Integrand size = 25, antiderivative size = 76 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=-\frac {c^3}{x}-a^2 c^3 x-2 a^3 c^3 x^2-\frac {1}{3} a^4 c^3 x^3+\frac {1}{2} a^5 c^3 x^4+\frac {1}{5} a^6 c^3 x^5+2 a c^3 \log (x) \] Output:
-c^3/x-a^2*c^3*x-2*a^3*c^3*x^2-1/3*a^4*c^3*x^3+1/2*a^5*c^3*x^4+1/5*a^6*c^3 *x^5+2*a*c^3*ln(x)
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=c^3 \left (-\frac {1}{x}-a^2 x-2 a^3 x^2-\frac {a^4 x^3}{3}+\frac {a^5 x^4}{2}+\frac {a^6 x^5}{5}+2 a \log (x)\right ) \] Input:
Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3)/x^2,x]
Output:
c^3*(-x^(-1) - a^2*x - 2*a^3*x^2 - (a^4*x^3)/3 + (a^5*x^4)/2 + (a^6*x^5)/5 + 2*a*Log[x])
Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle c^3 \int \frac {(1-a x)^2 (a x+1)^4}{x^2}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle c^3 \int \left (x^4 a^6+2 x^3 a^5-x^2 a^4-4 x a^3-a^2+\frac {2 a}{x}+\frac {1}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c^3 \left (\frac {a^6 x^5}{5}+\frac {a^5 x^4}{2}-\frac {a^4 x^3}{3}-2 a^3 x^2-a^2 x+2 a \log (x)-\frac {1}{x}\right )\) |
Input:
Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3)/x^2,x]
Output:
c^3*(-x^(-1) - a^2*x - 2*a^3*x^2 - (a^4*x^3)/3 + (a^5*x^4)/2 + (a^6*x^5)/5 + 2*a*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 = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.71
method | result | size |
default | \(c^{3} \left (\frac {a^{6} x^{5}}{5}+\frac {a^{5} x^{4}}{2}-\frac {a^{4} x^{3}}{3}-2 a^{3} x^{2}-a^{2} x -\frac {1}{x}+2 \ln \left (x \right ) a \right )\) | \(54\) |
risch | \(-\frac {c^{3}}{x}-a^{2} c^{3} x -2 a^{3} c^{3} x^{2}-\frac {a^{4} c^{3} x^{3}}{3}+\frac {a^{5} c^{3} x^{4}}{2}+\frac {a^{6} c^{3} x^{5}}{5}+2 a \,c^{3} \ln \left (x \right )\) | \(71\) |
norman | \(\frac {-c^{3}-a^{2} c^{3} x^{2}-2 a^{3} c^{3} x^{3}-\frac {1}{3} a^{4} c^{3} x^{4}+\frac {1}{2} a^{5} c^{3} x^{5}+\frac {1}{5} a^{6} c^{3} x^{6}}{x}+2 a \,c^{3} \ln \left (x \right )\) | \(75\) |
parallelrisch | \(\frac {6 a^{6} c^{3} x^{6}+15 a^{5} c^{3} x^{5}-10 a^{4} c^{3} x^{4}-60 a^{3} c^{3} x^{3}-30 a^{2} c^{3} x^{2}+60 a \,c^{3} \ln \left (x \right ) x -30 c^{3}}{30 x}\) | \(76\) |
meijerg | \(\frac {a^{2} c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {7}{2}} \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{105 a^{6}}+\frac {2 \left (-a^{2}\right )^{\frac {7}{2}} \operatorname {arctanh}\left (a x \right )}{a^{7}}\right )}{2 \sqrt {-a^{2}}}+\frac {a^{2} c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{\sqrt {-a^{2}}}-2 a \,c^{3} \operatorname {arctanh}\left (a x \right )+a \,c^{3} \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )+3 a \,c^{3} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )+3 a \,c^{3} \ln \left (-a^{2} x^{2}+1\right )+a \,c^{3} \left (2 \ln \left (x \right )+\ln \left (-a^{2}\right )-\ln \left (-a^{2} x^{2}+1\right )\right )-\frac {a^{2} c^{3} \left (-\frac {2}{x \sqrt {-a^{2}}}+\frac {2 a \,\operatorname {arctanh}\left (a x \right )}{\sqrt {-a^{2}}}\right )}{2 \sqrt {-a^{2}}}\) | \(280\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^2,x,method=_RETURNVERBOSE)
