Integrand size = 25, antiderivative size = 134 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {x^{1+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{c^2 (1+m) \sqrt {c-a^2 c x^2}}+\frac {a x^{2+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (3,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{c^2 (2+m) \sqrt {c-a^2 c x^2}} \]
x^(1+m)*hypergeom([3, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)*(-a^2*x^2+1)^(1/2)/c ^2/(1+m)/(-a^2*c*x^2+c)^(1/2)+a*x^(2+m)*hypergeom([3, 1+1/2*m],[2+1/2*m],a ^2*x^2)*(-a^2*x^2+1)^(1/2)/c^2/(2+m)/(-a^2*c*x^2+c)^(1/2)
Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},1+\frac {1+m}{2},a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\left (3,\frac {2+m}{2},1+\frac {2+m}{2},a^2 x^2\right )}{2+m}\right )}{c^2 \sqrt {c-a^2 c x^2}} \]
(Sqrt[1 - a^2*x^2]*((x^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, 1 + (1 + m) /2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[3, (2 + m)/2, 1 + ( 2 + m)/2, a^2*x^2])/(2 + m)))/(c^2*Sqrt[c - a^2*c*x^2])
Time = 0.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.77, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6703, 6700, 92, 82, 278}
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.
\(\displaystyle \int \frac {x^m e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^m}{(1-a x)^3 (a x+1)^2}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 92 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (a \int \frac {x^{m+1}}{(1-a x)^3 (a x+1)^3}dx+\int \frac {x^m}{(1-a x)^3 (a x+1)^3}dx\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (a \int \frac {x^{m+1}}{\left (1-a^2 x^2\right )^3}dx+\int \frac {x^m}{\left (1-a^2 x^2\right )^3}dx\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (3,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (3,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*((x^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[3, (2 + m)/2, (4 + m)/2 , a^2*x^2])/(2 + m)))/(c^2*Sqrt[c - a^2*c*x^2])
3.11.3.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> Simp[a Int[(a + b*x)^n*(c + d*x)^n*(f*x)^p, x], x] + Simp[b/f In t[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] && !RationalQ[p] && !IGtQ[m, 0] && NeQ[m + n + p + 2, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 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]
\[\int \frac {\left (a x +1\right ) x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^m/(a^7*c^3*x^7 - a^6*c^ 3*x^6 - 3*a^5*c^3*x^5 + 3*a^4*c^3*x^4 + 3*a^3*c^3*x^3 - 3*a^2*c^3*x^2 - a* c^3*x + c^3), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{m} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
\[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^m\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \]