Integrand size = 24, antiderivative size = 139 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {(1-a x)^4 \left (c-a^2 c x^2\right )^{5/2}}{a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}-\frac {4 (1-a x)^5 \left (c-a^2 c x^2\right )^{5/2}}{5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}+\frac {(1-a x)^6 \left (c-a^2 c x^2\right )^{5/2}}{6 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5} \] Output:
(-a*x+1)^4*(-a^2*c*x^2+c)^(5/2)/a^6/(1-1/a^2/x^2)^(5/2)/x^5-4/5*(-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 = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.45 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 (-1+a x)^4 \left (11+14 a x+5 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{30 a^2 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[(c - a^2*c*x^2)^(5/2)/E^ArcCoth[a*x],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^2*S qrt[1 - 1/(a^2*x^2)]*x)
Time = 0.76 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.58, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6746, 6747, 25, 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.
\(\displaystyle \int \left (c-a^2 c x^2\right )^{5/2} e^{-\coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{5/2} \int e^{-\coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5dx}{x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{5/2} \int -(1-a x)^3 (a x+1)^2dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\left (c-a^2 c x^2\right )^{5/2} \int (1-a x)^3 (a x+1)^2dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\left (c-a^2 c x^2\right )^{5/2} \int \left ((1-a x)^5-4 (1-a x)^4+4 (1-a x)^3\right )dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (-\frac {(1-a x)^6}{6 a}+\frac {4 (1-a x)^5}{5 a}-\frac {(1-a x)^4}{a}\right ) \left (c-a^2 c x^2\right )^{5/2}}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
Input:
Int[(c - a^2*c*x^2)^(5/2)/E^ArcCoth[a*x],x]
Output:
-(((c - a^2*c*x^2)^(5/2)*(-((1 - a*x)^4/a) + (4*(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 = 81, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {\left (5 a^{5} x^{5}-6 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}+15 a x -30\right ) x \,c^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {\frac {a x -1}{a x +1}}}{30 a x -30}\) | \(81\) |
gosper | \(\frac {x \left (5 a^{5} x^{5}-6 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}+15 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \sqrt {\frac {a x -1}{a x +1}}}{30 \left (a x +1\right )^{2} \left (a x -1\right )^{3}}\) | \(84\) |
orering | \(\frac {x \left (5 a^{5} x^{5}-6 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}+15 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \sqrt {\frac {a x -1}{a x +1}}}{30 \left (a x +1\right )^{2} \left (a x -1\right )^{3}}\) | \(84\) |
Input:
int((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x,method=_RETURNVERBOSE)
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)*x*c^2*(-c*(a^2* x^2-1))^(1/2)*((a*x-1)/(a*x+1))^(1/2)/(a*x-1)
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.53 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {{\left (5 \, a^{5} c^{2} x^{6} - 6 \, a^{4} c^{2} x^{5} - 15 \, a^{3} c^{2} x^{4} + 20 \, a^{2} c^{2} x^{3} + 15 \, a c^{2} x^{2} - 30 \, c^{2} x\right )} \sqrt {-a^{2} c}}{30 \, a} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="frica s")
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)/a
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((-a**2*c*x**2+c)**(5/2)*((a*x-1)/(a*x+1))**(1/2),x)
Output:
Timed out
\[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int { {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="maxim a")
Output:
integrate((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1)), x)
Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.84 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {1}{30} \, {\left (5 \, a^{5} c^{2} x^{6} \mathrm {sgn}\left (a x + 1\right ) - 6 \, a^{4} c^{2} x^{5} \mathrm {sgn}\left (a x + 1\right ) - 15 \, a^{3} c^{2} x^{4} \mathrm {sgn}\left (a x + 1\right ) + 20 \, a^{2} c^{2} x^{3} \mathrm {sgn}\left (a x + 1\right ) + 15 \, a c^{2} x^{2} \mathrm {sgn}\left (a x + 1\right ) - 30 \, c^{2} x \mathrm {sgn}\left (a x + 1\right ) - \frac {21 \, c^{2} \mathrm {sgn}\left (a x + 1\right )}{a}\right )} \sqrt {-c} \] Input:
integrate((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="giac" )
Output:
1/30*(5*a^5*c^2*x^6*sgn(a*x + 1) - 6*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) + 15*a*c^2*x^2*sgn(a*x + 1 ) - 30*c^2*x*sgn(a*x + 1) - 21*c^2*sgn(a*x + 1)/a)*sqrt(-c)
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int {\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}} \,d x \] Input:
int((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2),x)
Output:
int((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2), x)
Time = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.41 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} i \left (-5 a^{6} x^{6}+6 a^{5} x^{5}+15 a^{4} x^{4}-20 a^{3} x^{3}-15 a^{2} x^{2}+30 a x -11\right )}{30 a} \] Input:
int((-a^2*c*x^2+c)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x)
Output:
(sqrt(c)*c**2*i*( - 5*a**6*x**6 + 6*a**5*x**5 + 15*a**4*x**4 - 20*a**3*x** 3 - 15*a**2*x**2 + 30*a*x - 11))/(30*a)