Integrand size = 24, antiderivative size = 142 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {2 c^3 (1-a x)^6 \sqrt {c-a^2 c x^2}}{3 a \sqrt {1-a^2 x^2}}+\frac {4 c^3 (1-a x)^7 \sqrt {c-a^2 c x^2}}{7 a \sqrt {1-a^2 x^2}}-\frac {c^3 (1-a x)^8 \sqrt {c-a^2 c x^2}}{8 a \sqrt {1-a^2 x^2}} \] Output:
-2/3*c^3*(-a*x+1)^6*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)+4/7*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 = 60, normalized size of antiderivative = 0.42 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {c^3 (-1+a x)^6 \left (37+54 a x+21 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{168 a \sqrt {1-a^2 x^2}} \] Input:
Integrate[(c - a^2*c*x^2)^(7/2)/E^(3*ArcTanh[a*x]),x]
Output:
-1/168*(c^3*(-1 + a*x)^6*(37 + 54*a*x + 21*a^2*x^2)*Sqrt[c - a^2*c*x^2])/( a*Sqrt[1 - a^2*x^2])
Time = 0.40 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.56, 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.
\(\displaystyle \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {c^3 \sqrt {c-a^2 c x^2} \int e^{-3 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^{7/2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {c^3 \sqrt {c-a^2 c x^2} \int (1-a x)^5 (a x+1)^2dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^3 \sqrt {c-a^2 c x^2} \int \left ((1-a x)^7-4 (1-a x)^6+4 (1-a x)^5\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^3 \left (-\frac {(1-a x)^8}{8 a}+\frac {4 (1-a x)^7}{7 a}-\frac {2 (1-a x)^6}{3 a}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[(c - a^2*c*x^2)^(7/2)/E^(3*ArcTanh[a*x]),x]
Output:
(c^3*Sqrt[c - a^2*c*x^2]*((-2*(1 - a*x)^6)/(3*a) + (4*(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 = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.61
method | result | size |
default | \(-\frac {\left (21 a^{7} x^{7}-72 x^{6} a^{6}+28 a^{5} x^{5}+168 a^{4} x^{4}-210 a^{3} x^{3}-56 a^{2} x^{2}+252 a x -168\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c^{3} x}{168 \sqrt {-a^{2} x^{2}+1}}\) | \(87\) |
gosper | \(\frac {x \left (21 a^{7} x^{7}-72 x^{6} a^{6}+28 a^{5} x^{5}+168 a^{4} x^{4}-210 a^{3} x^{3}-56 a^{2} x^{2}+252 a x -168\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{168 \left (a x +1\right )^{5} \left (a x -1\right )^{5}}\) | \(97\) |
orering | \(\frac {x \left (21 a^{7} x^{7}-72 x^{6} a^{6}+28 a^{5} x^{5}+168 a^{4} x^{4}-210 a^{3} x^{3}-56 a^{2} x^{2}+252 a x -168\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{168 \left (a x +1\right )^{5} \left (a x -1\right )^{5}}\) | \(97\) |
Input:
int((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x,method=_RETURNVERB OSE)
Output:
-1/168*(21*a^7*x^7-72*a^6*x^6+28*a^5*x^5+168*a^4*x^4-210*a^3*x^3-56*a^2*x^ 2+252*a*x-168)/(-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 = 0.85 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {{\left (21 \, a^{7} c^{3} x^{8} - 72 \, a^{6} c^{3} x^{7} + 28 \, a^{5} c^{3} x^{6} + 168 \, a^{4} c^{3} x^{5} - 210 \, a^{3} c^{3} x^{4} - 56 \, a^{2} c^{3} x^{3} + 252 \, a c^{3} x^{2} - 168 \, c^{3} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{168 \, {\left (a^{2} x^{2} - 1\right )}} \] Input:
integrate((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm=" fricas")
Output:
1/168*(21*a^7*c^3*x^8 - 72*a^6*c^3*x^7 + 28*a^5*c^3*x^6 + 168*a^4*c^3*x^5 - 210*a^3*c^3*x^4 - 56*a^2*c^3*x^3 + 252*a*c^3*x^2 - 168*c^3*x)*sqrt(-a^2* c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}}}{\left (a x + 1\right )^{3}}\, dx \] Input:
integrate((-a**2*c*x**2+c)**(7/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)
Output:
Integral((-(a*x - 1)*(a*x + 1))**(3/2)*(-c*(a*x - 1)*(a*x + 1))**(7/2)/(a* x + 1)**3, x)
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm=" maxima")
Output:
integrate((-a^2*c*x^2 + c)^(7/2)*(-a^2*x^2 + 1)^(3/2)/(a*x + 1)^3, x)
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.61 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {1}{168} \, {\left (21 \, a^{7} c^{3} x^{8} - 72 \, a^{6} c^{3} x^{7} + 28 \, a^{5} c^{3} x^{6} + 168 \, a^{4} c^{3} x^{5} - 210 \, a^{3} c^{3} x^{4} - 56 \, a^{2} c^{3} x^{3} + 252 \, a c^{3} x^{2} - 168 \, c^{3} x\right )} \sqrt {c} \] Input:
integrate((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm=" giac")
Output:
-1/168*(21*a^7*c^3*x^8 - 72*a^6*c^3*x^7 + 28*a^5*c^3*x^6 + 168*a^4*c^3*x^5 - 210*a^3*c^3*x^4 - 56*a^2*c^3*x^3 + 252*a*c^3*x^2 - 168*c^3*x)*sqrt(c)
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^{7/2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \] Input:
int(((c - a^2*c*x^2)^(7/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3,x)
Output:
int(((c - a^2*c*x^2)^(7/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3, x)
Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.44 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {\sqrt {c}\, c^{3} x \left (-21 a^{7} x^{7}+72 a^{6} x^{6}-28 a^{5} x^{5}-168 a^{4} x^{4}+210 a^{3} x^{3}+56 a^{2} x^{2}-252 a x +168\right )}{168} \] Input:
int((-a^2*c*x^2+c)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
Output:
(sqrt(c)*c**3*x*( - 21*a**7*x**7 + 72*a**6*x**6 - 28*a**5*x**5 - 168*a**4* x**4 + 210*a**3*x**3 + 56*a**2*x**2 - 252*a*x + 168))/168