Integrand size = 25, antiderivative size = 78 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=-\frac {c^3}{2 x^2}-\frac {2 a c^3}{x}-4 a^3 c^3 x-\frac {1}{2} a^4 c^3 x^2+\frac {2}{3} a^5 c^3 x^3+\frac {1}{4} a^6 c^3 x^4-a^2 c^3 \log (x) \] Output:
-1/2*c^3/x^2-2*a*c^3/x-4*a^3*c^3*x-1/2*a^4*c^3*x^2+2/3*a^5*c^3*x^3+1/4*a^6 *c^3*x^4-a^2*c^3*ln(x)
Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=\frac {c^3 \left (-6-24 a x-48 a^3 x^3-6 a^4 x^4+8 a^5 x^5+3 a^6 x^6-12 a^2 x^2 \log (x)\right )}{12 x^2} \] Input:
Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3)/x^3,x]
Output:
(c^3*(-6 - 24*a*x - 48*a^3*x^3 - 6*a^4*x^4 + 8*a^5*x^5 + 3*a^6*x^6 - 12*a^ 2*x^2*Log[x]))/(12*x^2)
Time = 0.32 (sec) , antiderivative size = 61, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle c^3 \int \frac {(1-a x)^2 (a x+1)^4}{x^3}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle c^3 \int \left (x^3 a^6+2 x^2 a^5-x a^4-4 a^3-\frac {a^2}{x}+\frac {2 a}{x^2}+\frac {1}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c^3 \left (\frac {a^6 x^4}{4}+\frac {2 a^5 x^3}{3}-\frac {a^4 x^2}{2}-4 a^3 x-a^2 \log (x)-\frac {2 a}{x}-\frac {1}{2 x^2}\right )\) |
Input:
Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^3)/x^3,x]
Output:
c^3*(-1/2*1/x^2 - (2*a)/x - 4*a^3*x - (a^4*x^2)/2 + (2*a^5*x^3)/3 + (a^6*x ^4)/4 - a^2*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 = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.69
method | result | size |
default | \(c^{3} \left (\frac {a^{6} x^{4}}{4}+\frac {2 a^{5} x^{3}}{3}-\frac {a^{4} x^{2}}{2}-4 a^{3} x -\frac {1}{2 x^{2}}-\frac {2 a}{x}-a^{2} \ln \left (x \right )\right )\) | \(54\) |
risch | \(\frac {a^{6} c^{3} x^{4}}{4}+\frac {2 a^{5} c^{3} x^{3}}{3}-\frac {a^{4} c^{3} x^{2}}{2}-4 a^{3} c^{3} x +\frac {-2 a \,c^{3} x -\frac {1}{2} c^{3}}{x^{2}}-a^{2} c^{3} \ln \left (x \right )\) | \(71\) |
norman | \(\frac {-\frac {1}{2} c^{3}-2 a \,c^{3} x -4 a^{3} c^{3} x^{3}-\frac {1}{2} a^{4} c^{3} x^{4}+\frac {2}{3} a^{5} c^{3} x^{5}+\frac {1}{4} a^{6} c^{3} x^{6}}{x^{2}}-a^{2} c^{3} \ln \left (x \right )\) | \(73\) |
parallelrisch | \(-\frac {-3 a^{6} c^{3} x^{6}-8 a^{5} c^{3} x^{5}+6 a^{4} c^{3} x^{4}+48 a^{3} c^{3} x^{3}+12 c^{3} \ln \left (x \right ) a^{2} x^{2}+24 a \,c^{3} x +6 c^{3}}{12 x^{2}}\) | \(76\) |
meijerg | \(\frac {a^{2} c^{3} \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{6}+\ln \left (-a^{2} x^{2}+1\right )\right )}{2}+a^{2} c^{3} \left (-a^{2} x^{2}-\ln \left (-a^{2} x^{2}+1\right )\right )-a^{2} c^{3} \left (2 \ln \left (x \right )+\ln \left (-a^{2}\right )-\ln \left (-a^{2} x^{2}+1\right )\right )-\frac {a^{3} c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{\sqrt {-a^{2}}}-\frac {3 a^{3} c^{3} \left (-\frac {2 x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2}}+\frac {2 \left (-a^{2}\right )^{\frac {3}{2}} \operatorname {arctanh}\left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}-6 a^{2} c^{3} \operatorname {arctanh}\left (a x \right )-\frac {a^{3} c^{3} \left (-\frac {2}{x \sqrt {-a^{2}}}+\frac {2 a \,\operatorname {arctanh}\left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}-\frac {a^{2} c^{3} \left (\frac {1}{a^{2} x^{2}}-2 \ln \left (x \right )-\ln \left (-a^{2}\right )+\ln \left (-a^{2} x^{2}+1\right )\right )}{2}\) | \(294\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^3,x,method=_RETURNVERBOSE)
