Integrand size = 25, antiderivative size = 87 \[ \int e^{2 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^3 \, dx=\frac {c^3 x^4}{4}+\frac {2}{5} a c^3 x^5-\frac {1}{6} a^2 c^3 x^6-\frac {4}{7} a^3 c^3 x^7-\frac {1}{8} a^4 c^3 x^8+\frac {2}{9} a^5 c^3 x^9+\frac {1}{10} a^6 c^3 x^{10} \]
1/4*c^3*x^4+2/5*a*c^3*x^5-1/6*a^2*c^3*x^6-4/7*a^3*c^3*x^7-1/8*a^4*c^3*x^8+ 2/9*a^5*c^3*x^9+1/10*a^6*c^3*x^10
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int e^{2 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^3 \, dx=c^3 \left (\frac {x^4}{4}+\frac {2 a x^5}{5}-\frac {a^2 x^6}{6}-\frac {4 a^3 x^7}{7}-\frac {a^4 x^8}{8}+\frac {2 a^5 x^9}{9}+\frac {a^6 x^{10}}{10}\right ) \]
c^3*(x^4/4 + (2*a*x^5)/5 - (a^2*x^6)/6 - (4*a^3*x^7)/7 - (a^4*x^8)/8 + (2* a^5*x^9)/9 + (a^6*x^10)/10)
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6700, 99, 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.
\(\displaystyle \int x^3 e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3 \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle c^3 \int x^3 (1-a x)^2 (a x+1)^4dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle c^3 \int \left (a^6 x^9+2 a^5 x^8-a^4 x^7-4 a^3 x^6-a^2 x^5+2 a x^4+x^3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c^3 \left (\frac {a^6 x^{10}}{10}+\frac {2 a^5 x^9}{9}-\frac {a^4 x^8}{8}-\frac {4 a^3 x^7}{7}-\frac {a^2 x^6}{6}+\frac {2 a x^5}{5}+\frac {x^4}{4}\right )\) |
c^3*(x^4/4 + (2*a*x^5)/5 - (a^2*x^6)/6 - (4*a^3*x^7)/7 - (a^4*x^8)/8 + (2* a^5*x^9)/9 + (a^6*x^10)/10)
3.11.40.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {c^{3} x^{4} \left (252 a^{6} x^{6}+560 a^{5} x^{5}-315 a^{4} x^{4}-1440 a^{3} x^{3}-420 a^{2} x^{2}+1008 a x +630\right )}{2520}\) | \(55\) |
default | \(c^{3} \left (\frac {1}{10} a^{6} x^{10}+\frac {2}{9} a^{5} x^{9}-\frac {1}{8} a^{4} x^{8}-\frac {4}{7} a^{3} x^{7}-\frac {1}{6} a^{2} x^{6}+\frac {2}{5} a \,x^{5}+\frac {1}{4} x^{4}\right )\) | \(57\) |
norman | \(\frac {1}{4} c^{3} x^{4}+\frac {2}{5} a \,c^{3} x^{5}-\frac {1}{6} a^{2} c^{3} x^{6}-\frac {4}{7} a^{3} c^{3} x^{7}-\frac {1}{8} a^{4} c^{3} x^{8}+\frac {2}{9} a^{5} c^{3} x^{9}+\frac {1}{10} a^{6} c^{3} x^{10}\) | \(74\) |
risch | \(\frac {1}{4} c^{3} x^{4}+\frac {2}{5} a \,c^{3} x^{5}-\frac {1}{6} a^{2} c^{3} x^{6}-\frac {4}{7} a^{3} c^{3} x^{7}-\frac {1}{8} a^{4} c^{3} x^{8}+\frac {2}{9} a^{5} c^{3} x^{9}+\frac {1}{10} a^{6} c^{3} x^{10}\) | \(74\) |
parallelrisch | \(\frac {1}{4} c^{3} x^{4}+\frac {2}{5} a \,c^{3} x^{5}-\frac {1}{6} a^{2} c^{3} x^{6}-\frac {4}{7} a^{3} c^{3} x^{7}-\frac {1}{8} a^{4} c^{3} x^{8}+\frac {2}{9} a^{5} c^{3} x^{9}+\frac {1}{10} a^{6} c^{3} x^{10}\) | \(74\) |
meijerg | \(-\frac {c^{3} \left (-\frac {x^{2} a^{2} \left (12 a^{8} x^{8}+15 a^{6} x^{6}+20 a^{4} x^{4}+30 a^{2} x^{2}+60\right )}{60}-\ln \left (-a^{2} x^{2}+1\right )\right )}{2 a^{4}}-\frac {c^{3} \left (\frac {x^{2} a^{2} \left (15 a^{6} x^{6}+20 a^{4} x^{4}+30 a^{2} x^{2}+60\right )}{60}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a^{4}}+\frac {c^{3} \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a^{4}}+\frac {c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {11}{2}} \left (385 a^{8} x^{8}+495 