Output:
c^3*(1/5*a^6*x^5+1/2*a^5*x^4-1/3*a^4*x^3-2*a^3*x^2-a^2*x-1/x+2*ln(x)*a)
Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=\frac {6 \, a^{6} c^{3} x^{6} + 15 \, a^{5} c^{3} x^{5} - 10 \, a^{4} c^{3} x^{4} - 60 \, a^{3} c^{3} x^{3} - 30 \, a^{2} c^{3} x^{2} + 60 \, a c^{3} x \log \left (x\right ) - 30 \, c^{3}}{30 \, x} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^2,x, algorithm="fricas ")
Output:
1/30*(6*a^6*c^3*x^6 + 15*a^5*c^3*x^5 - 10*a^4*c^3*x^4 - 60*a^3*c^3*x^3 - 3 0*a^2*c^3*x^2 + 60*a*c^3*x*log(x) - 30*c^3)/x
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=\frac {a^{6} c^{3} x^{5}}{5} + \frac {a^{5} c^{3} x^{4}}{2} - \frac {a^{4} c^{3} x^{3}}{3} - 2 a^{3} c^{3} x^{2} - a^{2} c^{3} x + 2 a c^{3} \log {\left (x \right )} - \frac {c^{3}}{x} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)**3/x**2,x)
Output:
a**6*c**3*x**5/5 + a**5*c**3*x**4/2 - a**4*c**3*x**3/3 - 2*a**3*c**3*x**2 - a**2*c**3*x + 2*a*c**3*log(x) - c**3/x
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=\frac {1}{5} \, a^{6} c^{3} x^{5} + \frac {1}{2} \, a^{5} c^{3} x^{4} - \frac {1}{3} \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - a^{2} c^{3} x + 2 \, a c^{3} \log \left (x\right ) - \frac {c^{3}}{x} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^2,x, algorithm="maxima ")
Output:
1/5*a^6*c^3*x^5 + 1/2*a^5*c^3*x^4 - 1/3*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - a^2* c^3*x + 2*a*c^3*log(x) - c^3/x
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=\frac {1}{5} \, a^{6} c^{3} x^{5} + \frac {1}{2} \, a^{5} c^{3} x^{4} - \frac {1}{3} \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - a^{2} c^{3} x + 2 \, a c^{3} \log \left ({\left | x \right |}\right ) - \frac {c^{3}}{x} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^2,x, algorithm="giac")
Output:
1/5*a^6*c^3*x^5 + 1/2*a^5*c^3*x^4 - 1/3*a^4*c^3*x^3 - 2*a^3*c^3*x^2 - a^2* c^3*x + 2*a*c^3*log(abs(x)) - c^3/x
Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=2\,a\,c^3\,\ln \left (x\right )-a^2\,c^3\,x-\frac {c^3}{x}-2\,a^3\,c^3\,x^2-\frac {a^4\,c^3\,x^3}{3}+\frac {a^5\,c^3\,x^4}{2}+\frac {a^6\,c^3\,x^5}{5} \] Input:
int(-((c - a^2*c*x^2)^3*(a*x + 1)^2)/(x^2*(a^2*x^2 - 1)),x)
Output:
2*a*c^3*log(x) - a^2*c^3*x - c^3/x - 2*a^3*c^3*x^2 - (a^4*c^3*x^3)/3 + (a^ 5*c^3*x^4)/2 + (a^6*c^3*x^5)/5
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^2} \, dx=\frac {c^{3} \left (60 \,\mathrm {log}\left (x \right ) a x +6 a^{6} x^{6}+15 a^{5} x^{5}-10 a^{4} x^{4}-60 a^{3} x^{3}-30 a^{2} x^{2}-30\right )}{30 x} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^2,x)
Output:
(c**3*(60*log(x)*a*x + 6*a**6*x**6 + 15*a**5*x**5 - 10*a**4*x**4 - 60*a**3 *x**3 - 30*a**2*x**2 - 30))/(30*x)