Integrand size = 14, antiderivative size = 83 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {2 (1-a)^3 x}{b^4}-\frac {(1-a)^2 x^2}{b^3}-\frac {2 (1-a) x^3}{3 b^2}-\frac {x^4}{2 b}-\frac {x^5}{5}-\frac {2 (1-a)^4 \log (1-a-b x)}{b^5} \] Output:
-2*(1-a)^3*x/b^4-(1-a)^2*x^2/b^3-2/3*(1-a)*x^3/b^2-1/2*x^4/b-1/5*x^5-2*(1- a)^4*ln(-b*x-a+1)/b^5
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=\frac {2 (-1+a)^3 x}{b^4}-\frac {(-1+a)^2 x^2}{b^3}+\frac {2 (-1+a) x^3}{3 b^2}-\frac {x^4}{2 b}-\frac {x^5}{5}-\frac {2 (-1+a)^4 \log (1-a-b x)}{b^5} \] Input:
Integrate[E^(2*ArcTanh[a + b*x])*x^4,x]
Output:
(2*(-1 + a)^3*x)/b^4 - ((-1 + a)^2*x^2)/b^3 + (2*(-1 + a)*x^3)/(3*b^2) - x ^4/(2*b) - x^5/5 - (2*(-1 + a)^4*Log[1 - a - b*x])/b^5
Time = 0.32 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6713, 86, 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.
\(\displaystyle \int x^4 e^{2 \text {arctanh}(a+b x)} \, dx\) |
\(\Big \downarrow \) 6713 |
\(\displaystyle \int \frac {x^4 (a+b x+1)}{-a-b x+1}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {2 (a-1)^4}{b^4 (a+b x-1)}+\frac {2 (a-1)^3}{b^4}-\frac {2 (a-1)^2 x}{b^3}+\frac {2 (a-1) x^2}{b^2}-\frac {2 x^3}{b}-x^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 (1-a)^4 \log (-a-b x+1)}{b^5}-\frac {2 (1-a)^3 x}{b^4}-\frac {(1-a)^2 x^2}{b^3}-\frac {2 (1-a) x^3}{3 b^2}-\frac {x^4}{2 b}-\frac {x^5}{5}\) |
Input:
Int[E^(2*ArcTanh[a + b*x])*x^4,x]
Output:
(-2*(1 - a)^3*x)/b^4 - ((1 - a)^2*x^2)/b^3 - (2*(1 - a)*x^3)/(3*b^2) - x^4 /(2*b) - x^5/5 - (2*(1 - a)^4*Log[1 - a - b*x])/b^5
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.) , x_Symbol] :> Int[(d + e*x)^m*((1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^( n/2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10
method | result | size |
norman | \(-\frac {x^{5}}{5}-\frac {x^{4}}{2 b}-\frac {\left (a^{2}-2 a +1\right ) x^{2}}{b^{3}}+\frac {2 \left (-1+a \right ) x^{3}}{3 b^{2}}+\frac {2 \left (a^{3}-3 a^{2}+3 a -1\right ) x}{b^{4}}-\frac {2 \left (a^{4}-4 a^{3}+6 a^{2}-4 a +1\right ) \ln \left (b x +a -1\right )}{b^{5}}\) | \(91\) |
default | \(\frac {-\frac {1}{5} b^{4} x^{5}-\frac {1}{2} b^{3} x^{4}+\frac {2}{3} a \,b^{2} x^{3}-a^{2} b \,x^{2}-\frac {2}{3} b^{2} x^{3}+2 a^{3} x +2 a b \,x^{2}-6 a^{2} x -b \,x^{2}+6 a x -2 x}{b^{4}}+\frac {\left (-2 a^{4}+8 a^{3}-12 a^{2}+8 a -2\right ) \ln \left (b x +a -1\right )}{b^{5}}\) | \(112\) |
parallelrisch | \(-\frac {6 x^{5} b^{5}+15 b^{4} x^{4}-20 a \,b^{3} x^{3}+20 b^{3} x^{3}+30 a^{2} b^{2} x^{2}+60 \ln \left (b x +a -1\right ) a^{4}-60 a \,b^{2} x^{2}-60 a^{3} b x -240 \ln \left (b x +a -1\right ) a^{3}+30 b^{2} x^{2}+180 a^{2} b x +360 \ln \left (b x +a -1\right ) a^{2}-180 a b x -240 \ln \left (b x +a -1\right ) a +60 b x +60 \ln \left (b x +a -1\right )}{30 b^{5}}\) | \(146\) |