a^{6} x^{6}+693 a^{4} x^{4}+1155 a^{2} x^{2}+3465\right )}{3465 a^{10}}+\frac {2 \left (-a^{2}\right )^{\frac {11}{2}} \operatorname {arctanh}\left (a x \right )}{a^{11}}\right )}{a^{3} \sqrt {-a^{2}}}+\frac {3 c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {9}{2}} \left (45 a^{6} x^{6}+63 a^{4} x^{4}+105 a^{2} x^{2}+315\right )}{315 a^{8}}+\frac {2 \left (-a^{2}\right )^{\frac {9}{2}} \operatorname {arctanh}\left (a x \right )}{a^{9}}\right )}{a^{3} \sqrt {-a^{2}}}+\frac {3 c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {7}{2}} \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{105 a^{6}}+\frac {2 \left (-a^{2}\right )^{\frac {7}{2}} \operatorname {arctanh}\left (a x \right )}{a^{7}}\right )}{a^{3} \sqrt {-a^{2}}}+\frac {c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{a^{3} \sqrt {-a^{2}}}+\frac {c^{3} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{2 a^{4}}\) | \(453\) |
Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int e^{2 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^3 \, dx=\frac {1}{10} \, a^{6} c^{3} x^{10} + \frac {2}{9} \, a^{5} c^{3} x^{9} - \frac {1}{8} \, a^{4} c^{3} x^{8} - \frac {4}{7} \, a^{3} c^{3} x^{7} - \frac {1}{6} \, a^{2} c^{3} x^{6} + \frac {2}{5} \, a c^{3} x^{5} + \frac {1}{4} \, c^{3} x^{4} \]
1/10*a^6*c^3*x^10 + 2/9*a^5*c^3*x^9 - 1/8*a^4*c^3*x^8 - 4/7*a^3*c^3*x^7 - 1/6*a^2*c^3*x^6 + 2/5*a*c^3*x^5 + 1/4*c^3*x^4
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int e^{2 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^3 \, dx=\frac {a^{6} c^{3} x^{10}}{10} + \frac {2 a^{5} c^{3} x^{9}}{9} - \frac {a^{4} c^{3} x^{8}}{8} - \frac {4 a^{3} c^{3} x^{7}}{7} - \frac {a^{2} c^{3} x^{6}}{6} + \frac {2 a c^{3} x^{5}}{5} + \frac {c^{3} x^{4}}{4} \]
a**6*c**3*x**10/10 + 2*a**5*c**3*x**9/9 - a**4*c**3*x**8/8 - 4*a**3*c**3*x **7/7 - a**2*c**3*x**6/6 + 2*a*c**3*x**5/5 + c**3*x**4/4
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int e^{2 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^3 \, dx=\frac {1}{10} \, a^{6} c^{3} x^{10} + \frac {2}{9} \, a^{5} c^{3} x^{9} - \frac {1}{8} \, a^{4} c^{3} x^{8} - \frac {4}{7} \, a^{3} c^{3} x^{7} - \frac {1}{6} \, a^{2} c^{3} x^{6} + \frac {2}{5} \, a c^{3} x^{5} + \frac {1}{4} \, c^{3} x^{4} \]
1/10*a^6*c^3*x^10 + 2/9*a^5*c^3*x^9 - 1/8*a^4*c^3*x^8 - 4/7*a^3*c^3*x^7 - 1/6*a^2*c^3*x^6 + 2/5*a*c^3*x^5 + 1/4*c^3*x^4
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int e^{2 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^3 \, dx=\frac {1}{10} \, a^{6} c^{3} x^{10} + \frac {2}{9} \, a^{5} c^{3} x^{9} - \frac {1}{8} \, a^{4} c^{3} x^{8} - \frac {4}{7} \, a^{3} c^{3} x^{7} - \frac {1}{6} \, a^{2} c^{3} x^{6} + \frac {2}{5} \, a c^{3} x^{5} + \frac {1}{4} \, c^{3} x^{4} \]
1/10*a^6*c^3*x^10 + 2/9*a^5*c^3*x^9 - 1/8*a^4*c^3*x^8 - 4/7*a^3*c^3*x^7 - 1/6*a^2*c^3*x^6 + 2/5*a*c^3*x^5 + 1/4*c^3*x^4
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84 \[ \int e^{2 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^3 \, dx=\frac {a^6\,c^3\,x^{10}}{10}+\frac {2\,a^5\,c^3\,x^9}{9}-\frac {a^4\,c^3\,x^8}{8}-\frac {4\,a^3\,c^3\,x^7}{7}-\frac {a^2\,c^3\,x^6}{6}+\frac {2\,a\,c^3\,x^5}{5}+\frac {c^3\,x^4}{4} \]