Integrand size = 24, antiderivative size = 93 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {2 (1+a x)^5 \left (c-a^2 c x^2\right )^{5/2}}{5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}+\frac {(1+a x)^6 \left (c-a^2 c x^2\right )^{5/2}}{6 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5} \] 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 = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.59 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 (1+a x)^5 (-7+5 a x) \sqrt {c-a^2 c x^2}}{30 a^2 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^(5/2),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 = 0.74 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.71, 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.
\(\displaystyle \int \left (c-a^2 c x^2\right )^{5/2} e^{3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{5/2} \int -\left ((1-a x) (a x+1)^4\right )dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\left (c-a^2 c x^2\right )^{5/2} \int (1-a x) (a x+1)^4dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\left (c-a^2 c x^2\right )^{5/2} \int \left (2 (a x+1)^4-(a x+1)^5\right )dx}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (\frac {2 (a x+1)^5}{5 a}-\frac {(a x+1)^6}{6 a}\right ) \left (c-a^2 c x^2\right )^{5/2}}{a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\) |
Input:
Int[E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^(5/2),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 = 0.90
method | result | size |
gosper | \(\frac {x \left (5 a^{5} x^{5}+18 a^{4} x^{4}+15 a^{3} x^{3}-20 a^{2} x^{2}-45 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{30 \left (a x -1\right ) \left (a x +1\right )^{4} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(84\) |
orering | \(\frac {x \left (5 a^{5} x^{5}+18 a^{4} x^{4}+15 a^{3} x^{3}-20 a^{2} x^{2}-45 a x -30\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{30 \left (a x -1\right ) \left (a x +1\right )^{4} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(84\) |
default | \(\frac {\left (5 a^{5} x^{5}+18 a^{4} x^{4}+15 a^{3} x^{3}-20 a^{2} x^{2}-45 a x -30\right ) x \,c^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x -1\right )}{30 \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(86\) |
Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(5/2),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)^4/((a*x-1)/(a*x+1))^(3/2)
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {{\left (5 \, a^{5} c^{2} x^{6} + 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} - 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} - 30 \, c^{2} x\right )} \sqrt {-a^{2} c}}{30 \, a} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="fri cas")
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(1/((a*x-1)/(a*x+1))**(3/2)*(-a**2*c*x**2+c)**(5/2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.51 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {{\left (5 \, a^{7} \sqrt {-c} c^{2} x^{7} + 13 \, a^{6} \sqrt {-c} c^{2} x^{6} - 3 \, a^{5} \sqrt {-c} c^{2} x^{5} - 35 \, a^{4} \sqrt {-c} c^{2} x^{4} - 25 \, a^{3} \sqrt {-c} c^{2} x^{3} + 15 \, a^{2} \sqrt {-c} c^{2} x^{2} + 30 \, \sqrt {-c} c^{2}\right )} {\left (a x + 1\right )}^{2}}{30 \, {\left (a^{3} x^{2} + 2 \, a^{2} x + a\right )} {\left (a x - 1\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="max ima")
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 ))
Exception generated. \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="gia c")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[0,5,5,2,0]%%%}+%%%{3,[0,4,4,2,0]%%%}+%%%{2,[0,3,3,2,0]% %%}+%%%{-
Timed out. \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^{5/2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \] 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 = 0.61 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} i \left (-5 a^{6} x^{6}-18 a^{5} x^{5}-15 a^{4} x^{4}+20 a^{3} x^{3}+45 a^{2} x^{2}+30 a x -57\right )}{30 a} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(5/2),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 - 57))/(30*a)