risch | \(-\frac {x^{5}}{5}-\frac {x^{4}}{2 b}+\frac {2 a \,x^{3}}{3 b^{2}}-\frac {a^{2} x^{2}}{b^{3}}-\frac {2 x^{3}}{3 b^{2}}+\frac {2 a^{3} x}{b^{4}}+\frac {2 a \,x^{2}}{b^{3}}-\frac {6 x \,a^{2}}{b^{4}}-\frac {x^{2}}{b^{3}}+\frac {6 a x}{b^{4}}-\frac {2 x}{b^{4}}-\frac {2 \ln \left (b x +a -1\right ) a^{4}}{b^{5}}+\frac {8 \ln \left (b x +a -1\right ) a^{3}}{b^{5}}-\frac {12 \ln \left (b x +a -1\right ) a^{2}}{b^{5}}+\frac {8 \ln \left (b x +a -1\right ) a}{b^{5}}-\frac {2 \ln \left (b x +a -1\right )}{b^{5}}\) | \(161\) |
Input:
int((b*x+a+1)^2/(1-(b*x+a)^2)*x^4,x,method=_RETURNVERBOSE)
Output:
-1/5*x^5-1/2*x^4/b-1/b^3*(a^2-2*a+1)*x^2+2/3/b^2*(-1+a)*x^3+2*(a^3-3*a^2+3 *a-1)/b^4*x-2*(a^4-4*a^3+6*a^2-4*a+1)/b^5*ln(b*x+a-1)
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {6 \, b^{5} x^{5} + 15 \, b^{4} x^{4} - 20 \, {\left (a - 1\right )} b^{3} x^{3} + 30 \, {\left (a^{2} - 2 \, a + 1\right )} b^{2} x^{2} - 60 \, {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} b x + 60 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{30 \, b^{5}} \] Input:
integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^4,x, algorithm="fricas")
Output:
-1/30*(6*b^5*x^5 + 15*b^4*x^4 - 20*(a - 1)*b^3*x^3 + 30*(a^2 - 2*a + 1)*b^ 2*x^2 - 60*(a^3 - 3*a^2 + 3*a - 1)*b*x + 60*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1 )*log(b*x + a - 1))/b^5
Time = 0.18 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=- \frac {x^{5}}{5} - x^{3} \left (- \frac {2 a}{3 b^{2}} + \frac {2}{3 b^{2}}\right ) - x^{2} \left (\frac {a^{2}}{b^{3}} - \frac {2 a}{b^{3}} + \frac {1}{b^{3}}\right ) - x \left (- \frac {2 a^{3}}{b^{4}} + \frac {6 a^{2}}{b^{4}} - \frac {6 a}{b^{4}} + \frac {2}{b^{4}}\right ) - \frac {x^{4}}{2 b} - \frac {2 \left (a - 1\right )^{4} \log {\left (a + b x - 1 \right )}}{b^{5}} \] Input:
integrate((b*x+a+1)**2/(1-(b*x+a)**2)*x**4,x)
Output:
-x**5/5 - x**3*(-2*a/(3*b**2) + 2/(3*b**2)) - x**2*(a**2/b**3 - 2*a/b**3 + b**(-3)) - x*(-2*a**3/b**4 + 6*a**2/b**4 - 6*a/b**4 + 2/b**4) - x**4/(2*b ) - 2*(a - 1)**4*log(a + b*x - 1)/b**5
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {6 \, b^{4} x^{5} + 15 \, b^{3} x^{4} - 20 \, {\left (a - 1\right )} b^{2} x^{3} + 30 \, {\left (a^{2} - 2 \, a + 1\right )} b x^{2} - 60 \, {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} x}{30 \, b^{4}} - \frac {2 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}} \] Input:
integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^4,x, algorithm="maxima")
Output:
