Integrand size = 24, antiderivative size = 187 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {8 (1-a x)^5 \left (c-a^2 c x^2\right )^{7/2}}{5 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}+\frac {2 (1-a x)^6 \left (c-a^2 c x^2\right )^{7/2}}{a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}-\frac {6 (1-a x)^7 \left (c-a^2 c x^2\right )^{7/2}}{7 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}+\frac {(1-a x)^8 \left (c-a^2 c x^2\right )^{7/2}}{8 a^8 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7} \] Output:
-8/5*(-a*x+1)^5*(-a^2*c*x^2+c)^(7/2)/a^8/(1-1/a^2/x^2)^(7/2)/x^7+2*(-a*x+1 )^6*(-a^2*c*x^2+c)^(7/2)/a^8/(1-1/a^2/x^2)^(7/2)/x^7-6/7*(-a*x+1)^7*(-a^2* c*x^2+c)^(7/2)/a^8/(1-1/a^2/x^2)^(7/2)/x^7+1/8*(-a*x+1)^8*(-a^2*c*x^2+c)^( 7/2)/a^8/(1-1/a^2/x^2)^(7/2)/x^7
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.38 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {c^3 (-1+a x)^5 \sqrt {c-a^2 c x^2} \left (93+185 a x+135 a^2 x^2+35 a^3 x^3\right )}{280 a^2 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[(c - a^2*c*x^2)^(7/2)/E^ArcCoth[a*x],x]
Output:
-1/280*(c^3*(-1 + a*x)^5*Sqrt[c - a^2*c*x^2]*(93 + 185*a*x + 135*a^2*x^2 + 35*a^3*x^3))/(a^2*Sqrt[1 - 1/(a^2*x^2)]*x)
Time = 0.80 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.51, 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.
\(\displaystyle \int \left (c-a^2 c x^2\right )^{7/2} e^{-\coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{7/2} \int e^{-\coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7dx}{x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{7/2} \int (1-a x)^4 (a x+1)^3dx}{a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{7/2} \int \left (-(1-a x)^7+6 (1-a x)^6-12 (1-a x)^5+8 (1-a x)^4\right )dx}{a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (\frac {(1-a x)^8}{8 a}-\frac {6 (1-a x)^7}{7 a}+\frac {2 (1-a x)^6}{a}-\frac {8 (1-a x)^5}{5 a}\right ) \left (c-a^2 c x^2\right )^{7/2}}{a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}\) |
Input:
Int[(c - a^2*c*x^2)^(7/2)/E^ArcCoth[a*x],x]
Output:
((c - a^2*c*x^2)^(7/2)*((-8*(1 - a*x)^5)/(5*a) + (2*(1 - a*x)^6)/a - (6*(1 - a*x)^7)/(7*a) + (1 - a*x)^8/(8*a)))/(a^7*(1 - 1/(a^2*x^2))^(7/2)*x^7)
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 = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.52
method | result | size |
default | \(-\frac {\left (35 a^{7} x^{7}-40 x^{6} a^{6}-140 a^{5} x^{5}+168 a^{4} x^{4}+210 a^{3} x^{3}-280 a^{2} x^{2}-140 a x +280\right ) x \,c^{3} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {\frac {a x -1}{a x +1}}}{280 \left (a x -1\right )}\) | \(97\) |
gosper | \(\frac {x \left (35 a^{7} x^{7}-40 x^{6} a^{6}-140 a^{5} x^{5}+168 a^{4} x^{4}+210 a^{3} x^{3}-280 a^{2} x^{2}-140 a x +280\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}} \sqrt {\frac {a x -1}{a x +1}}}{280 \left (a x +1\right )^{3} \left (a x -1\right )^{4}}\) | \(100\) |
orering | \(\frac {x \left (35 a^{7} x^{7}-40 x^{6} a^{6}-140 a^{5} x^{5}+168 a^{4} x^{4}+210 a^{3} x^{3}-280 a^{2} x^{2}-140 a x +280\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}} \sqrt {\frac {a x -1}{a x +1}}}{280 \left (a x +1\right )^{3} \left (a x -1\right )^{4}}\) | \(100\) |
