Integrand size = 22, antiderivative size = 183 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {8 c^3 (1+a x)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}}-\frac {2 c^3 (1+a x)^6 \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}+\frac {6 c^3 (1+a x)^7 \sqrt {c-a^2 c x^2}}{7 a \sqrt {1-a^2 x^2}}-\frac {c^3 (1+a x)^8 \sqrt {c-a^2 c x^2}}{8 a \sqrt {1-a^2 x^2}} \] Output:
8/5*c^3*(a*x+1)^5*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-2*c^3*(a*x+1)^ 6*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)+6/7*c^3*(a*x+1)^7*(-a^2*c*x^2+ c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-1/8*c^3*(a*x+1)^8*(-a^2*c*x^2+c)^(1/2)/a/(-a ^2*x^2+1)^(1/2)
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.37 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {c^3 (1+a x)^5 \sqrt {c-a^2 c x^2} \left (-93+185 a x-135 a^2 x^2+35 a^3 x^3\right )}{280 a \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(7/2),x]
Output:
-1/280*(c^3*(1 + a*x)^5*Sqrt[c - a^2*c*x^2]*(-93 + 185*a*x - 135*a^2*x^2 + 35*a^3*x^3))/(a*Sqrt[1 - a^2*x^2])
Time = 0.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, 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.
\(\displaystyle \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {c^3 \sqrt {c-a^2 c x^2} \int e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^{7/2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {c^3 \sqrt {c-a^2 c x^2} \int (1-a x)^3 (a x+1)^4dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^3 \sqrt {c-a^2 c x^2} \int \left (-(a x+1)^7+6 (a x+1)^6-12 (a x+1)^5+8 (a x+1)^4\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^3 \left (-\frac {(a x+1)^8}{8 a}+\frac {6 (a x+1)^7}{7 a}-\frac {2 (a x+1)^6}{a}+\frac {8 (a x+1)^5}{5 a}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(7/2),x]
Output:
(c^3*Sqrt[c - a^2*c*x^2]*((8*(1 + a*x)^5)/(5*a) - (2*(1 + a*x)^6)/a + (6*( 1 + a*x)^7)/(7*a) - (1 + a*x)^8/(8*a)))/Sqrt[1 - a^2*x^2]
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 = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.48
method | result | size |
default | \(-\frac {\left (35 a^{7} x^{7}+40 x^{6} a^{6}-140 a^{5} x^{5}-168 a^{4} x^{4}+210 a^{3} x^{3}+280 a^{2} x^{2}-140 a x -280\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c^{3} x}{280 \sqrt {-a^{2} x^{2}+1}}\) | \(87\) |
gosper | \(\frac {x \left (35 a^{7} x^{7}+40 x^{6} a^{6}-140 a^{5} x^{5}-168 a^{4} x^{4}+210 a^{3} x^{3}+280 a^{2} x^{2}-140 a x -280\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{280 \left (a x +1\right )^{3} \left (a x -1\right )^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(97\) |
orering | \(\frac {x \left (35 a^{7} x^{7}+40 x^{6} a^{6}-140 a^{5} x^{5}-168 a^{4} x^{4}+210 a^{3} x^{3}+280 a^{2} x^{2}-140 a x -280\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}{280 \left (a x +1\right )^{3} \left (a x -1\right )^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(97\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x,method=_RETURNVERBOS E)
Output:
-1/280*(35*a^7*x^7+40*a^6*x^6-140*a^5*x^5-168*a^4*x^4+210*a^3*x^3+280*a^2* x^2-140*a*x-280)/(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*c^3*x
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.66 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {{\left (35 \, a^{7} c^{3} x^{8} + 40 \, a^{6} c^{3} x^{7} - 140 \, a^{5} c^{3} x^{6} - 168 \, a^{4} c^{3} x^{5} + 210 \, a^{3} c^{3} x^{4} + 280 \, a^{2} c^{3} x^{3} - 140 \, a c^{3} x^{2} - 280 \, c^{3} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{280 \, {\left (a^{2} x^{2} - 1\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="fr icas")
Output:
1/280*(35*a^7*c^3*x^8 + 40*a^6*c^3*x^7 - 140*a^5*c^3*x^6 - 168*a^4*c^3*x^5 + 210*a^3*c^3*x^4 + 280*a^2*c^3*x^3 - 140*a*c^3*x^2 - 280*c^3*x)*sqrt(-a^ 2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)
\[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a**2*c*x**2+c)**(7/2),x)
Output:
Integral((-c*(a*x - 1)*(a*x + 1))**(7/2)*(a*x + 1)/sqrt(-(a*x - 1)*(a*x + 1)), x)
Time = 0.05 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.84 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {1}{7} \, a^{6} c^{\frac {7}{2}} x^{7} + \frac {3}{5} \, a^{4} c^{\frac {7}{2}} x^{5} - a^{2} c^{\frac {7}{2}} x^{3} + c^{\frac {7}{2}} x + \frac {1}{8} \, {\left (\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{4} c^{3} x^{6} - 3 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{2} c^{3} x^{4} + 3 \, a^{2} c^{\frac {7}{2}} x^{4} + 6 \, c^{\frac {7}{2}} x^{2} - \frac {10 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} c^{3}}{a^{2}}\right )} a \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="ma xima")
Output:
-1/7*a^6*c^(7/2)*x^7 + 3/5*a^4*c^(7/2)*x^5 - a^2*c^(7/2)*x^3 + c^(7/2)*x + 1/8*(sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*a^4*c^3*x^6 - 3*sqrt(a^4*c*x^4 - 2 *a^2*c*x^2 + c)*a^2*c^3*x^4 + 3*a^2*c^(7/2)*x^4 + 6*c^(7/2)*x^2 - 10*sqrt( a^4*c*x^4 - 2*a^2*c*x^2 + c)*c^3/a^2)*a
\[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} {\left (a x + 1\right )}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x, algorithm="gi ac")
Output:
integrate((-a^2*c*x^2 + c)^(7/2)*(a*x + 1)/sqrt(-a^2*x^2 + 1), x)
Time = 14.08 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.58 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {\sqrt {c-a^2\,c\,x^2}\,\left (-\frac {a^7\,c^3\,x^8}{8}-\frac {a^6\,c^3\,x^7}{7}+\frac {a^5\,c^3\,x^6}{2}+\frac {3\,a^4\,c^3\,x^5}{5}-\frac {3\,a^3\,c^3\,x^4}{4}-a^2\,c^3\,x^3+\frac {a\,c^3\,x^2}{2}+c^3\,x\right )}{\sqrt {1-a^2\,x^2}} \] Input:
int(((c - a^2*c*x^2)^(7/2)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
((c - a^2*c*x^2)^(1/2)*(c^3*x + (a*c^3*x^2)/2 - a^2*c^3*x^3 - (3*a^3*c^3*x ^4)/4 + (3*a^4*c^3*x^5)/5 + (a^5*c^3*x^6)/2 - (a^6*c^3*x^7)/7 - (a^7*c^3*x ^8)/8))/(1 - a^2*x^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.34 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {\sqrt {c}\, c^{3} x \left (-35 a^{7} x^{7}-40 a^{6} x^{6}+140 a^{5} x^{5}+168 a^{4} x^{4}-210 a^{3} x^{3}-280 a^{2} x^{2}+140 a x +280\right )}{280} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(7/2),x)
Output:
(sqrt(c)*c**3*x*( - 35*a**7*x**7 - 40*a**6*x**6 + 140*a**5*x**5 + 168*a**4 *x**4 - 210*a**3*x**3 - 280*a**2*x**2 + 140*a*x + 280))/280