Integrand size = 22, antiderivative size = 66 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=-\frac {c^4}{3 a^4 x^3}+\frac {3 c^4}{a^3 x^2}-\frac {16 c^4}{a^2 x}-c^4 x-\frac {26 c^4 \log (x)}{a}+\frac {32 c^4 \log (1+a x)}{a} \] Output:
-1/3*c^4/a^4/x^3+3*c^4/a^3/x^2-16*c^4/a^2/x-c^4*x-26*c^4*ln(x)/a+32*c^4*ln (a*x+1)/a
Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {c^4 \left (-\frac {1}{3 x^3}+\frac {3 a}{x^2}-\frac {16 a^2}{x}-a^4 x-26 a^3 \log (x)+32 a^3 \log (1+a x)\right )}{a^4} \] Input:
Integrate[(c - c/(a*x))^4/E^(2*ArcTanh[a*x]),x]
Output:
(c^4*(-1/3*1/x^3 + (3*a)/x^2 - (16*a^2)/x - a^4*x - 26*a^3*Log[x] + 32*a^3 *Log[1 + a*x]))/a^4
Time = 0.40 (sec) , antiderivative size = 53, 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 = {6681, 6679, 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.
\(\displaystyle \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle \frac {c^4 \int \frac {e^{-2 \text {arctanh}(a x)} (1-a x)^4}{x^4}dx}{a^4}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \frac {c^4 \int \frac {(1-a x)^5}{x^4 (a x+1)}dx}{a^4}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {c^4 \int \left (\frac {32 a^4}{a x+1}-a^4-\frac {26 a^3}{x}+\frac {16 a^2}{x^2}-\frac {6 a}{x^3}+\frac {1}{x^4}\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^4 \left (a^4 (-x)-26 a^3 \log (x)+32 a^3 \log (a x+1)-\frac {16 a^2}{x}+\frac {3 a}{x^2}-\frac {1}{3 x^3}\right )}{a^4}\) |
Input:
Int[(c - c/(a*x))^4/E^(2*ArcTanh[a*x]),x]
Output:
(c^4*(-1/3*1/x^3 + (3*a)/x^2 - (16*a^2)/x - a^4*x - 26*a^3*Log[x] + 32*a^3 *Log[1 + a*x]))/a^4
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_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Simp[c^p Int[u*(1 + d*(x/c))^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] , x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol ] :> Simp[d^p Int[u*(1 + c*(x/d))^p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; F reeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.79
method | result | size |
default | \(\frac {c^{4} \left (-x \,a^{4}+32 a^{3} \ln \left (a x +1\right )-\frac {1}{3 x^{3}}+\frac {3 a}{x^{2}}-\frac {16 a^{2}}{x}-26 a^{3} \ln \left (x \right )\right )}{a^{4}}\) | \(52\) |
risch | \(-c^{4} x +\frac {-16 a^{2} c^{4} x^{2}+3 a \,c^{4} x -\frac {1}{3} c^{4}}{a^{4} x^{3}}-\frac {26 c^{4} \ln \left (x \right )}{a}+\frac {32 c^{4} \ln \left (-a x -1\right )}{a}\) | \(64\) |
parallelrisch | \(-\frac {3 a^{4} c^{4} x^{4}+78 c^{4} \ln \left (x \right ) a^{3} x^{3}-96 c^{4} \ln \left (a x +1\right ) a^{3} x^{3}+48 a^{2} c^{4} x^{2}-9 a \,c^{4} x +c^{4}}{3 a^{4} x^{3}}\) | \(72\) |
norman | \(\frac {15 a^{3} c^{4} x^{4}-\frac {c^{4}}{3 a}+\frac {8 c^{4} x}{3}-13 a \,c^{4} x^{2}-a^{4} c^{4} x^{5}}{\left (a x +1\right ) a^{3} x^{3}}-\frac {26 c^{4} \ln \left (x \right )}{a}+\frac {32 c^{4} \ln \left (a x +1\right )}{a}\) | \(86\) |
meijerg | \(-\frac {c^{4} \left (\frac {a x \left (3 a x +6\right )}{3 a x +3}-2 \ln \left (a x +1\right )\right )}{a}-\frac {5 c^{4} x}{a x +1}+\frac {4 c^{4} \left (-\frac {a x}{a x +1}+\ln \left (a x +1\right )\right )}{a}+\frac {5 c^{4} \left (-\frac {1}{a x}-1-2 \ln \left (x \right )-2 \ln \left (a \right )+\frac {3 a x}{3 a x +3}+2 \ln \left (a x +1\right )\right )}{a}-\frac {4 c^{4} \left (-\frac {1}{2 a^{2} x^{2}}+\frac {2}{a x}+1+3 \ln \left (x \right )+3 \ln \left (a \right )-\frac {4 a x}{4 a x +4}-3 \ln \left (a x +1\right )\right )}{a}+\frac {c^{4} \left (-\frac {1}{3 x^{3} a^{3}}+\frac {1}{a^{2} x^{2}}-\frac {3}{a x}-1-4 \ln \left (x \right )-4 \ln \left (a \right )+\frac {5 a x}{5 a x +5}+4 \ln \left (a x +1\right )\right )}{a}\) | \(235\) |
