Integrand size = 22, antiderivative size = 52 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {4 c^3 (1+a x)^5}{5 a}+\frac {2 c^3 (1+a x)^6}{3 a}-\frac {c^3 (1+a x)^7}{7 a} \] Output:
-4/5*c^3*(a*x+1)^5/a+2/3*c^3*(a*x+1)^6/a-1/7*c^3*(a*x+1)^7/a
Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {c^3 (1+a x)^5 \left (29-40 a x+15 a^2 x^2\right )}{105 a} \] Input:
Integrate[E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^3,x]
Output:
-1/105*(c^3*(1 + a*x)^5*(29 - 40*a*x + 15*a^2*x^2))/a
Time = 0.55 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.92, 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.
\(\displaystyle \int \left (c-a^2 c x^2\right )^3 e^{2 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int c^3 e^{2 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^3dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -c^3 \int e^{2 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^3dx\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle -c^3 \int (1-a x)^2 (a x+1)^4dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -c^3 \int \left ((a x+1)^6-4 (a x+1)^5+4 (a x+1)^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -c^3 \left (\frac {(a x+1)^7}{7 a}-\frac {2 (a x+1)^6}{3 a}+\frac {4 (a x+1)^5}{5 a}\right )\) |
Input:
Int[E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^3,x]
Output:
-(c^3*((4*(1 + a*x)^5)/(5*a) - (2*(1 + a*x)^6)/(3*a) + (1 + a*x)^7/(7*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 = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02
method | result | size |
gosper | \(-\frac {c^{3} x \left (15 x^{6} a^{6}+35 a^{5} x^{5}-21 a^{4} x^{4}-105 a^{3} x^{3}-35 a^{2} x^{2}+105 a x +105\right )}{105}\) | \(53\) |
default | \(c^{3} \left (-\frac {1}{7} a^{6} x^{7}-\frac {1}{3} a^{5} x^{6}+\frac {1}{5} x^{5} a^{4}+a^{3} x^{4}+\frac {1}{3} a^{2} x^{3}-a \,x^{2}-x \right )\) | \(54\) |
norman | \(a^{3} c^{3} x^{4}-c^{3} x -a \,c^{3} x^{2}+\frac {1}{3} a^{2} c^{3} x^{3}-\frac {1}{3} a^{5} c^{3} x^{6}-\frac {1}{7} a^{6} c^{3} x^{7}+\frac {1}{5} c^{3} a^{4} x^{5}\) | \(71\) |
risch | \(a^{3} c^{3} x^{4}-c^{3} x -a \,c^{3} x^{2}+\frac {1}{3} a^{2} c^{3} x^{3}-\frac {1}{3} a^{5} c^{3} x^{6}-\frac {1}{7} a^{6} c^{3} x^{7}+\frac {1}{5} c^{3} a^{4} x^{5}\) | \(71\) |
parallelrisch | \(a^{3} c^{3} x^{4}-c^{3} x -a \,c^{3} x^{2}+\frac {1}{3} a^{2} c^{3} x^{3}-\frac {1}{3} a^{5} c^{3} x^{6}-\frac {1}{7} a^{6} c^{3} x^{7}+\frac {1}{5} c^{3} a^{4} x^{5}\) | \(71\) |
orering | \(\frac {x \left (15 x^{6} a^{6}+35 a^{5} x^{5}-21 a^{4} x^{4}-105 a^{3} x^{3}-35 a^{2} x^{2}+105 a x +105\right ) \left (-a^{2} c \,x^{2}+c \right )^{3}}{105 \left (a x +1\right )^{3} \left (a x -1\right )^{3}}\) | \(77\) |
