Integrand size = 24, antiderivative size = 95 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {2 (1-a x)^5 \left (c-a^2 c x^2\right )^{5/2}}{5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}+\frac {(1-a x)^6 \left (c-a^2 c x^2\right )^{5/2}}{6 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5} \] Output:
-2/5*(-a*x+1)^5*(-a^2*c*x^2+c)^(5/2)/a^6/(1-1/a^2/x^2)^(5/2)/x^5+1/6*(-a*x +1)^6*(-a^2*c*x^2+c)^(5/2)/a^6/(1-1/a^2/x^2)^(5/2)/x^5
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.58 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 (-1+a x)^5 (7+5 a x) \sqrt {c-a^2 c x^2}}{30 a^2 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[(c - a^2*c*x^2)^(5/2)/E^(3*ArcCoth[a*x]),x]
Output:
(c^2*(-1 + a*x)^5*(7 + 5*a*x)*Sqrt[c - a^2*c*x^2])/(30*a^2*Sqrt[1 - 1/(a^2 *x^2)]*x)
Time = 0.76 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6746, 6747, 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.
\(\displaystyle \int \left (c-a^2 c x^2\right )^{5/2} e^{-3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{5/2} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5dx}{x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{5/2} \int (1-a x)^4 (a x+1)dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{5/2} \int \left (2 (1-a x)^4-(1-a x)^5\right )dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (\frac {(1-a x)^6}{6 a}-\frac {2 (1-a x)^5}{5 a}\right ) \left (c-a^2 c x^2\right )^{5/2}}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
Input:
Int[(c - a^2*c*x^2)^(5/2)/E^(3*ArcCoth[a*x]),x]
Output:
((c - a^2*c*x^2)^(5/2)*((-2*(1 - a*x)^5)/(5*a) + (1 - a*x)^6/(6*a)))/(a^5* (1 - 1/(a^2*x^2))^(5/2)*x^5)
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^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && !IntegerQ[p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[c^p/a^(2*p) Int[(u/x^(2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88
method | result | size |
gosper | \(\frac {x \left (5 a^{5} x^{5}-18 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}-45 a x +30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{30 \left (a x +1\right ) \left (a x -1\right )^{4}}\) | \(84\) |
orering | \(\frac {x \left (5 a^{5} x^{5}-18 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}-45 a x +30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{30 \left (a x +1\right ) \left (a x -1\right )^{4}}\) | \(84\) |
default | \(\frac {\left (5 a^{5} x^{5}-18 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}-45 a x +30\right ) x \,c^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{30 \left (a x -1\right )^{2}}\) | \(86\) |
Input:
int((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/30*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)*(-a^2*c*x^2+ c)^(5/2)*((a*x-1)/(a*x+1))^(3/2)/(a*x+1)/(a*x-1)^4
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.77 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {{\left (5 \, a^{5} c^{2} x^{6} - 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} + 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} + 30 \, c^{2} x\right )} \sqrt {-a^{2} c}}{30 \, a} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="frica s")
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)/a
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((-a**2*c*x**2+c)**(5/2)*((a*x-1)/(a*x+1))**(3/2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.47 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {{\left (5 \, a^{7} \sqrt {-c} c^{2} x^{7} - 13 \, a^{6} \sqrt {-c} c^{2} x^{6} - 3 \, a^{5} \sqrt {-c} c^{2} x^{5} + 35 \, a^{4} \sqrt {-c} c^{2} x^{4} - 25 \, a^{3} \sqrt {-c} c^{2} x^{3} - 15 \, a^{2} \sqrt {-c} c^{2} x^{2} - 30 \, \sqrt {-c} c^{2}\right )} {\left (a x - 1\right )}^{2}}{30 \, {\left (a^{3} x^{2} - 2 \, a^{2} x + a\right )} {\left (a x + 1\right )}} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxim a")
Output:
1/30*(5*a^7*sqrt(-c)*c^2*x^7 - 13*a^6*sqrt(-c)*c^2*x^6 - 3*a^5*sqrt(-c)*c^ 2*x^5 + 35*a^4*sqrt(-c)*c^2*x^4 - 25*a^3*sqrt(-c)*c^2*x^3 - 15*a^2*sqrt(-c )*c^2*x^2 - 30*sqrt(-c)*c^2)*(a*x - 1)^2/((a^3*x^2 - 2*a^2*x + a)*(a*x + 1 ))
Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {1}{30} \, {\left (5 \, a^{5} c^{2} x^{6} \mathrm {sgn}\left (a x + 1\right ) - 18 \, a^{4} c^{2} x^{5} \mathrm {sgn}\left (a x + 1\right ) + 15 \, a^{3} c^{2} x^{4} \mathrm {sgn}\left (a x + 1\right ) + 20 \, a^{2} c^{2} x^{3} \mathrm {sgn}\left (a x + 1\right ) - 45 \, a c^{2} x^{2} \mathrm {sgn}\left (a x + 1\right ) + 30 \, c^{2} x \mathrm {sgn}\left (a x + 1\right ) + \frac {57 \, c^{2} \mathrm {sgn}\left (a x + 1\right )}{a}\right )} \sqrt {-c} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac" )
Output:
1/30*(5*a^5*c^2*x^6*sgn(a*x + 1) - 18*a^4*c^2*x^5*sgn(a*x + 1) + 15*a^3*c^ 2*x^4*sgn(a*x + 1) + 20*a^2*c^2*x^3*sgn(a*x + 1) - 45*a*c^2*x^2*sgn(a*x + 1) + 30*c^2*x*sgn(a*x + 1) + 57*c^2*sgn(a*x + 1)/a)*sqrt(-c)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int {\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:
int((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.60 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} i \left (-5 a^{6} x^{6}+18 a^{5} x^{5}-15 a^{4} x^{4}-20 a^{3} x^{3}+45 a^{2} x^{2}-30 a x +7\right )}{30 a} \] Input:
int((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x)
Output:
(sqrt(c)*c**2*i*( - 5*a**6*x**6 + 18*a**5*x**5 - 15*a**4*x**4 - 20*a**3*x* *3 + 45*a**2*x**2 - 30*a*x + 7))/(30*a)