Integrand size = 22, antiderivative size = 69 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {4 c^4 (1+a x)^6}{3 a}-\frac {12 c^4 (1+a x)^7}{7 a}+\frac {3 c^4 (1+a x)^8}{4 a}-\frac {c^4 (1+a x)^9}{9 a} \] 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 = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {c^4 (1+a x)^6 \left (-65+138 a x-105 a^2 x^2+28 a^3 x^3\right )}{252 a} \] Input:
Integrate[E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^4,x]
Output:
-1/252*(c^4*(1 + a*x)^6*(-65 + 138*a*x - 105*a^2*x^2 + 28*a^3*x^3))/a
Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {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.
\(\displaystyle \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx\) |
\(\Big \downarrow \) 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*ArcTanh[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_.) + (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])
Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86
method | result | size |
default | \(c^{4} \left (-\frac {1}{9} a^{8} x^{9}-\frac {1}{4} a^{7} x^{8}+\frac {2}{7} a^{6} x^{7}+a^{5} x^{6}-\frac {3}{2} a^{3} x^{4}-\frac {2}{3} a^{2} x^{3}+a \,x^{2}+x \right )\) | \(59\) |
gosper | \(-\frac {c^{4} x \left (28 a^{8} x^{8}+63 a^{7} x^{7}-72 x^{6} a^{6}-252 a^{5} x^{5}+378 a^{3} x^{3}+168 a^{2} x^{2}-252 a x -252\right )}{252}\) | \(61\) |
norman | \(c^{4} x +a \,c^{4} x^{2}+a^{5} c^{4} x^{6}-\frac {2}{3} a^{2} c^{4} x^{3}-\frac {3}{2} a^{3} c^{4} x^{4}+\frac {2}{7} a^{6} c^{4} x^{7}-\frac {1}{4} a^{7} c^{4} x^{8}-\frac {1}{9} a^{8} c^{4} x^{9}\) | \(80\) |
risch | \(c^{4} x +a \,c^{4} x^{2}+a^{5} c^{4} x^{6}-\frac {2}{3} a^{2} c^{4} x^{3}-\frac {3}{2} a^{3} c^{4} x^{4}+\frac {2}{7} a^{6} c^{4} x^{7}-\frac {1}{4} a^{7} c^{4} x^{8}-\frac {1}{9} a^{8} c^{4} x^{9}\) | \(80\) |
parallelrisch | \(c^{4} x +a \,c^{4} x^{2}+a^{5} c^{4} x^{6}-\frac {2}{3} a^{2} c^{4} x^{3}-\frac {3}{2} a^{3} c^{4} x^{4}+\frac {2}{7} a^{6} c^{4} x^{7}-\frac {1}{4} a^{7} c^{4} x^{8}-\frac {1}{9} a^{8} c^{4} x^{9}\) | \(80\) |
orering | \(\frac {x \left (28 a^{8} x^{8}+63 a^{7} x^{7}-72 x^{6} a^{6}-252 a^{5} x^{5}+378 a^{3} x^{3}+168 a^{2} x^{2}-252 a x -252\right ) \left (-a^{2} c \,x^{2}+c \right )^{4}}{252 \left (a x +1\right )^{3} \left (a x -1\right )^{3} \left (-a^{2} x^{2}+1\right )}\) | \(97\) |
meijerg | \(-\frac {c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {11}{2}} \left (385 a^{8} x^{8}+495 x^{6} a^{6}+693 a^{4} x^{4}+1155 a^{2} x^{2}+3465\right )}{3465 a^{10}}+\frac {2 \left (-a^{2}\right )^{\frac {11}{2}} \operatorname {arctanh}\left (a x \right )}{a^{11}}\right )}{2 \sqrt {-a^{2}}}-\frac {3 c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {9}{2}} \left (45 x^{6} a^{6}+63 a^{4} x^{4}+105 a^{2} x^{2}+315\right )}{315 a^{8}}+\frac {2 \left (-a^{2}\right )^{\frac {9}{2}} \operatorname {arctanh}\left (a x \right )}{a^{9}}\right )}{2 \sqrt {-a^{2}}}-\frac {c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {7}{2}} \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{105 a^{6}}+\frac {2 \left (-a^{2}\right )^{\frac {7}{2}} \operatorname {arctanh}\left (a x \right )}{a^{7}}\right )}{\sqrt {-a^{2}}}+\frac {c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{\sqrt {-a^{2}}}+\frac {3 c^{4} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \operatorname {arctanh}\left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}-\frac {c^{4} \left (\frac {a^{2} x^{2} \left (15 x^{6} a^{6}+20 a^{4} x^{4}+30 a^{2} x^{2}+60\right )}{60}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {4 c^{4} \left (-\frac {a^{2} x^{2} \left (4 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{12}-\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {6 c^{4} \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {4 c^{4} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {c^{4} \ln \left (-a^{2} x^{2}+1\right )}{a}+\frac {c^{4} \operatorname {arctanh}\left (a x \right )}{a}\) | \(499\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^4,x,method=_RETURNVERBOSE)
Output:
c^4*(-1/9*a^8*x^9-1/4*a^7*x^8+2/7*a^6*x^7+a^5*x^6-3/2*a^3*x^4-2/3*a^2*x^3+ a*x^2+x)
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {1}{9} \, a^{8} c^{4} x^{9} - \frac {1}{4} \, a^{7} c^{4} x^{8} + \frac {2}{7} \, a^{6} c^{4} x^{7} + a^{5} c^{4} x^{6} - \frac {3}{2} \, a^{3} c^{4} x^{4} - \frac {2}{3} \, a^{2} c^{4} x^{3} + a c^{4} x^{2} + c^{4} x \] Input:
integrate((a*x+1)^2/(-a^2*x^2+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 = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=- \frac {a^{8} c^{4} x^{9}}{9} - \frac {a^{7} c^{4} x^{8}}{4} + \frac {2 a^{6} c^{4} x^{7}}{7} + a^{5} c^{4} x^{6} - \frac {3 a^{3} c^{4} x^{4}}{2} - \frac {2 a^{2} c^{4} x^{3}}{3} + a c^{4} x^{2} + c^{4} x \] Input:
integrate((a*x+1)**2/(-a**2*x**2+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 = 79, normalized size of antiderivative = 1.14 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {1}{9} \, a^{8} c^{4} x^{9} - \frac {1}{4} \, a^{7} c^{4} x^{8} + \frac {2}{7} \, a^{6} c^{4} x^{7} + a^{5} c^{4} x^{6} - \frac {3}{2} \, a^{3} c^{4} x^{4} - \frac {2}{3} \, a^{2} c^{4} x^{3} + a c^{4} x^{2} + c^{4} x \] Input:
integrate((a*x+1)^2/(-a^2*x^2+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 = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {1}{9} \, a^{8} c^{4} x^{9} - \frac {1}{4} \, a^{7} c^{4} x^{8} + \frac {2}{7} \, a^{6} c^{4} x^{7} + a^{5} c^{4} x^{6} - \frac {3}{2} \, a^{3} c^{4} x^{4} - \frac {2}{3} \, a^{2} c^{4} x^{3} + a c^{4} x^{2} + c^{4} x \] Input:
integrate((a*x+1)^2/(-a^2*x^2+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 = 79, normalized size of antiderivative = 1.14 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=-\frac {a^8\,c^4\,x^9}{9}-\frac {a^7\,c^4\,x^8}{4}+\frac {2\,a^6\,c^4\,x^7}{7}+a^5\,c^4\,x^6-\frac {3\,a^3\,c^4\,x^4}{2}-\frac {2\,a^2\,c^4\,x^3}{3}+a\,c^4\,x^2+c^4\,x \] Input:
int(-((c - a^2*c*x^2)^4*(a*x + 1)^2)/(a^2*x^2 - 1),x)
Output:
c^4*x + a*c^4*x^2 - (2*a^2*c^4*x^3)/3 - (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 = 0.15 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^4 \, dx=\frac {c^{4} x \left (-28 a^{8} x^{8}-63 a^{7} x^{7}+72 a^{6} x^{6}+252 a^{5} x^{5}-378 a^{3} x^{3}-168 a^{2} x^{2}+252 a x +252\right )}{252} \] Input:
int((a*x+1)^2/(-a^2*x^2+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 - 3 78*a**3*x**3 - 168*a**2*x**2 + 252*a*x + 252))/252