meijerg | \(\frac {c^{3} \left (-\frac {a x \left (120 x^{6} a^{6}+140 a^{5} x^{5}+168 a^{4} x^{4}+210 a^{3} x^{3}+280 a^{2} x^{2}+420 a x +840\right )}{840}-\ln \left (-a x +1\right )\right )}{a}-\frac {3 c^{3} \left (-\frac {x a \left (12 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{60}-\ln \left (-a x +1\right )\right )}{a}+\frac {3 c^{3} \left (-\frac {a x \left (4 a^{2} x^{2}+6 a x +12\right )}{12}-\ln \left (-a x +1\right )\right )}{a}-\frac {c^{3} \left (-a x -\ln \left (-a x +1\right )\right )}{a}-\frac {c^{3} \left (\frac {a x \left (70 a^{5} x^{5}+84 a^{4} x^{4}+105 a^{3} x^{3}+140 a^{2} x^{2}+210 a x +420\right )}{420}+\ln \left (-a x +1\right )\right )}{a}+\frac {3 c^{3} \left (\frac {a x \left (15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{60}+\ln \left (-a x +1\right )\right )}{a}-\frac {3 c^{3} \left (\frac {a x \left (3 a x +6\right )}{6}+\ln \left (-a x +1\right )\right )}{a}+\frac {c^{3} \ln \left (-a x +1\right )}{a}\) | \(319\) |
Input:
int(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/105*c^3*x*(15*a^6*x^6+35*a^5*x^5-21*a^4*x^4-105*a^3*x^3-35*a^2*x^2+105* a*x+105)
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {1}{7} \, a^{6} c^{3} x^{7} - \frac {1}{3} \, a^{5} c^{3} x^{6} + \frac {1}{5} \, a^{4} c^{3} x^{5} + a^{3} c^{3} x^{4} + \frac {1}{3} \, a^{2} c^{3} x^{3} - a c^{3} x^{2} - c^{3} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
-1/7*a^6*c^3*x^7 - 1/3*a^5*c^3*x^6 + 1/5*a^4*c^3*x^5 + a^3*c^3*x^4 + 1/3*a ^2*c^3*x^3 - a*c^3*x^2 - c^3*x
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=- \frac {a^{6} c^{3} x^{7}}{7} - \frac {a^{5} c^{3} x^{6}}{3} + \frac {a^{4} c^{3} x^{5}}{5} + a^{3} c^{3} x^{4} + \frac {a^{2} c^{3} x^{3}}{3} - a c^{3} x^{2} - c^{3} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a**2*c*x**2+c)**3,x)
Output:
-a**6*c**3*x**7/7 - a**5*c**3*x**6/3 + a**4*c**3*x**5/5 + a**3*c**3*x**4 + a**2*c**3*x**3/3 - a*c**3*x**2 - c**3*x
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {1}{7} \, a^{6} c^{3} x^{7} - \frac {1}{3} \, a^{5} c^{3} x^{6} + \frac {1}{5} \, a^{4} c^{3} x^{5} + a^{3} c^{3} x^{4} + \frac {1}{3} \, a^{2} c^{3} x^{3} - a c^{3} x^{2} - c^{3} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
-1/7*a^6*c^3*x^7 - 1/3*a^5*c^3*x^6 + 1/5*a^4*c^3*x^5 + a^3*c^3*x^4 + 1/3*a ^2*c^3*x^3 - a*c^3*x^2 - c^3*x
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {1}{7} \, a^{6} c^{3} x^{7} - \frac {1}{3} \, a^{5} c^{3} x^{6} + \frac {1}{5} \, a^{4} c^{3} x^{5} + a^{3} c^{3} x^{4} + \frac {1}{3} \, a^{2} c^{3} x^{3} - a c^{3} x^{2} - c^{3} x \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
-1/7*a^6*c^3*x^7 - 1/3*a^5*c^3*x^6 + 1/5*a^4*c^3*x^5 + a^3*c^3*x^4 + 1/3*a ^2*c^3*x^3 - a*c^3*x^2 - c^3*x
Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=-\frac {a^6\,c^3\,x^7}{7}-\frac {a^5\,c^3\,x^6}{3}+\frac {a^4\,c^3\,x^5}{5}+a^3\,c^3\,x^4+\frac {a^2\,c^3\,x^3}{3}-a\,c^3\,x^2-c^3\,x \] Input:
int(((c - a^2*c*x^2)^3*(a*x + 1))/(a*x - 1),x)
Output:
(a^2*c^3*x^3)/3 - a*c^3*x^2 - c^3*x + a^3*c^3*x^4 + (a^4*c^3*x^5)/5 - (a^5 *c^3*x^6)/3 - (a^6*c^3*x^7)/7
Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx=\frac {c^{3} x \left (-15 a^{6} x^{6}-35 a^{5} x^{5}+21 a^{4} x^{4}+105 a^{3} x^{3}+35 a^{2} x^{2}-105 a x -105\right )}{105} \] Input:
int(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^3,x)
Output:
(c**3*x*( - 15*a**6*x**6 - 35*a**5*x**5 + 21*a**4*x**4 + 105*a**3*x**3 + 3 5*a**2*x**2 - 105*a*x - 105))/105