Integrand size = 24, antiderivative size = 93 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {2 c^2 (1+a x)^5 \sqrt {c-a^2 c x^2}}{5 a \sqrt {1-a^2 x^2}}-\frac {c^2 (1+a x)^6 \sqrt {c-a^2 c x^2}}{6 a \sqrt {1-a^2 x^2}} \] 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 = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.56 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {c^2 (1+a x)^5 (-7+5 a x) \sqrt {c-a^2 c x^2}}{30 a \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2),x]
Output:
-1/30*(c^2*(1 + a*x)^5*(-7 + 5*a*x)*Sqrt[c - a^2*c*x^2])/(a*Sqrt[1 - a^2*x ^2])
Time = 0.58 (sec) , antiderivative size = 62, 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 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int (1-a x) (a x+1)^4dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^2 \sqrt {c-a^2 c x^2} \int \left (2 (a x+1)^4-(a x+1)^5\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \left (\frac {2 (a x+1)^5}{5 a}-\frac {(a x+1)^6}{6 a}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2),x]
Output:
(c^2*Sqrt[c - a^2*c*x^2]*((2*(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 = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76
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 | \(\frac {x \left (5 a^{5} x^{5}+18 a^{4} x^{4}+15 a^{3} x^{3}-20 a^{2} x^{2}-45 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{30 \left (a x -1\right ) \left (a x +1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(81\) |
orering | \(\frac {x \left (5 a^{5} x^{5}+18 a^{4} x^{4}+15 a^{3} x^{3}-20 a^{2} x^{2}-45 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{30 \left (a x -1\right ) \left (a x +1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(81\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),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 = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {{\left (5 \, a^{5} c^{2} x^{6} + 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} - 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} - 30 \, c^{2} x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{30 \, {\left (a^{2} x^{2} - 1\right )}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),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 - 4 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)
\[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral((-c*(a*x - 1)*(a*x + 1))**(5/2)*(a*x + 1)**3/(-(a*x - 1)*(a*x + 1 ))**(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (81) = 162\).
Time = 0.07 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.55 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {1}{3} \, a^{2} c^{\frac {5}{2}} x^{3} + \frac {1}{12} \, {\left (\frac {2 \, a^{4} c^{3} x^{8}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {5 \, a^{2} c^{3} x^{6}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {3 \, c^{3} x^{4}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}}\right )} a^{3} + c^{\frac {5}{2}} x - \frac {1}{5} \, {\left (3 \, a^{2} c^{\frac {5}{2}} x^{5} - 5 \, c^{\frac {5}{2}} x^{3}\right )} a^{2} + \frac {3}{4} \, {\left (\frac {a^{4} c^{3} x^{6}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} - \frac {3 \, a^{2} c^{3} x^{4}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {2 \, c^{3}}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} a^{2}}\right )} a \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm=" maxima")
Output:
-1/3*a^2*c^(5/2)*x^3 + 1/12*(2*a^4*c^3*x^8/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) - 5*a^2*c^3*x^6/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) + 3*c^3*x^4/sqrt(a^4* c*x^4 - 2*a^2*c*x^2 + c))*a^3 + c^(5/2)*x - 1/5*(3*a^2*c^(5/2)*x^5 - 5*c^( 5/2)*x^3)*a^2 + 3/4*(a^4*c^3*x^6/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) - 3*a^2 *c^3*x^4/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) + 2*c^3/(sqrt(a^4*c*x^4 - 2*a^2 *c*x^2 + c)*a^2))*a
\[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int { \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm=" giac")
Output:
integrate((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)^3/(-a^2*x^2 + 1)^(3/2), x)
Time = 23.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c-a^2\,c\,x^2}\,\left (-\frac {a^5\,c^2\,x^6}{6}-\frac {3\,a^4\,c^2\,x^5}{5}-\frac {a^3\,c^2\,x^4}{2}+\frac {2\,a^2\,c^2\,x^3}{3}+\frac {3\,a\,c^2\,x^2}{2}+c^2\,x\right )}{\sqrt {1-a^2\,x^2}} \] Input:
int(((c - a^2*c*x^2)^(5/2)*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
Output:
((c - a^2*c*x^2)^(1/2)*(c^2*x + (3*a*c^2*x^2)/2 + (2*a^2*c^2*x^3)/3 - (a^3 *c^2*x^4)/2 - (3*a^4*c^2*x^5)/5 - (a^5*c^2*x^6)/6))/(1 - a^2*x^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.49 \[ \int e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} x \left (-5 a^{5} x^{5}-18 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}+45 a x +30\right )}{30} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(5/2),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