Integrand size = 25, antiderivative size = 79 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=2 a c^3 x-\frac {1}{2} a^2 c^3 x^2-\frac {4}{3} a^3 c^3 x^3-\frac {1}{4} a^4 c^3 x^4+\frac {2}{5} a^5 c^3 x^5+\frac {1}{6} a^6 c^3 x^6+c^3 \log (x) \] Output:
2*a*c^3*x-1/2*a^2*c^3*x^2-4/3*a^3*c^3*x^3-1/4*a^4*c^3*x^4+2/5*a^5*c^3*x^5+ 1/6*a^6*c^3*x^6+c^3*ln(x)
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=c^3 \left (\frac {1}{60} a x \left (120-30 a x-80 a^2 x^2-15 a^3 x^3+24 a^4 x^4+10 a^5 x^5\right )+\log (x)\right ) \] Input:
Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3)/x,x]
Output:
c^3*((a*x*(120 - 30*a*x - 80*a^2*x^2 - 15*a^3*x^3 + 24*a^4*x^4 + 10*a^5*x^ 5))/60 + Log[x])
Time = 0.47 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.77, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle c^3 \int \frac {(1-a x)^2 (a x+1)^4}{x}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle c^3 \int \left (x^5 a^6+2 x^4 a^5-x^3 a^4-4 x^2 a^3-x a^2+2 a+\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c^3 \left (\frac {a^6 x^6}{6}+\frac {2 a^5 x^5}{5}-\frac {a^4 x^4}{4}-\frac {4 a^3 x^3}{3}-\frac {a^2 x^2}{2}+2 a x+\log (x)\right )\) |
Input:
Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3)/x,x]
Output:
c^3*(2*a*x - (a^2*x^2)/2 - (4*a^3*x^3)/3 - (a^4*x^4)/4 + (2*a^5*x^5)/5 + ( a^6*x^6)/6 + Log[x])
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 = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66
method | result | size |
default | \(c^{3} \left (\frac {x^{6} a^{6}}{6}+\frac {2 a^{5} x^{5}}{5}-\frac {a^{4} x^{4}}{4}-\frac {4 a^{3} x^{3}}{3}-\frac {a^{2} x^{2}}{2}+2 a x +\ln \left (x \right )\right )\) | \(52\) |
norman | \(2 a \,c^{3} x -\frac {a^{2} c^{3} x^{2}}{2}-\frac {4 a^{3} c^{3} x^{3}}{3}-\frac {a^{4} c^{3} x^{4}}{4}+\frac {2 a^{5} c^{3} x^{5}}{5}+\frac {a^{6} c^{3} x^{6}}{6}+c^{3} \ln \left (x \right )\) | \(70\) |
risch | \(2 a \,c^{3} x -\frac {a^{2} c^{3} x^{2}}{2}-\frac {4 a^{3} c^{3} x^{3}}{3}-\frac {a^{4} c^{3} x^{4}}{4}+\frac {2 a^{5} c^{3} x^{5}}{5}+\frac {a^{6} c^{3} x^{6}}{6}+c^{3} \ln \left (x \right )\) | \(70\) |
parallelrisch | \(2 a \,c^{3} x -\frac {a^{2} c^{3} x^{2}}{2}-\frac {4 a^{3} c^{3} x^{3}}{3}-\frac {a^{4} c^{3} x^{4}}{4}+\frac {2 a^{5} c^{3} x^{5}}{5}+\frac {a^{6} c^{3} x^{6}}{6}+c^{3} \ln \left (x \right )\) | \(70\) |
meijerg | \(-\frac {c^{3} \left (-\frac {a^{2} x^{2} \left (4 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{12}-\ln \left (-a^{2} x^{2}+1\right )\right )}{2}-c^{3} \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )+c^{3} \ln \left (-a^{2} x^{2}+1\right )+\frac {a \,c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {7}{2}} \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{105 a^{6}}+\frac {2 \left (-a^{2}\right )^{\frac {7}{2}} \operatorname {arctanh}\left (a x \right )}{a^{7}}\right )}{\sqrt {-a^{2}}}+\frac {3 a \,c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{\sqrt {-a^{2}}}+\frac {3 a \,c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \operatorname {arctanh}\left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}+2 c^{3} \operatorname {arctanh}\left (a x \right )+\frac {c^{3} \left (2 \ln \left (x \right )+\ln \left (-a^{2}\right )-\ln \left (-a^{2} x^{2}+1\right )\right )}{2}\) | \(291\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x,x,method=_RETURNVERBOSE)
