Integrand size = 22, antiderivative size = 136 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 (1+a x)^4 \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}-\frac {4 c^2 (1+a x)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}}+\frac {c^2 (1+a x)^6 \sqrt {c-a^2 c x^2}}{6 a \sqrt {1-a^2 x^2}} \] 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 = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.44 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 (1+a x)^4 \left (11-14 a x+5 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{30 a \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(5/2),x]
Output:
(c^2*(1 + a*x)^4*(11 - 14*a*x + 5*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(30*a*Sqrt [1 - a^2*x^2])
Time = 0.60 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.54, 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.
\(\displaystyle \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int (1-a x)^2 (a x+1)^3dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int \left ((a x+1)^5-4 (a x+1)^4+4 (a x+1)^3\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \left (\frac {(a x+1)^6}{6 a}-\frac {4 (a x+1)^5}{5 a}+\frac {(a x+1)^4}{a}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[E^ArcTanh[a*x]*(c - a^2*c*x^2)^(5/2),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 = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.52
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 | \(\frac {x \left (5 a^{5} x^{5}+6 a^{4} x^{4}-15 a^{3} x^{3}-20 a^{2} x^{2}+15 a x +30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{30 \left (a x +1\right )^{2} \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}\) | \(81\) |
orering | \(\frac {x \left (5 a^{5} x^{5}+6 a^{4} x^{4}-15 a^{3} x^{3}-20 a^{2} x^{2}+15 a x +30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{30 \left (a x +1\right )^{2} \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}\) | \(81\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(5/2),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 = 0.72 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {{\left (5 \, a^{5} c^{2} x^{6} + 6 \, a^{4} c^{2} x^{5} - 15 \, a^{3} c^{2} x^{4} - 20 \, a^{2} c^{2} x^{3} + 15 \, a c^{2} x^{2} + 30 \, c^{2} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{30 \, {\left (a^{2} x^{2} - 1\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(5/2),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 + 1 5*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)
\[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral((-c*(a*x - 1)*(a*x + 1))**(5/2)*(a*x + 1)/sqrt(-(a*x - 1)*(a*x + 1)), x)
Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {1}{5} \, a^{4} c^{\frac {5}{2}} x^{5} - \frac {2}{3} \, a^{2} c^{\frac {5}{2}} x^{3} + c^{\frac {5}{2}} x - \frac {1}{6} \, {\left (\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{2} c^{2} x^{4} - 2 \, a^{2} c^{\frac {5}{2}} x^{4} - 4 \, c^{\frac {5}{2}} x^{2} + \frac {7 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} c^{2}}{a^{2}}\right )} a \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="ma xima")
Output:
1/5*a^4*c^(5/2)*x^5 - 2/3*a^2*c^(5/2)*x^3 + c^(5/2)*x - 1/6*(sqrt(a^4*c*x^ 4 - 2*a^2*c*x^2 + c)*a^2*c^2*x^4 - 2*a^2*c^(5/2)*x^4 - 4*c^(5/2)*x^2 + 7*s qrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*c^2/a^2)*a
\[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="gi ac")
Output:
integrate((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)/sqrt(-a^2*x^2 + 1), x)
Time = 23.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c-a^2\,c\,x^2}\,\left (\frac {a^5\,c^2\,x^6}{6}+\frac {a^4\,c^2\,x^5}{5}-\frac {a^3\,c^2\,x^4}{2}-\frac {2\,a^2\,c^2\,x^3}{3}+\frac {a\,c^2\,x^2}{2}+c^2\,x\right )}{\sqrt {1-a^2\,x^2}} \] Input:
int(((c - a^2*c*x^2)^(5/2)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
((c - a^2*c*x^2)^(1/2)*(c^2*x + (a*c^2*x^2)/2 - (2*a^2*c^2*x^3)/3 - (a^3*c ^2*x^4)/2 + (a^4*c^2*x^5)/5 + (a^5*c^2*x^6)/6))/(1 - a^2*x^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.34 \[ \int e^{\text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} x \left (5 a^{5} x^{5}+6 a^{4} x^{4}-15 a^{3} x^{3}-20 a^{2} x^{2}+15 a x +30\right )}{30} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*c*x^2+c)^(5/2),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