Integrand size = 27, antiderivative size = 256 \[ \int e^{n \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=-\frac {x^2 (1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \sqrt {c-a^2 c x^2}}{5 a^2 \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \left (8+n^2+3 a n x\right ) \sqrt {c-a^2 c x^2}}{60 a^4 \sqrt {1-a^2 x^2}}-\frac {2^{\frac {1}{2} (-1+n)} n \left (11+n^2\right ) (1-a x)^{\frac {3-n}{2}} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-n),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{15 a^4 (3-n) \sqrt {1-a^2 x^2}} \]
-1/5*x^2*(-a*x+1)^(3/2-1/2*n)*(a*x+1)^(3/2+1/2*n)*(-a^2*c*x^2+c)^(1/2)/a^2 /(-a^2*x^2+1)^(1/2)-1/60*(-a*x+1)^(3/2-1/2*n)*(a*x+1)^(3/2+1/2*n)*(3*a*n*x +n^2+8)*(-a^2*c*x^2+c)^(1/2)/a^4/(-a^2*x^2+1)^(1/2)-1/15*2^(-1/2+1/2*n)*n* (n^2+11)*(-a*x+1)^(3/2-1/2*n)*hypergeom([3/2-1/2*n, -1/2-1/2*n],[5/2-1/2*n ],-1/2*a*x+1/2)*(-a^2*c*x^2+c)^(1/2)/a^4/(3-n)/(-a^2*x^2+1)^(1/2)
Time = 0.37 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.92 \[ \int e^{n \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {(1-a x)^{\frac {3}{2}-\frac {n}{2}} \sqrt {c-a^2 c x^2} \left (a^2 (-3+n) x^2 (1+a x)^{\frac {3+n}{2}}-2^{\frac {7+n}{2}} n \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5-n),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )+2^{\frac {7+n}{2}} (-1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3-n),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )-2^{\frac {3+n}{2}} (-2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-n),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )\right )}{5 a^4 (3-n) \sqrt {1-a^2 x^2}} \]
((1 - a*x)^(3/2 - n/2)*Sqrt[c - a^2*c*x^2]*(a^2*(-3 + n)*x^2*(1 + a*x)^((3 + n)/2) - 2^((7 + n)/2)*n*Hypergeometric2F1[(-5 - n)/2, (3 - n)/2, (5 - n )/2, (1 - a*x)/2] + 2^((7 + n)/2)*(-1 + n)*Hypergeometric2F1[(-3 - n)/2, ( 3 - n)/2, (5 - n)/2, (1 - a*x)/2] - 2^((3 + n)/2)*(-2 + n)*Hypergeometric2 F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2]))/(5*a^4*(3 - n)*Sqrt[1 - a^2*x^2])
Time = 0.58 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6703, 6700, 111, 25, 164, 79}
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.
| \(\displaystyle \int x^3 \sqrt {c-a^2 c x^2} e^{n \text {arctanh}(a x)} \, dx\) |
| \(\Big \downarrow \) 6703 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{n \text {arctanh}(a x)} x^3 \sqrt {1-a^2 x^2}dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int x^3 (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n+1}{2}}dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (-\frac {\int -x (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n+1}{2}} (a n x+2)dx}{5 a^2}-\frac {x^2 (a x+1)^{\frac {n+3}{2}} (1-a x)^{\frac {3-n}{2}}}{5 a^2}\right )}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\frac {\int x (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n+1}{2}} (a n x+2)dx}{5 a^2}-\frac {x^2 (1-a x)^{\frac {3-n}{2}} (a x+1)^{\frac {n+3}{2}}}{5 a^2}\right )}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\frac {\frac {n \left (n^2+11\right ) \int (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n+1}{2}}dx}{12 a}-\frac {(1-a x)^{\frac {3-n}{2}} (a x+1)^{\frac {n+3}{2}} \left (3 a n x+n^2+8\right )}{12 a^2}}{5 a^2}-\frac {x^2 (1-a x)^{\frac {3-n}{2}} (a x+1)^{\frac {n+3}{2}}}{5 a^2}\right )}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\frac {-\frac {2^{\frac {n+3}{2}-2} n \left (n^2+11\right ) (1-a x)^{\frac {3-n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-n-1),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{3 a^2 (3-n)}-\frac {(a x+1)^{\frac {n+3}{2}} \left (3 a n x+n^2+8\right ) (1-a x)^{\frac {3-n}{2}}}{12 a^2}}{5 a^2}-\frac {x^2 (1-a x)^{\frac {3-n}{2}} (a x+1)^{\frac {n+3}{2}}}{5 a^2}\right )}{\sqrt {1-a^2 x^2}}\) |
(Sqrt[c - a^2*c*x^2]*(-1/5*(x^2*(1 - a*x)^((3 - n)/2)*(1 + a*x)^((3 + n)/2 ))/a^2 + (-1/12*((1 - a*x)^((3 - n)/2)*(1 + a*x)^((3 + n)/2)*(8 + n^2 + 3* a*n*x))/a^2 - (2^(-2 + (3 + n)/2)*n*(11 + n^2)*(1 - a*x)^((3 - n)/2)*Hyper geometric2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2])/(3*a^2*(3 - n )))/(5*a^2)))/Sqrt[1 - a^2*x^2]
3.14.27.3.1 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_))^(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 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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]
\[\int {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )} x^{3} \sqrt {-a^{2} c \,x^{2}+c}d x\]
\[ \int e^{n \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\int { \sqrt {-a^{2} c x^{2} + c} x^{3} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n} \,d x } \]
\[ \int e^{n \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\int x^{3} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} e^{n \operatorname {atanh}{\left (a x \right )}}\, dx \]
\[ \int e^{n \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\int { \sqrt {-a^{2} c x^{2} + c} x^{3} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n} \,d x } \]
Exception generated. \[ \int e^{n \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int e^{n \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\int x^3\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}\,\sqrt {c-a^2\,c\,x^2} \,d x \]