Integrand size = 27, antiderivative size = 136 \[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=-\frac {3 x^{1+m} \sqrt {c-a^2 c x^2}}{(1+m) \sqrt {1-a^2 x^2}}+\frac {a x^{2+m} \sqrt {c-a^2 c x^2}}{(2+m) \sqrt {1-a^2 x^2}}+\frac {4 x^{1+m} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}(1,1+m,2+m,-a x)}{(1+m) \sqrt {1-a^2 x^2}} \] Output:
-3*x^(1+m)*(-a^2*c*x^2+c)^(1/2)/(1+m)/(-a^2*x^2+1)^(1/2)+a*x^(2+m)*(-a^2*c *x^2+c)^(1/2)/(2+m)/(-a^2*x^2+1)^(1/2)+4*x^(1+m)*(-a^2*c*x^2+c)^(1/2)*hype rgeom([1, 1+m],[2+m],-a*x)/(1+m)/(-a^2*x^2+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54 \[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\frac {x^{1+m} \sqrt {c-a^2 c x^2} (-6+a x+m (-3+a x)+4 (2+m) \operatorname {Hypergeometric2F1}(1,1+m,2+m,-a x))}{(1+m) (2+m) \sqrt {1-a^2 x^2}} \] Input:
Integrate[(x^m*Sqrt[c - a^2*c*x^2])/E^(3*ArcTanh[a*x]),x]
Output:
(x^(1 + m)*Sqrt[c - a^2*c*x^2]*(-6 + a*x + m*(-3 + a*x) + 4*(2 + m)*Hyperg eometric2F1[1, 1 + m, 2 + m, -(a*x)]))/((1 + m)*(2 + m)*Sqrt[1 - a^2*x^2])
Time = 0.52 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.58, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6703, 6700, 99, 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.
| \(\displaystyle \int x^m e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2} \, dx\) |
| \(\Big \downarrow \) 6703 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {1-a^2 x^2}dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {x^m (1-a x)^2}{a x+1}dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \left (\frac {4 x^m}{a x+1}-3 x^m+a x^{m+1}\right )dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\frac {4 x^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,-a x)}{m+1}+\frac {a x^{m+2}}{m+2}-\frac {3 x^{m+1}}{m+1}\right )}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[(x^m*Sqrt[c - a^2*c*x^2])/E^(3*ArcTanh[a*x]),x]
Output:
(Sqrt[c - a^2*c*x^2]*((-3*x^(1 + m))/(1 + m) + (a*x^(2 + m))/(2 + m) + (4* x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -(a*x)])/(1 + m)))/Sqrt[1 - a ^2*x^2]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]
\[\int \frac {x^{m} \sqrt {-a^{2} c \,x^{2}+c}\, \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{\left (a x +1\right )^{3}}d x\]
Input:
int(x^m*(-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
Output:
int(x^m*(-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
\[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{m}}{{\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate(x^m*(-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorit hm="fricas")
Output:
integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(a*x - 1)*x^m/(a^2*x^2 + 2*a*x + 1), x)
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\text {Timed out} \] Input:
integrate(x**m*(-a**2*c*x**2+c)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{m}}{{\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate(x^m*(-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorit hm="maxima")
Output:
integrate(sqrt(-a^2*c*x^2 + c)*(-a^2*x^2 + 1)^(3/2)*x^m/(a*x + 1)^3, x)
Exception generated. \[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^m*(-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorit hm="giac")
Output:
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. \[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\int \frac {x^m\,\sqrt {c-a^2\,c\,x^2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \] Input:
int((x^m*(c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3,x)
Output:
int((x^m*(c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3, x)
\[ \int e^{-3 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c}\, \left (x^{m} a^{2} m^{2} x^{2}+x^{m} a^{2} m \,x^{2}-3 x^{m} a \,m^{2} x -6 x^{m} a m x +4 x^{m} m^{2}+12 x^{m} m +8 x^{m}-4 \left (\int \frac {x^{m}}{a \,x^{2}+x}d x \right ) m^{3}-12 \left (\int \frac {x^{m}}{a \,x^{2}+x}d x \right ) m^{2}-8 \left (\int \frac {x^{m}}{a \,x^{2}+x}d x \right ) m \right )}{a m \left (m^{2}+3 m +2\right )} \] Input:
int(x^m*(-a^2*c*x^2+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
Output:
(sqrt(c)*(x**m*a**2*m**2*x**2 + x**m*a**2*m*x**2 - 3*x**m*a*m**2*x - 6*x** m*a*m*x + 4*x**m*m**2 + 12*x**m*m + 8*x**m - 4*int(x**m/(a*x**2 + x),x)*m* *3 - 12*int(x**m/(a*x**2 + x),x)*m**2 - 8*int(x**m/(a*x**2 + x),x)*m))/(a* m*(m**2 + 3*m + 2))