Integrand size = 27, antiderivative size = 269 \[ \int \frac {e^{n \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=-\frac {(1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {c-a^2 c x^2}}{(1-n) \sqrt {1-a^2 x^2}}+\frac {2 (1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+n),\frac {1+n}{2},\frac {1+a x}{1-a x}\right )}{(1-n) \sqrt {1-a^2 x^2}}+\frac {2^{\frac {1+n}{2}} n (1-a x)^{\frac {3-n}{2}} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{\left (3-4 n+n^2\right ) \sqrt {1-a^2 x^2}} \]
-(-a*x+1)^(3/2-1/2*n)*(a*x+1)^(-1/2+1/2*n)*(-a^2*c*x^2+c)^(1/2)/(1-n)/(-a^ 2*x^2+1)^(1/2)+2*(-a*x+1)^(1/2-1/2*n)*(a*x+1)^(-1/2+1/2*n)*hypergeom([1, - 1/2+1/2*n],[1/2+1/2*n],(a*x+1)/(-a*x+1))*(-a^2*c*x^2+c)^(1/2)/(1-n)/(-a^2* x^2+1)^(1/2)+2^(1/2+1/2*n)*n*(-a*x+1)^(3/2-1/2*n)*hypergeom([1/2-1/2*n, 3/ 2-1/2*n],[5/2-1/2*n],-1/2*a*x+1/2)*(-a^2*c*x^2+c)^(1/2)/(n^2-4*n+3)/(-a^2* x^2+1)^(1/2)
Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.77 \[ \int \frac {e^{n \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\frac {2 (1-a x)^{\frac {1}{2} (-1-n)} \sqrt {c-a^2 c x^2} \left ((-1+n) (1+a x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {n}{2},\frac {1-a x}{1+a x}\right )+2^{\frac {1+n}{2}} \left (-\left ((-1+n) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-\frac {n}{2},-\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {a x}{2}\right )\right )+(1+n) (-1+a x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},\frac {1}{2}-\frac {a x}{2}\right )\right )\right )}{\left (-1+n^2\right ) \sqrt {1-a^2 x^2}} \]
(2*(1 - a*x)^((-1 - n)/2)*Sqrt[c - a^2*c*x^2]*((-1 + n)*(1 + a*x)^((1 + n) /2)*Hypergeometric2F1[1, -1/2 - n/2, 1/2 - n/2, (1 - a*x)/(1 + a*x)] + 2^( (1 + n)/2)*(-((-1 + n)*Hypergeometric2F1[-1/2 - n/2, -1/2 - n/2, 1/2 - n/2 , 1/2 - (a*x)/2]) + (1 + n)*(-1 + a*x)*Hypergeometric2F1[-1/2 - n/2, 1/2 - n/2, 3/2 - n/2, 1/2 - (a*x)/2])))/((-1 + n^2)*Sqrt[1 - a^2*x^2])
Time = 0.59 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6703, 6700, 139, 88, 79, 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.
\(\displaystyle \int \frac {\sqrt {c-a^2 c x^2} e^{n \text {arctanh}(a x)}}{x} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {e^{n \text {arctanh}(a x)} \sqrt {1-a^2 x^2}}{x}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {(1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n+1}{2}}}{x}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 139 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\int \frac {(1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n-3}{2}}}{x}dx+a \int (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n-3}{2}} (a x+2)dx\right )}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 88 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\int \frac {(1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n-3}{2}}}{x}dx+a \left (-\frac {n \int (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n-1}{2}}dx}{1-n}-\frac {(a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {3-n}{2}}}{a (1-n)}\right )\right )}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\int \frac {(1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n-3}{2}}}{x}dx+a \left (\frac {2^{\frac {n+1}{2}} n (1-a x)^{\frac {3-n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{a (1-n) (3-n)}-\frac {(1-a x)^{\frac {3-n}{2}} (a x+1)^{\frac {n-1}{2}}}{a (1-n)}\right )\right )}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 141 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\frac {2 (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1-n}{2}} \operatorname {Hypergeometric2F1}\left (1,\frac {n-1}{2},\frac {n+1}{2},\frac {a x+1}{1-a x}\right )}{1-n}+a \left (\frac {2^{\frac {n+1}{2}} n (1-a x)^{\frac {3-n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{a (1-n) (3-n)}-\frac {(1-a x)^{\frac {3-n}{2}} (a x+1)^{\frac {n-1}{2}}}{a (1-n)}\right )\right )}{\sqrt {1-a^2 x^2}}\) |
(Sqrt[c - a^2*c*x^2]*((2*(1 - a*x)^((1 - n)/2)*(1 + a*x)^((-1 + n)/2)*Hype rgeometric2F1[1, (-1 + n)/2, (1 + n)/2, (1 + a*x)/(1 - a*x)])/(1 - n) + a* (-(((1 - a*x)^((3 - n)/2)*(1 + a*x)^((-1 + n)/2))/(a*(1 - n))) + (2^((1 + n)/2)*n*(1 - a*x)^((3 - n)/2)*Hypergeometric2F1[(1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2])/(a*(1 - n)*(3 - n)))))/Sqrt[1 - a^2*x^2]
3.14.31.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && !RationalQ[p] && SumSimpl erQ[p, 1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[f^(p - 1)/d^p Int[(a + b*x)^m*((d*e*p - c*f*(p - 1) + d*f*x)/(c + d*x)^(m + 1)), x], x] + Simp[f^(p - 1) Int[(a + b*x)^m*((e + f*x)^p/(c + d*x)^(m + 1))*ExpandToSum[f^(-p + 1)*(c + d*x)^(-p + 1) - (d*e *p - c*f*(p - 1) + d*f*x)/(d^p*(e + f*x)^p), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m + n + p, 0] && ILtQ[p, 0] && (LtQ[m, 0] || SumS implerQ[m, 1] || !(LtQ[n, 0] || SumSimplerQ[n, 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]
\[\int \frac {{\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )} \sqrt {-a^{2} c \,x^{2}+c}}{x}d x\]
\[ \int \frac {e^{n \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x} \,d x } \]
\[ \int \frac {e^{n \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\int \frac {\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} e^{n \operatorname {atanh}{\left (a x \right )}}}{x}\, dx \]
\[ \int \frac {e^{n \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x} \,d x } \]
Exception generated. \[ \int \frac {e^{n \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {e^{n \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}\,\sqrt {c-a^2\,c\,x^2}}{x} \,d x \]