Integrand size = 24, antiderivative size = 95 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {2 c^2 (1-a x)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}}+\frac {c^2 (1-a x)^6 \sqrt {c-a^2 c x^2}}{6 a \sqrt {1-a^2 x^2}} \] Output:
-2/5*c^2*(-a*x+1)^5*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)+1/6*c^2*(-a* x+1)^6*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)
Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.55 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 (-1+a x)^5 (7+5 a x) \sqrt {c-a^2 c x^2}}{30 a \sqrt {1-a^2 x^2}} \] Input:
Integrate[(c - a^2*c*x^2)^(5/2)/E^(3*ArcTanh[a*x]),x]
Output:
(c^2*(-1 + a*x)^5*(7 + 5*a*x)*Sqrt[c - a^2*c*x^2])/(30*a*Sqrt[1 - a^2*x^2] )
Time = 0.60 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67, 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.
\(\displaystyle \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int e^{-3 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^{5/2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int (1-a x)^4 (a x+1)dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int \left (2 (1-a x)^4-(1-a x)^5\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \left (\frac {(1-a x)^6}{6 a}-\frac {2 (1-a x)^5}{5 a}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[(c - a^2*c*x^2)^(5/2)/E^(3*ArcTanh[a*x]),x]
Output:
(c^2*Sqrt[c - a^2*c*x^2]*((-2*(1 - a*x)^5)/(5*a) + (1 - a*x)^6/(6*a)))/Sqr t[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 = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75
method | result | size |
default | \(\frac {\left (5 a^{5} x^{5}-18 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}-45 a x +30\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c^{2} x}{30 \sqrt {-a^{2} x^{2}+1}}\) | \(71\) |
gosper | \(\frac {x \left (5 a^{5} x^{5}-18 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}-45 a x +30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{30 \left (a x +1\right )^{4} \left (a x -1\right )^{4}}\) | \(81\) |
orering | \(\frac {x \left (5 a^{5} x^{5}-18 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}-45 a x +30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{30 \left (a x +1\right )^{4} \left (a x -1\right )^{4}}\) | \(81\) |
Input:
int((-a^2*c*x^2+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x,method=_RETURNVERB OSE)
Output:
1/30*(5*a^5*x^5-18*a^4*x^4+15*a^3*x^3+20*a^2*x^2-45*a*x+30)/(-a^2*x^2+1)^( 1/2)*(-c*(a^2*x^2-1))^(1/2)*c^2*x
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.03 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {{\left (5 \, a^{5} c^{2} x^{6} - 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} + 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} + 30 \, c^{2} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{30 \, {\left (a^{2} x^{2} - 1\right )}} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm=" fricas")
Output:
-1/30*(5*a^5*c^2*x^6 - 18*a^4*c^2*x^5 + 15*a^3*c^2*x^4 + 20*a^2*c^2*x^3 - 45*a*c^2*x^2 + 30*c^2*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}{\left (a x + 1\right )^{3}}\, dx \] Input:
integrate((-a**2*c*x**2+c)**(5/2)/(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))**(5/2)/(a* x + 1)**3, x)
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm=" maxima")
Output:
integrate((-a^2*c*x^2 + c)^(5/2)*(-a^2*x^2 + 1)^(3/2)/(a*x + 1)^3, x)
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {1}{30} \, {\left (5 \, a^{5} c^{2} x^{6} - 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} + 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} + 30 \, c^{2} x\right )} \sqrt {c} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm=" giac")
Output:
1/30*(5*a^5*c^2*x^6 - 18*a^4*c^2*x^5 + 15*a^3*c^2*x^4 + 20*a^2*c^2*x^3 - 4 5*a*c^2*x^2 + 30*c^2*x)*sqrt(c)
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \] Input:
int(((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3,x)
Output:
int(((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3, x)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.48 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} x \left (5 a^{5} x^{5}-18 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}-45 a x +30\right )}{30} \] Input:
int((-a^2*c*x^2+c)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
Output:
(sqrt(c)*c**2*x*(5*a**5*x**5 - 18*a**4*x**4 + 15*a**3*x**3 + 20*a**2*x**2 - 45*a*x + 30))/30