Input:
int((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/280*(35*a^7*x^7-40*a^6*x^6-140*a^5*x^5+168*a^4*x^4+210*a^3*x^3-280*a^2* x^2-140*a*x+280)*x*c^3*(-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 = 95, normalized size of antiderivative = 0.51 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {{\left (35 \, a^{7} c^{3} x^{8} - 40 \, a^{6} c^{3} x^{7} - 140 \, a^{5} c^{3} x^{6} + 168 \, a^{4} c^{3} x^{5} + 210 \, a^{3} c^{3} x^{4} - 280 \, a^{2} c^{3} x^{3} - 140 \, a c^{3} x^{2} + 280 \, c^{3} x\right )} \sqrt {-a^{2} c}}{280 \, a} \] Input:
integrate((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="frica s")
Output:
-1/280*(35*a^7*c^3*x^8 - 40*a^6*c^3*x^7 - 140*a^5*c^3*x^6 + 168*a^4*c^3*x^ 5 + 210*a^3*c^3*x^4 - 280*a^2*c^3*x^3 - 140*a*c^3*x^2 + 280*c^3*x)*sqrt(-a ^2*c)/a
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\text {Timed out} \] Input:
integrate((-a**2*c*x**2+c)**(7/2)*((a*x-1)/(a*x+1))**(1/2),x)
Output:
Timed out
\[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int { {\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} \sqrt {\frac {a x - 1}{a x + 1}} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="maxim a")
Output:
integrate((-a^2*c*x^2 + c)^(7/2)*sqrt((a*x - 1)/(a*x + 1)), x)
Time = 0.12 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.81 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=-\frac {1}{280} \, {\left (35 \, a^{7} c^{3} x^{8} \mathrm {sgn}\left (a x + 1\right ) - 40 \, a^{6} c^{3} x^{7} \mathrm {sgn}\left (a x + 1\right ) - 140 \, a^{5} c^{3} x^{6} \mathrm {sgn}\left (a x + 1\right ) + 168 \, a^{4} c^{3} x^{5} \mathrm {sgn}\left (a x + 1\right ) + 210 \, a^{3} c^{3} x^{4} \mathrm {sgn}\left (a x + 1\right ) - 280 \, a^{2} c^{3} x^{3} \mathrm {sgn}\left (a x + 1\right ) - 140 \, a c^{3} x^{2} \mathrm {sgn}\left (a x + 1\right ) + 280 \, c^{3} x \mathrm {sgn}\left (a x + 1\right ) + \frac {163 \, c^{3} \mathrm {sgn}\left (a x + 1\right )}{a}\right )} \sqrt {-c} \] Input:
integrate((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="giac" )
Output:
-1/280*(35*a^7*c^3*x^8*sgn(a*x + 1) - 40*a^6*c^3*x^7*sgn(a*x + 1) - 140*a^ 5*c^3*x^6*sgn(a*x + 1) + 168*a^4*c^3*x^5*sgn(a*x + 1) + 210*a^3*c^3*x^4*sg n(a*x + 1) - 280*a^2*c^3*x^3*sgn(a*x + 1) - 140*a*c^3*x^2*sgn(a*x + 1) + 2 80*c^3*x*sgn(a*x + 1) + 163*c^3*sgn(a*x + 1)/a)*sqrt(-c)
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\int {\left (c-a^2\,c\,x^2\right )}^{7/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}} \,d x \] Input:
int((c - a^2*c*x^2)^(7/2)*((a*x - 1)/(a*x + 1))^(1/2),x)
Output:
int((c - a^2*c*x^2)^(7/2)*((a*x - 1)/(a*x + 1))^(1/2), x)
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.39 \[ \int e^{-\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{7/2} \, dx=\frac {\sqrt {c}\, c^{3} i \left (35 a^{8} x^{8}-40 a^{7} x^{7}-140 a^{6} x^{6}+168 a^{5} x^{5}+210 a^{4} x^{4}-280 a^{3} x^{3}-140 a^{2} x^{2}+280 a x -93\right )}{280 a} \] Input:
int((-a^2*c*x^2+c)^(7/2)*((a*x-1)/(a*x+1))^(1/2),x)
Output:
(sqrt(c)*c**3*i*(35*a**8*x**8 - 40*a**7*x**7 - 140*a**6*x**6 + 168*a**5*x* *5 + 210*a**4*x**4 - 280*a**3*x**3 - 140*a**2*x**2 + 280*a*x - 93))/(280*a )