Integrand size = 22, antiderivative size = 69 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {4 c^4 (1+a x)^6}{3 a}+\frac {12 c^4 (1+a x)^7}{7 a}-\frac {3 c^4 (1+a x)^8}{4 a}+\frac {c^4 (1+a x)^9}{9 a} \] Output:
-4/3*c^4*(a*x+1)^6/a+12/7*c^4*(a*x+1)^7/a-3/4*c^4*(a*x+1)^8/a+1/9*c^4*(a*x +1)^9/a
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {c^4 (1+a x)^6 \left (-65+138 a x-105 a^2 x^2+28 a^3 x^3\right )}{252 a} \] Input:
Integrate[E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^4,x]
Output:
(c^4*(1 + a*x)^6*(-65 + 138*a*x - 105*a^2*x^2 + 28*a^3*x^3))/(252*a)
Time = 0.59 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6717, 27, 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.
\(\displaystyle \int \left (c-a^2 c x^2\right )^4 e^{2 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int c^4 e^{2 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^4dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -c^4 \int e^{2 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^4dx\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle -c^4 \int (1-a x)^3 (a x+1)^5dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -c^4 \int \left (-(a x+1)^8+6 (a x+1)^7-12 (a x+1)^6+8 (a x+1)^5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -c^4 \left (-\frac {(a x+1)^9}{9 a}+\frac {3 (a x+1)^8}{4 a}-\frac {12 (a x+1)^7}{7 a}+\frac {4 (a x+1)^6}{3 a}\right )\) |
Input:
Int[E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^4,x]
Output:
-(c^4*((4*(1 + a*x)^6)/(3*a) - (12*(1 + a*x)^7)/(7*a) + (3*(1 + a*x)^8)/(4 *a) - (1 + a*x)^9/(9*a)))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88
method | result | size |
gosper | \(\frac {c^{4} x \left (28 a^{8} x^{8}+63 a^{7} x^{7}-72 x^{6} a^{6}-252 a^{5} x^{5}+378 a^{3} x^{3}+168 a^{2} x^{2}-252 a x -252\right )}{252}\) | \(61\) |
default | \(c^{4} \left (\frac {1}{9} a^{8} x^{9}+\frac {1}{4} a^{7} x^{8}-\frac {2}{7} a^{6} x^{7}-a^{5} x^{6}+\frac {3}{2} a^{3} x^{4}+\frac {2}{3} a^{2} x^{3}-a \,x^{2}-x \right )\) | \(63\) |
norman | \(-c^{4} x -a \,c^{4} x^{2}+\frac {2}{3} a^{2} c^{4} x^{3}+\frac {3}{2} a^{3} c^{4} x^{4}-a^{5} c^{4} x^{6}-\frac {2}{7} a^{6} c^{4} x^{7}+\frac {1}{4} a^{7} c^{4} x^{8}+\frac {1}{9} a^{8} c^{4} x^{9}\) | \(83\) |
risch | \(-c^{4} x -a \,c^{4} x^{2}+\frac {2}{3} a^{2} c^{4} x^{3}+\frac {3}{2} a^{3} c^{4} x^{4}-a^{5} c^{4} x^{6}-\frac {2}{7} a^{6} c^{4} x^{7}+\frac {1}{4} a^{7} c^{4} x^{8}+\frac {1}{9} a^{8} c^{4} x^{9}\) | \(83\) |
parallelrisch | \(-c^{4} x -a \,c^{4} x^{2}+\frac {2}{3} a^{2} c^{4} x^{3}+\frac {3}{2} a^{3} c^{4} x^{4}-a^{5} c^{4} x^{6}-\frac {2}{7} a^{6} c^{4} x^{7}+\frac {1}{4} a^{7} c^{4} x^{8}+\frac {1}{9} a^{8} c^{4} x^{9}\) | \(83\) |
orering | \(\frac {x \left (28 a^{8} x^{8}+63 a^{7} x^{7}-72 x^{6} a^{6}-252 a^{5} x^{5}+378 a^{3} x^{3}+168 a^{2} x^{2}-252 a x -252\right ) \left (-a^{2} c \,x^{2}+c \right )^{4}}{252 \left (a x +1\right )^{4} \left (a x -1\right )^{4}}\) | \(85\) |
meijerg | \(-\frac {c^{4} \left (-\frac {a x \left (280 a^{8} x^{8}+315 a^{7} x^{7}+360 x^{6} a^{6}+420 a^{5} x^{5}+504 a^{4} x^{4}+630 a^{3} x^{3}+840 a^{2} x^{2}+1260 a x +2520\right )}{2520}-\ln \left (-a x +1\right )\right )}{a}+\frac {4 c^{4} \left (-\frac {a x \left (120 x^{6} a^{6}+140 a^{5} x^{5}+168 a^{4} x^{4}+210 a^{3} x^{3}+280 a^{2} x^{2}+420 a x +840\right )}{840}-\ln \left (-a x +1\right )\right )}{a}-\frac {6 c^{4} \left (-\frac {x a \left (12 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{60}-\ln \left (-a x +1\right )\right )}{a}+\frac {4 c^{4} \left (-\frac {a x \left (4 a^{2} x^{2}+6 a x +12\right )}{12}-\ln \left (-a x +1\right )\right )}{a}-\frac {c^{4} \left (-a x -\ln \left (-a x +1\right )\right )}{a}+\frac {c^{4} \left (\frac {a x \left (315 a^{7} x^{7}+360 x^{6} a^{6}+420 a^{5} x^{5}+504 a^{4} x^{4}+630 a^{3} x^{3}+840 a^{2} x^{2}+1260 a x +2520\right )}{2520}+\ln \left (-a x +1\right )\right )}{a}-\frac {4 c^{4} \left (\frac {a x \left (70 a^{5} x^{5}+84 a^{4} x^{4}+105 a^{3} x^{3}+140 a^{2} x^{2}+210 a x +420\right )}{420}+\ln \left (-a x +1\right )\right )}{a}+\frac {6 c^{4} \left (\frac {a x \left (15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{60}+\ln \left (-a x +1\right )\right )}{a}-\frac {4 c^{4} \left (\frac {a x \left (3 a x +6\right )}{6}+\ln \left (-a x +1\right )\right )}{a}+\frac {c^{4} \ln \left (-a x +1\right )}{a}\) | \(477\) |
Input:
int(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^4,x,method=_RETURNVERBOSE)
Output:
1/252*c^4*x*(28*a^8*x^8+63*a^7*x^7-72*a^6*x^6-252*a^5*x^5+378*a^3*x^3+168* a^2*x^2-252*a*x-252)
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {1}{9} \, a^{8} c^{4} x^{9} + \frac {1}{4} \, a^{7} c^{4} x^{8} - \frac {2}{7} \, a^{6} c^{4} x^{7} - a^{5} c^{4} x^{6} + \frac {3}{2} \, a^{3} c^{4} x^{4} + \frac {2}{3} \, a^{2} c^{4} x^{3} - a c^{4} x^{2} - c^{4} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^4,x, algorithm="fricas")
Output:
1/9*a^8*c^4*x^9 + 1/4*a^7*c^4*x^8 - 2/7*a^6*c^4*x^7 - a^5*c^4*x^6 + 3/2*a^ 3*c^4*x^4 + 2/3*a^2*c^4*x^3 - a*c^4*x^2 - c^4*x
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {a^{8} c^{4} x^{9}}{9} + \frac {a^{7} c^{4} x^{8}}{4} - \frac {2 a^{6} c^{4} x^{7}}{7} - a^{5} c^{4} x^{6} + \frac {3 a^{3} c^{4} x^{4}}{2} + \frac {2 a^{2} c^{4} x^{3}}{3} - a c^{4} x^{2} - c^{4} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a**2*c*x**2+c)**4,x)
Output:
a**8*c**4*x**9/9 + a**7*c**4*x**8/4 - 2*a**6*c**4*x**7/7 - a**5*c**4*x**6 + 3*a**3*c**4*x**4/2 + 2*a**2*c**4*x**3/3 - a*c**4*x**2 - c**4*x
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {1}{9} \, a^{8} c^{4} x^{9} + \frac {1}{4} \, a^{7} c^{4} x^{8} - \frac {2}{7} \, a^{6} c^{4} x^{7} - a^{5} c^{4} x^{6} + \frac {3}{2} \, a^{3} c^{4} x^{4} + \frac {2}{3} \, a^{2} c^{4} x^{3} - a c^{4} x^{2} - c^{4} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^4,x, algorithm="maxima")
Output:
1/9*a^8*c^4*x^9 + 1/4*a^7*c^4*x^8 - 2/7*a^6*c^4*x^7 - a^5*c^4*x^6 + 3/2*a^ 3*c^4*x^4 + 2/3*a^2*c^4*x^3 - a*c^4*x^2 - c^4*x
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {1}{9} \, a^{8} c^{4} x^{9} + \frac {1}{4} \, a^{7} c^{4} x^{8} - \frac {2}{7} \, a^{6} c^{4} x^{7} - a^{5} c^{4} x^{6} + \frac {3}{2} \, a^{3} c^{4} x^{4} + \frac {2}{3} \, a^{2} c^{4} x^{3} - a c^{4} x^{2} - c^{4} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^4,x, algorithm="giac")
Output:
1/9*a^8*c^4*x^9 + 1/4*a^7*c^4*x^8 - 2/7*a^6*c^4*x^7 - a^5*c^4*x^6 + 3/2*a^ 3*c^4*x^4 + 2/3*a^2*c^4*x^3 - a*c^4*x^2 - c^4*x
Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {a^8\,c^4\,x^9}{9}+\frac {a^7\,c^4\,x^8}{4}-\frac {2\,a^6\,c^4\,x^7}{7}-a^5\,c^4\,x^6+\frac {3\,a^3\,c^4\,x^4}{2}+\frac {2\,a^2\,c^4\,x^3}{3}-a\,c^4\,x^2-c^4\,x \] Input:
int(((c - a^2*c*x^2)^4*(a*x + 1))/(a*x - 1),x)
Output:
(2*a^2*c^4*x^3)/3 - a*c^4*x^2 - c^4*x + (3*a^3*c^4*x^4)/2 - a^5*c^4*x^6 - (2*a^6*c^4*x^7)/7 + (a^7*c^4*x^8)/4 + (a^8*c^4*x^9)/9
Time = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {c^{4} x \left (28 a^{8} x^{8}+63 a^{7} x^{7}-72 a^{6} x^{6}-252 a^{5} x^{5}+378 a^{3} x^{3}+168 a^{2} x^{2}-252 a x -252\right )}{252} \] Input:
int(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^4,x)
Output:
(c**4*x*(28*a**8*x**8 + 63*a**7*x**7 - 72*a**6*x**6 - 252*a**5*x**5 + 378* a**3*x**3 + 168*a**2*x**2 - 252*a*x - 252))/252