1 - n)/2)*(1 + a*x)^((1 + n)/2))/x) - ( 2*a*n*(1 - a*x)^((1 - n)/2)*(1 + a*x)^((-1 + n)/2)*Hypergeometric2F1[1, (1 - n)/2, (3 - n)/2, (1 - a*x)/(1 + a*x)])/(1 - n)))/Sqrt[c - a^2*c*x^2]
3.14.39.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f ))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] || !Su mSimplerQ[p, 1]) && !ILtQ[m, 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]
\[\int \frac {{\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{x^{2} \sqrt {-a^{2} c \,x^{2}+c}}d x\]
\[ \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{\sqrt {-a^{2} c x^{2} + c} x^{2}} \,d x } \]
\[ \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{x^{2} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]
\[ \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{\sqrt {-a^{2} c x^{2} + c} x^{2}} \,d x } \]
\[ \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{\sqrt {-a^{2} c x^{2} + c} x^{2}} \,d x } \]
Timed out. \[ \int \frac {e^{n \text {arctanh}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{x^2\,\sqrt {c-a^2\,c\,x^2}} \,d x \]