Input:
int((c-c/a/x)^4/(a*x+1)^2*(-a^2*x^2+1),x,method=_RETURNVERBOSE)
Output:
c^4/a^4*(-x*a^4+32*a^3*ln(a*x+1)-1/3/x^3+3*a/x^2-16*a^2/x-26*a^3*ln(x))
Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=-\frac {3 \, a^{4} c^{4} x^{4} - 96 \, a^{3} c^{4} x^{3} \log \left (a x + 1\right ) + 78 \, a^{3} c^{4} x^{3} \log \left (x\right ) + 48 \, a^{2} c^{4} x^{2} - 9 \, a c^{4} x + c^{4}}{3 \, a^{4} x^{3}} \] Input:
integrate((c-c/a/x)^4/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")
Output:
-1/3*(3*a^4*c^4*x^4 - 96*a^3*c^4*x^3*log(a*x + 1) + 78*a^3*c^4*x^3*log(x) + 48*a^2*c^4*x^2 - 9*a*c^4*x + c^4)/(a^4*x^3)
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=- c^{4} x - \frac {2 c^{4} \cdot \left (13 \log {\left (x \right )} - 16 \log {\left (x + \frac {1}{a} \right )}\right )}{a} - \frac {48 a^{2} c^{4} x^{2} - 9 a c^{4} x + c^{4}}{3 a^{4} x^{3}} \] Input:
integrate((c-c/a/x)**4/(a*x+1)**2*(-a**2*x**2+1),x)
Output:
-c**4*x - 2*c**4*(13*log(x) - 16*log(x + 1/a))/a - (48*a**2*c**4*x**2 - 9* a*c**4*x + c**4)/(3*a**4*x**3)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=-c^{4} x + \frac {32 \, c^{4} \log \left (a x + 1\right )}{a} - \frac {26 \, c^{4} \log \left (x\right )}{a} - \frac {48 \, a^{2} c^{4} x^{2} - 9 \, a c^{4} x + c^{4}}{3 \, a^{4} x^{3}} \] Input:
integrate((c-c/a/x)^4/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")
Output:
-c^4*x + 32*c^4*log(a*x + 1)/a - 26*c^4*log(x)/a - 1/3*(48*a^2*c^4*x^2 - 9 *a*c^4*x + c^4)/(a^4*x^3)
Time = 0.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.70 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=-\frac {6 \, c^{4} \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {26 \, c^{4} \log \left ({\left | -\frac {1}{a x + 1} + 1 \right |}\right )}{a} + \frac {{\left (3 \, c^{4} + \frac {49 \, c^{4}}{a x + 1} - \frac {117 \, c^{4}}{{\left (a x + 1\right )}^{2}} + \frac {66 \, c^{4}}{{\left (a x + 1\right )}^{3}}\right )} {\left (a x + 1\right )}}{3 \, a {\left (\frac {1}{a x + 1} - 1\right )}^{3}} \] Input:
integrate((c-c/a/x)^4/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="giac")
Output:
-6*c^4*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a - 26*c^4*log(abs(-1/(a*x + 1) + 1))/a + 1/3*(3*c^4 + 49*c^4/(a*x + 1) - 117*c^4/(a*x + 1)^2 + 66*c^4 /(a*x + 1)^3)*(a*x + 1)/(a*(1/(a*x + 1) - 1)^3)
Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {32\,c^4\,\ln \left (a\,x+1\right )}{a}-\frac {16\,a^2\,c^4\,x^2-3\,a\,c^4\,x+\frac {c^4}{3}}{a^4\,x^3}-\frac {26\,c^4\,\ln \left (x\right )}{a}-c^4\,x \] Input:
int(-((c - c/(a*x))^4*(a^2*x^2 - 1))/(a*x + 1)^2,x)
Output:
(32*c^4*log(a*x + 1))/a - (c^4/3 + 16*a^2*c^4*x^2 - 3*a*c^4*x)/(a^4*x^3) - (26*c^4*log(x))/a - c^4*x
Time = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^4 \, dx=\frac {c^{4} \left (96 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-78 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-3 a^{4} x^{4}-48 a^{2} x^{2}+9 a x -1\right )}{3 a^{4} x^{3}} \] Input:
int((c-c/a/x)^4/(a*x+1)^2*(-a^2*x^2+1),x)
Output:
(c**4*(96*log(a*x + 1)*a**3*x**3 - 78*log(x)*a**3*x**3 - 3*a**4*x**4 - 48* a**2*x**2 + 9*a*x - 1))/(3*a**4*x**3)