Output:
c^3*(1/6*x^6*a^6+2/5*a^5*x^5-1/4*a^4*x^4-4/3*a^3*x^3-1/2*a^2*x^2+2*a*x+ln( x))
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=\frac {1}{6} \, a^{6} c^{3} x^{6} + \frac {2}{5} \, a^{5} c^{3} x^{5} - \frac {1}{4} \, a^{4} c^{3} x^{4} - \frac {4}{3} \, a^{3} c^{3} x^{3} - \frac {1}{2} \, a^{2} c^{3} x^{2} + 2 \, a c^{3} x + c^{3} \log \left (x\right ) \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x,x, algorithm="fricas")
Output:
1/6*a^6*c^3*x^6 + 2/5*a^5*c^3*x^5 - 1/4*a^4*c^3*x^4 - 4/3*a^3*c^3*x^3 - 1/ 2*a^2*c^3*x^2 + 2*a*c^3*x + c^3*log(x)
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=\frac {a^{6} c^{3} x^{6}}{6} + \frac {2 a^{5} c^{3} x^{5}}{5} - \frac {a^{4} c^{3} x^{4}}{4} - \frac {4 a^{3} c^{3} x^{3}}{3} - \frac {a^{2} c^{3} x^{2}}{2} + 2 a c^{3} x + c^{3} \log {\left (x \right )} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)**3/x,x)
Output:
a**6*c**3*x**6/6 + 2*a**5*c**3*x**5/5 - a**4*c**3*x**4/4 - 4*a**3*c**3*x** 3/3 - a**2*c**3*x**2/2 + 2*a*c**3*x + c**3*log(x)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=\frac {1}{6} \, a^{6} c^{3} x^{6} + \frac {2}{5} \, a^{5} c^{3} x^{5} - \frac {1}{4} \, a^{4} c^{3} x^{4} - \frac {4}{3} \, a^{3} c^{3} x^{3} - \frac {1}{2} \, a^{2} c^{3} x^{2} + 2 \, a c^{3} x + c^{3} \log \left (x\right ) \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x,x, algorithm="maxima")
Output:
1/6*a^6*c^3*x^6 + 2/5*a^5*c^3*x^5 - 1/4*a^4*c^3*x^4 - 4/3*a^3*c^3*x^3 - 1/ 2*a^2*c^3*x^2 + 2*a*c^3*x + c^3*log(x)
Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=\frac {1}{6} \, a^{6} c^{3} x^{6} + \frac {2}{5} \, a^{5} c^{3} x^{5} - \frac {1}{4} \, a^{4} c^{3} x^{4} - \frac {4}{3} \, a^{3} c^{3} x^{3} - \frac {1}{2} \, a^{2} c^{3} x^{2} + 2 \, a c^{3} x + c^{3} \log \left ({\left | x \right |}\right ) \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x,x, algorithm="giac")
Output:
1/6*a^6*c^3*x^6 + 2/5*a^5*c^3*x^5 - 1/4*a^4*c^3*x^4 - 4/3*a^3*c^3*x^3 - 1/ 2*a^2*c^3*x^2 + 2*a*c^3*x + c^3*log(abs(x))
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=c^3\,\ln \left (x\right )-\frac {a^2\,c^3\,x^2}{2}-\frac {4\,a^3\,c^3\,x^3}{3}-\frac {a^4\,c^3\,x^4}{4}+\frac {2\,a^5\,c^3\,x^5}{5}+\frac {a^6\,c^3\,x^6}{6}+2\,a\,c^3\,x \] Input:
int(-((c - a^2*c*x^2)^3*(a*x + 1)^2)/(x*(a^2*x^2 - 1)),x)
Output:
c^3*log(x) - (a^2*c^3*x^2)/2 - (4*a^3*c^3*x^3)/3 - (a^4*c^3*x^4)/4 + (2*a^ 5*c^3*x^5)/5 + (a^6*c^3*x^6)/6 + 2*a*c^3*x
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x} \, dx=\frac {c^{3} \left (60 \,\mathrm {log}\left (x \right )+10 a^{6} x^{6}+24 a^{5} x^{5}-15 a^{4} x^{4}-80 a^{3} x^{3}-30 a^{2} x^{2}+120 a x \right )}{60} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x,x)
Output:
(c**3*(60*log(x) + 10*a**6*x**6 + 24*a**5*x**5 - 15*a**4*x**4 - 80*a**3*x* *3 - 30*a**2*x**2 + 120*a*x))/60