Integrand size = 24, antiderivative size = 185 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx=\frac {8 c^4 (1+a x)^7 \sqrt {c-a^2 c x^2}}{7 a \sqrt {1-a^2 x^2}}-\frac {3 c^4 (1+a x)^8 \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}+\frac {2 c^4 (1+a x)^9 \sqrt {c-a^2 c x^2}}{3 a \sqrt {1-a^2 x^2}}-\frac {c^4 (1+a x)^{10} \sqrt {c-a^2 c x^2}}{10 a \sqrt {1-a^2 x^2}} \]
8/7*c^4*(a*x+1)^7*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-3/2*c^4*(a*x+1 )^8*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)+2/3*c^4*(a*x+1)^9*(-a^2*c*x^ 2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-1/10*c^4*(a*x+1)^10*(-a^2*c*x^2+c)^(1/2)/a /(-a^2*x^2+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.37 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx=-\frac {c^4 (1+a x)^7 \sqrt {c-a^2 c x^2} \left (-44+98 a x-77 a^2 x^2+21 a^3 x^3\right )}{210 a \sqrt {1-a^2 x^2}} \]
-1/210*(c^4*(1 + a*x)^7*Sqrt[c - a^2*c*x^2]*(-44 + 98*a*x - 77*a^2*x^2 + 2 1*a^3*x^3))/(a*Sqrt[1 - a^2*x^2])
Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6693, 6690, 49, 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.
\(\displaystyle \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {c^4 \sqrt {c-a^2 c x^2} \int e^{3 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^{9/2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {c^4 \sqrt {c-a^2 c x^2} \int (1-a x)^3 (a x+1)^6dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^4 \sqrt {c-a^2 c x^2} \int \left (-(a x+1)^9+6 (a x+1)^8-12 (a x+1)^7+8 (a x+1)^6\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^4 \left (-\frac {(a x+1)^{10}}{10 a}+\frac {2 (a x+1)^9}{3 a}-\frac {3 (a x+1)^8}{2 a}+\frac {8 (a x+1)^7}{7 a}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}\) |
(c^4*Sqrt[c - a^2*c*x^2]*((8*(1 + a*x)^7)/(7*a) - (3*(1 + a*x)^8)/(2*a) + (2*(1 + a*x)^9)/(3*a) - (1 + a*x)^10/(10*a)))/Sqrt[1 - a^2*x^2]
3.12.68.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p Int[(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a , c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPart[p]) Int [(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(\frac {x \left (21 a^{9} x^{9}+70 a^{8} x^{8}-240 a^{6} x^{6}-210 a^{5} x^{5}+252 a^{4} x^{4}+420 a^{3} x^{3}-315 a x -210\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {9}{2}}}{210 \left (a x -1\right )^{3} \left (a x +1\right )^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(97\) |
default | \(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c^{4} x \left (21 a^{9} x^{9}+70 a^{8} x^{8}-240 a^{6} x^{6}-210 a^{5} x^{5}+252 a^{4} x^{4}+420 a^{3} x^{3}-315 a x -210\right )}{210 a^{2} x^{2}-210}\) | \(98\) |
1/210*x*(21*a^9*x^9+70*a^8*x^8-240*a^6*x^6-210*a^5*x^5+252*a^4*x^4+420*a^3 *x^3-315*a*x-210)*(-a^2*c*x^2+c)^(9/2)/(a*x-1)^3/(a*x+1)^3/(-a^2*x^2+1)^(3 /2)
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.65 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx=\frac {{\left (21 \, a^{9} c^{4} x^{10} + 70 \, a^{8} c^{4} x^{9} - 240 \, a^{6} c^{4} x^{7} - 210 \, a^{5} c^{4} x^{6} + 252 \, a^{4} c^{4} x^{5} + 420 \, a^{3} c^{4} x^{4} - 315 \, a c^{4} x^{2} - 210 \, c^{4} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{210 \, {\left (a^{2} x^{2} - 1\right )}} \]
1/210*(21*a^9*c^4*x^10 + 70*a^8*c^4*x^9 - 240*a^6*c^4*x^7 - 210*a^5*c^4*x^ 6 + 252*a^4*c^4*x^5 + 420*a^3*c^4*x^4 - 315*a*c^4*x^2 - 210*c^4*x)*sqrt(-a ^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)
\[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx=\int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {9}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (161) = 322\).
Time = 0.29 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.21 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx=-\frac {1}{7} \, a^{6} c^{\frac {9}{2}} x^{7} + \frac {3}{5} \, a^{4} c^{\frac {9}{2}} x^{5} - a^{2} c^{\frac {9}{2}} x^{3} + c^{\frac {9}{2}} x + \frac {1}{40} \, {\left (\frac {4 \, a^{8} c^{5} x^{12}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {19 \, a^{6} c^{5} x^{10}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {35 \, a^{4} c^{5} x^{8}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {30 \, a^{2} c^{5} x^{6}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {10 \, c^{5} x^{4}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}}\right )} a^{3} - \frac {1}{105} \, {\left (35 \, a^{6} c^{\frac {9}{2}} x^{9} - 135 \, a^{4} c^{\frac {9}{2}} x^{7} + 189 \, a^{2} c^{\frac {9}{2}} x^{5} - 105 \, c^{\frac {9}{2}} x^{3}\right )} a^{2} + \frac {3}{8} \, {\left (\frac {a^{8} c^{5} x^{10}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {5 \, a^{6} c^{5} x^{8}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {10 \, a^{4} c^{5} x^{6}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {10 \, a^{2} c^{5} x^{4}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {4 \, c^{5}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{2}}\right )} a \]
-1/7*a^6*c^(9/2)*x^7 + 3/5*a^4*c^(9/2)*x^5 - a^2*c^(9/2)*x^3 + c^(9/2)*x + 1/40*(4*a^8*c^5*x^12/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) - 19*a^6*c^5*x^10/ sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) + 35*a^4*c^5*x^8/sqrt(a^4*c*x^4 - 2*a^2* c*x^2 + c) - 30*a^2*c^5*x^6/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) + 10*c^5*x^4 /sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c))*a^3 - 1/105*(35*a^6*c^(9/2)*x^9 - 135* a^4*c^(9/2)*x^7 + 189*a^2*c^(9/2)*x^5 - 105*c^(9/2)*x^3)*a^2 + 3/8*(a^8*c^ 5*x^10/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) - 5*a^6*c^5*x^8/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) + 10*a^4*c^5*x^6/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) - 10*a ^2*c^5*x^4/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) + 4*c^5/(sqrt(a^4*c*x^4 - 2*a ^2*c*x^2 + c)*a^2))*a
\[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{2}} {\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Time = 3.73 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.57 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{9/2} \, dx=\frac {\sqrt {c-a^2\,c\,x^2}\,\left (-\frac {a^9\,c^4\,x^{10}}{10}-\frac {a^8\,c^4\,x^9}{3}+\frac {8\,a^6\,c^4\,x^7}{7}+a^5\,c^4\,x^6-\frac {6\,a^4\,c^4\,x^5}{5}-2\,a^3\,c^4\,x^4+\frac {3\,a\,c^4\,x^2}{2}+c^4\,x\right )}{\sqrt {1-a^2\,x^2}} \]