-1/30*(6*b^4*x^5 + 15*b^3*x^4 - 20*(a - 1)*b^2*x^3 + 30*(a^2 - 2*a + 1)*b* x^2 - 60*(a^3 - 3*a^2 + 3*a - 1)*x)/b^4 - 2*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1 )*log(b*x + a - 1)/b^5
Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.47 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {2 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left ({\left | b x + a - 1 \right |}\right )}{b^{5}} - \frac {6 \, b^{5} x^{5} + 15 \, b^{4} x^{4} - 20 \, a b^{3} x^{3} + 30 \, a^{2} b^{2} x^{2} + 20 \, b^{3} x^{3} - 60 \, a^{3} b x - 60 \, a b^{2} x^{2} + 180 \, a^{2} b x + 30 \, b^{2} x^{2} - 180 \, a b x + 60 \, b x}{30 \, b^{5}} \] Input:
integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^4,x, algorithm="giac")
Output:
-2*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(abs(b*x + a - 1))/b^5 - 1/30*(6*b^5 *x^5 + 15*b^4*x^4 - 20*a*b^3*x^3 + 30*a^2*b^2*x^2 + 20*b^3*x^3 - 60*a^3*b* x - 60*a*b^2*x^2 + 180*a^2*b*x + 30*b^2*x^2 - 180*a*b*x + 60*b*x)/b^5
Time = 14.01 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.71 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=x^4\,\left (\frac {a-1}{4\,b}-\frac {a+1}{4\,b}\right )-\frac {x^5}{5}-\frac {\ln \left (a+b\,x-1\right )\,\left (2\,a^4-8\,a^3+12\,a^2-8\,a+2\right )}{b^5}+\frac {x^2\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,{\left (a-1\right )}^2}{2\,b^2}-\frac {x^3\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,\left (a-1\right )}{3\,b}-\frac {x\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,{\left (a-1\right )}^3}{b^3} \] Input:
int(-(x^4*(a + b*x + 1)^2)/((a + b*x)^2 - 1),x)
Output:
x^4*((a - 1)/(4*b) - (a + 1)/(4*b)) - x^5/5 - (log(a + b*x - 1)*(12*a^2 - 8*a - 8*a^3 + 2*a^4 + 2))/b^5 + (x^2*((a - 1)/b - (a + 1)/b)*(a - 1)^2)/(2 *b^2) - (x^3*((a - 1)/b - (a + 1)/b)*(a - 1))/(3*b) - (x*((a - 1)/b - (a + 1)/b)*(a - 1)^3)/b^3
Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.75 \[ \int e^{2 \text {arctanh}(a+b x)} x^4 \, dx=\frac {-60 \,\mathrm {log}\left (b x +a -1\right ) a^{4}+240 \,\mathrm {log}\left (b x +a -1\right ) a^{3}-360 \,\mathrm {log}\left (b x +a -1\right ) a^{2}+240 \,\mathrm {log}\left (b x +a -1\right ) a -60 \,\mathrm {log}\left (b x +a -1\right )+60 a^{3} b x -30 a^{2} b^{2} x^{2}-180 a^{2} b x +20 a \,b^{3} x^{3}+60 a \,b^{2} x^{2}+180 a b x -6 b^{5} x^{5}-15 b^{4} x^{4}-20 b^{3} x^{3}-30 b^{2} x^{2}-60 b x}{30 b^{5}} \] Input:
int((b*x+a+1)^2/(1-(b*x+a)^2)*x^4,x)
Output:
( - 60*log(a + b*x - 1)*a**4 + 240*log(a + b*x - 1)*a**3 - 360*log(a + b*x - 1)*a**2 + 240*log(a + b*x - 1)*a - 60*log(a + b*x - 1) + 60*a**3*b*x - 30*a**2*b**2*x**2 - 180*a**2*b*x + 20*a*b**3*x**3 + 60*a*b**2*x**2 + 180*a *b*x - 6*b**5*x**5 - 15*b**4*x**4 - 20*b**3*x**3 - 30*b**2*x**2 - 60*b*x)/ (30*b**5)