Integrand size = 24, antiderivative size = 140 \[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {c^2 (1-a x)^4 \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}+\frac {4 c^2 (1-a x)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}}-\frac {c^2 (1-a x)^6 \sqrt {c-a^2 c x^2}}{6 a \sqrt {1-a^2 x^2}} \] Output:
-c^2*(-a*x+1)^4*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)+4/5*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 = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.43 \[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {c^2 (-1+a x)^4 \left (11+14 a x+5 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{30 a \sqrt {1-a^2 x^2}} \] Input:
Integrate[(c - a^2*c*x^2)^(5/2)/E^ArcTanh[a*x],x]
Output:
-1/30*(c^2*(-1 + a*x)^4*(11 + 14*a*x + 5*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(a* Sqrt[1 - a^2*x^2])
Time = 0.39 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.55, 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.
\(\displaystyle \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int e^{-\text {arctanh}(a x)} \left (1-a^2 x^2\right )^{5/2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int (1-a x)^3 (a x+1)^2dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int \left ((1-a x)^5-4 (1-a x)^4+4 (1-a x)^3\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \left (-\frac {(1-a x)^6}{6 a}+\frac {4 (1-a x)^5}{5 a}-\frac {(1-a x)^4}{a}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[(c - a^2*c*x^2)^(5/2)/E^ArcTanh[a*x],x]
Output:
(c^2*Sqrt[c - a^2*c*x^2]*(-((1 - a*x)^4/a) + (4*(1 - a*x)^5)/(5*a) - (1 - a*x)^6/(6*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 = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.51
method | result | size |
default | \(-\frac {\left (5 a^{5} x^{5}-6 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}+15 a x -30\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c^{2} x}{30 \sqrt {-a^{2} x^{2}+1}}\) | \(71\) |
gosper | \(\frac {x \left (5 a^{5} x^{5}-6 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}+15 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \sqrt {-a^{2} x^{2}+1}}{30 \left (a x -1\right )^{3} \left (a x +1\right )^{3}}\) | \(81\) |
orering | \(\frac {x \left (5 a^{5} x^{5}-6 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}+15 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \sqrt {-a^{2} x^{2}+1}}{30 \left (a x -1\right )^{3} \left (a x +1\right )^{3}}\) | \(81\) |
Input:
int((-a^2*c*x^2+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOS E)
Output:
-1/30*(5*a^5*x^5-6*a^4*x^4-15*a^3*x^3+20*a^2*x^2+15*a*x-30)/(-a^2*x^2+1)^( 1/2)*(-c*(a^2*x^2-1))^(1/2)*c^2*x
Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.70 \[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {{\left (5 \, a^{5} c^{2} x^{6} - 6 \, a^{4} c^{2} x^{5} - 15 \, a^{3} c^{2} x^{4} + 20 \, a^{2} c^{2} x^{3} + 15 \, a c^{2} x^{2} - 30 \, c^{2} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{30 \, {\left (a^{2} x^{2} - 1\right )}} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fr icas")
Output:
1/30*(5*a^5*c^2*x^6 - 6*a^4*c^2*x^5 - 15*a^3*c^2*x^4 + 20*a^2*c^2*x^3 + 15 *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)
\[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}{a x + 1}\, dx \] Input:
integrate((-a**2*c*x**2+c)**(5/2)/(a*x+1)*(-a**2*x**2+1)**(1/2),x)
Output:
Integral(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(5/2)/(a*x + 1), x)
\[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}}{a x + 1} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="ma xima")
Output:
integrate((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)/(a*x + 1), x)
Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.46 \[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {1}{30} \, {\left (5 \, a^{5} c^{2} x^{6} - 6 \, a^{4} c^{2} x^{5} - 15 \, a^{3} c^{2} x^{4} + 20 \, a^{2} c^{2} x^{3} + 15 \, a c^{2} x^{2} - 30 \, c^{2} x\right )} \sqrt {c} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="gi ac")
Output:
-1/30*(5*a^5*c^2*x^6 - 6*a^4*c^2*x^5 - 15*a^3*c^2*x^4 + 20*a^2*c^2*x^3 + 1 5*a*c^2*x^2 - 30*c^2*x)*sqrt(c)
Timed out. \[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \] Input:
int(((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2))/(a*x + 1),x)
Output:
int(((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2))/(a*x + 1), x)
Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.33 \[ \int e^{-\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} x \left (-5 a^{5} x^{5}+6 a^{4} x^{4}+15 a^{3} x^{3}-20 a^{2} x^{2}-15 a x +30\right )}{30} \] Input:
int((-a^2*c*x^2+c)^(5/2)/(a*x+1)*(-a^2*x^2+1)^(1/2),x)
Output:
(sqrt(c)*c**2*x*( - 5*a**5*x**5 + 6*a**4*x**4 + 15*a**3*x**3 - 20*a**2*x** 2 - 15*a*x + 30))/30