Integrand size = 22, antiderivative size = 51 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+c^2 x+\frac {4 c^2 \log (x)}{a} \] Output:
-1/3*c^2/a^4/x^3-2*c^2/a^3/x^2-6*c^2/a^2/x+c^2*x+4*c^2*ln(x)/a
Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+c^2 x+\frac {4 c^2 \log (x)}{a} \] Input:
Integrate[E^(4*ArcTanh[a*x])*(c - c/(a^2*x^2))^2,x]
Output:
-1/3*c^2/(a^4*x^3) - (2*c^2)/(a^3*x^2) - (6*c^2)/(a^2*x) + c^2*x + (4*c^2* Log[x])/a
Time = 0.38 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6707, 6700, 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.
\(\displaystyle \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle \frac {c^2 \int \frac {e^{4 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^2}{x^4}dx}{a^4}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {c^2 \int \frac {(a x+1)^4}{x^4}dx}{a^4}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c^2 \int \left (a^4+\frac {4 a^3}{x}+\frac {6 a^2}{x^2}+\frac {4 a}{x^3}+\frac {1}{x^4}\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \left (a^4 x+4 a^3 \log (x)-\frac {6 a^2}{x}-\frac {2 a}{x^2}-\frac {1}{3 x^3}\right )}{a^4}\) |
Input:
Int[E^(4*ArcTanh[a*x])*(c - c/(a^2*x^2))^2,x]
Output:
(c^2*(-1/3*1/x^3 - (2*a)/x^2 - (6*a^2)/x + a^4*x + 4*a^3*Log[x]))/a^4
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_.))*(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])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {c^{2} \left (x \,a^{4}-\frac {2 a}{x^{2}}-\frac {6 a^{2}}{x}+4 a^{3} \ln \left (x \right )-\frac {1}{3 x^{3}}\right )}{a^{4}}\) | \(40\) |
risch | \(c^{2} x +\frac {-6 c^{2} a^{2} x^{2}-2 a \,c^{2} x -\frac {1}{3} c^{2}}{a^{4} x^{3}}+\frac {4 c^{2} \ln \left (x \right )}{a}\) | \(48\) |
parallelrisch | \(\frac {3 a^{4} c^{2} x^{4}+12 c^{2} \ln \left (x \right ) a^{3} x^{3}-18 c^{2} a^{2} x^{2}-6 a \,c^{2} x -c^{2}}{3 a^{4} x^{3}}\) | \(57\) |
norman | \(\frac {-2 a^{4} c^{2} x^{5}+a^{5} c^{2} x^{6}+\frac {c^{2}}{3 a}+2 c^{2} x +\frac {17 a \,c^{2} x^{2}}{3}-7 a^{3} c^{2} x^{4}}{\left (a^{2} x^{2}-1\right ) a^{3} x^{3}}+\frac {4 c^{2} \ln \left (x \right )}{a}\) | \(86\) |
meijerg | \(\frac {c^{2} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}-\frac {2 c^{2} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \operatorname {arctanh}\left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}-\frac {5 c^{2} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \operatorname {arctanh}\left (a x \right )}{a}\right )}{\sqrt {-a^{2}}}+\frac {2 c^{2} \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {2 a \,c^{2} x^{2}}{-a^{2} x^{2}+1}-\frac {2 c^{2} \left (1+2 \ln \left (x \right )+\ln \left (-a^{2}\right )+\frac {2 a^{2} x^{2}}{-2 a^{2} x^{2}+2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {2 c^{2} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \,\operatorname {arctanh}\left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}-\frac {2 c^{2} \left (\frac {1}{a^{2} x^{2}}-1-4 \ln \left (x \right )-2 \ln \left (-a^{2}\right )-\frac {3 a^{2} x^{2}}{-3 a^{2} x^{2}+3}+2 \ln \left (-a^{2} x^{2}+1\right )\right )}{a}+\frac {c^{2} \left (-\frac {2 \left (-15 a^{4} x^{4}+10 a^{2} x^{2}+2\right )}{3 x^{3} \left (-a^{2}\right )^{\frac {3}{2}} \left (-2 a^{2} x^{2}+2\right )}+\frac {5 a^{3} \operatorname {arctanh}\left (a x \right )}{\left (-a^{2}\right )^{\frac {3}{2}}}\right )}{2 \sqrt {-a^{2}}}\) | \(476\) |
Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^2,x,method=_RETURNVERBOSE)
Output:
c^2/a^4*(x*a^4-2*a/x^2-6*a^2/x+4*a^3*ln(x)-1/3/x^3)
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {3 \, a^{4} c^{2} x^{4} + 12 \, a^{3} c^{2} x^{3} \log \left (x\right ) - 18 \, a^{2} c^{2} x^{2} - 6 \, a c^{2} x - c^{2}}{3 \, a^{4} x^{3}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^2,x, algorithm="fricas")
Output:
1/3*(3*a^4*c^2*x^4 + 12*a^3*c^2*x^3*log(x) - 18*a^2*c^2*x^2 - 6*a*c^2*x - c^2)/(a^4*x^3)
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {a^{4} c^{2} x + 4 a^{3} c^{2} \log {\left (x \right )} + \frac {- 18 a^{2} c^{2} x^{2} - 6 a c^{2} x - c^{2}}{3 x^{3}}}{a^{4}} \] Input:
integrate((a*x+1)**4/(-a**2*x**2+1)**2*(c-c/a**2/x**2)**2,x)
Output:
(a**4*c**2*x + 4*a**3*c**2*log(x) + (-18*a**2*c**2*x**2 - 6*a*c**2*x - c** 2)/(3*x**3))/a**4
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=c^{2} x + \frac {4 \, c^{2} \log \left (x\right )}{a} - \frac {18 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^2,x, algorithm="maxima")
Output:
c^2*x + 4*c^2*log(x)/a - 1/3*(18*a^2*c^2*x^2 + 6*a*c^2*x + c^2)/(a^4*x^3)
Time = 0.14 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=c^{2} x + \frac {4 \, c^{2} \log \left ({\left | x \right |}\right )}{a} - \frac {18 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^2,x, algorithm="giac")
Output:
c^2*x + 4*c^2*log(abs(x))/a - 1/3*(18*a^2*c^2*x^2 + 6*a*c^2*x + c^2)/(a^4* x^3)
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=-\frac {c^2\,\left (6\,a\,x+18\,a^2\,x^2-3\,a^4\,x^4-12\,a^3\,x^3\,\ln \left (x\right )+1\right )}{3\,a^4\,x^3} \] Input:
int(((c - c/(a^2*x^2))^2*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)
Output:
-(c^2*(6*a*x + 18*a^2*x^2 - 3*a^4*x^4 - 12*a^3*x^3*log(x) + 1))/(3*a^4*x^3 )
Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx=\frac {c^{2} \left (12 \,\mathrm {log}\left (x \right ) a^{3} x^{3}+3 a^{4} x^{4}-18 a^{2} x^{2}-6 a x -1\right )}{3 a^{4} x^{3}} \] Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^2,x)
Output:
(c**2*(12*log(x)*a**3*x**3 + 3*a**4*x**4 - 18*a**2*x**2 - 6*a*x - 1))/(3*a **4*x**3)