Output:
c^3*(1/4*a^6*x^4+2/3*a^5*x^3-1/2*a^4*x^2-4*a^3*x-1/2/x^2-2*a/x-a^2*ln(x))
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=\frac {3 \, a^{6} c^{3} x^{6} + 8 \, a^{5} c^{3} x^{5} - 6 \, a^{4} c^{3} x^{4} - 48 \, a^{3} c^{3} x^{3} - 12 \, a^{2} c^{3} x^{2} \log \left (x\right ) - 24 \, a c^{3} x - 6 \, c^{3}}{12 \, x^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^3,x, algorithm="fricas ")
Output:
1/12*(3*a^6*c^3*x^6 + 8*a^5*c^3*x^5 - 6*a^4*c^3*x^4 - 48*a^3*c^3*x^3 - 12* a^2*c^3*x^2*log(x) - 24*a*c^3*x - 6*c^3)/x^2
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=\frac {a^{6} c^{3} x^{4}}{4} + \frac {2 a^{5} c^{3} x^{3}}{3} - \frac {a^{4} c^{3} x^{2}}{2} - 4 a^{3} c^{3} x - a^{2} c^{3} \log {\left (x \right )} + \frac {- 4 a c^{3} x - c^{3}}{2 x^{2}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)**3/x**3,x)
Output:
a**6*c**3*x**4/4 + 2*a**5*c**3*x**3/3 - a**4*c**3*x**2/2 - 4*a**3*c**3*x - a**2*c**3*log(x) + (-4*a*c**3*x - c**3)/(2*x**2)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=\frac {1}{4} \, a^{6} c^{3} x^{4} + \frac {2}{3} \, a^{5} c^{3} x^{3} - \frac {1}{2} \, a^{4} c^{3} x^{2} - 4 \, a^{3} c^{3} x - a^{2} c^{3} \log \left (x\right ) - \frac {4 \, a c^{3} x + c^{3}}{2 \, x^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^3,x, algorithm="maxima ")
Output:
1/4*a^6*c^3*x^4 + 2/3*a^5*c^3*x^3 - 1/2*a^4*c^3*x^2 - 4*a^3*c^3*x - a^2*c^ 3*log(x) - 1/2*(4*a*c^3*x + c^3)/x^2
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=\frac {1}{4} \, a^{6} c^{3} x^{4} + \frac {2}{3} \, a^{5} c^{3} x^{3} - \frac {1}{2} \, a^{4} c^{3} x^{2} - 4 \, a^{3} c^{3} x - a^{2} c^{3} \log \left ({\left | x \right |}\right ) - \frac {4 \, a c^{3} x + c^{3}}{2 \, x^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^3,x, algorithm="giac")
Output:
1/4*a^6*c^3*x^4 + 2/3*a^5*c^3*x^3 - 1/2*a^4*c^3*x^2 - 4*a^3*c^3*x - a^2*c^ 3*log(abs(x)) - 1/2*(4*a*c^3*x + c^3)/x^2
Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=\frac {2\,a^5\,c^3\,x^3}{3}-4\,a^3\,c^3\,x-\frac {a^4\,c^3\,x^2}{2}-\frac {\frac {c^3}{2}+2\,a\,c^3\,x}{x^2}+\frac {a^6\,c^3\,x^4}{4}-a^2\,c^3\,\ln \left (x\right ) \] Input:
int(-((c - a^2*c*x^2)^3*(a*x + 1)^2)/(x^3*(a^2*x^2 - 1)),x)
Output:
(2*a^5*c^3*x^3)/3 - 4*a^3*c^3*x - (a^4*c^3*x^2)/2 - (c^3/2 + 2*a*c^3*x)/x^ 2 + (a^6*c^3*x^4)/4 - a^2*c^3*log(x)
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \frac {e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^3}{x^3} \, dx=\frac {c^{3} \left (-12 \,\mathrm {log}\left (x \right ) a^{2} x^{2}+3 a^{6} x^{6}+8 a^{5} x^{5}-6 a^{4} x^{4}-48 a^{3} x^{3}-24 a x -6\right )}{12 x^{2}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^3/x^3,x)
Output:
(c**3*( - 12*log(x)*a**2*x**2 + 3*a**6*x**6 + 8*a**5*x**5 - 6*a**4*x**4 - 48*a**3*x**3 - 24*a*x - 6))/(12*x**2)