Integrand size = 14, antiderivative size = 71 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {2 (1+a)^3 x}{b^4}+\frac {(1+a)^2 x^2}{b^3}-\frac {2 (1+a) x^3}{3 b^2}+\frac {x^4}{2 b}-\frac {x^5}{5}+\frac {2 (1+a)^4 \log (1+a+b x)}{b^5} \] 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 = 71, normalized size of antiderivative = 1.00 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {2 (1+a)^3 x}{b^4}+\frac {(1+a)^2 x^2}{b^3}-\frac {2 (1+a) x^3}{3 b^2}+\frac {x^4}{2 b}-\frac {x^5}{5}+\frac {2 (1+a)^4 \log (1+a+b x)}{b^5} \] Input:
Integrate[x^4/E^(2*ArcTanh[a + b*x]),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 = 0.30 (sec) , antiderivative size = 71, 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 |
\(\displaystyle \frac {2 (a+1)^4 \log (a+b x+1)}{b^5}-\frac {2 (a+1)^3 x}{b^4}+\frac {(a+1)^2 x^2}{b^3}-\frac {2 (a+1) x^3}{3 b^2}+\frac {x^4}{2 b}-\frac {x^5}{5}\) |
Input:
Int[x^4/E^(2*ArcTanh[a + b*x]),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 = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.59
method | result | size |
default | \(-\frac {\frac {1}{5} b^{4} x^{5}-\frac {1}{2} b^{3} x^{4}+\frac {2}{3} a \,b^{2} x^{3}-a^{2} b \,x^{2}+\frac {2}{3} b^{2} x^{3}+2 a^{3} x -2 a b \,x^{2}+6 a^{2} x -b \,x^{2}+6 a x +2 x}{b^{4}}+\frac {\left (2 a^{4}+8 a^{3}+12 a^{2}+8 a +2\right ) \ln \left (b x +a +1\right )}{b^{5}}\) | \(113\) |
norman | \(\frac {\left (-\frac {a}{5}+\frac {3}{10}\right ) x^{5}+\frac {2 a^{5}+10 a^{4}+20 a^{3}+20 a^{2}+10 a +2}{b^{5}}-\frac {b \,x^{6}}{5}-\frac {\left (a +1\right ) x^{4}}{6 b}+\frac {\left (a^{2}+2 a +1\right ) x^{3}}{3 b^{2}}-\frac {\left (a^{3}+3 a^{2}+3 a +1\right ) x^{2}}{b^{3}}}{b x +a +1}+\frac {2 \left (a^{4}+4 a^{3}+6 a^{2}+4 a +1\right ) \ln \left (b x +a +1\right )}{b^{5}}\) | \(134\) |
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 | \(-\frac {x^{5}}{5}+\frac {x^{4}}{2 b}-\frac {2 a \,x^{3}}{3 b^{2}}+\frac {a^{2} x^{2}}{b^{3}}-\frac {2 x^{3}}{3 b^{2}}-\frac {2 a^{3} x}{b^{4}}+\frac {2 a \,x^{2}}{b^{3}}-\frac {6 x \,a^{2}}{b^{4}}+\frac {x^{2}}{b^{3}}-\frac {6 a x}{b^{4}}-\frac {2 x}{b^{4}}+\frac {2 \ln \left (b x +a +1\right ) a^{4}}{b^{5}}+\frac {8 \ln \left (b x +a +1\right ) a^{3}}{b^{5}}+\frac {12 \ln \left (b x +a +1\right ) a^{2}}{b^{5}}+\frac {8 \ln \left (b x +a +1\right ) a}{b^{5}}+\frac {2 \ln \left (b x +a +1\right )}{b^{5}}\) | \(159\) |
Input:
int(x^4/(b*x+a+1)^2*(1-(b*x+a)^2),x,method=_RETURNVERBOSE)
Output:
-1/b^4*(1/5*b^4*x^5-1/2*b^3*x^4+2/3*a*b^2*x^3-a^2*b*x^2+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)+(2*a^4+8*a^3+12*a^2+8*a+2)/b^5*ln(b*x +a+1)
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.31 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {6 \, b^{5} x^{5} - 15 \, b^{4} x^{4} + 20 \, {\left (a + 1\right )} b^{3} x^{3} - 30 \, {\left (a^{2} + 2 \, a + 1\right )} b^{2} x^{2} + 60 \, {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} b x - 60 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )}{30 \, b^{5}} \] Input:
integrate(x^4/(b*x+a+1)^2*(1-(b*x+a)^2),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 = 0.17 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.44 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=- \frac {x^{5}}{5} - x^{3} \cdot \left (\frac {2 a}{3 b^{2}} + \frac {2}{3 b^{2}}\right ) - x^{2} \left (- \frac {a^{2}}{b^{3}} - \frac {2 a}{b^{3}} - \frac {1}{b^{3}}\right ) - x \left (\frac {2 a^{3}}{b^{4}} + \frac {6 a^{2}}{b^{4}} + \frac {6 a}{b^{4}} + \frac {2}{b^{4}}\right ) + \frac {x^{4}}{2 b} + \frac {2 \left (a + 1\right )^{4} \log {\left (a + b x + 1 \right )}}{b^{5}} \] Input:
integrate(x**4/(b*x+a+1)**2*(1-(b*x+a)**2),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 - 1/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 = 1.32 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=-\frac {6 \, b^{4} x^{5} - 15 \, b^{3} x^{4} + 20 \, {\left (a + 1\right )} b^{2} x^{3} - 30 \, {\left (a^{2} + 2 \, a + 1\right )} b x^{2} + 60 \, {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} x}{30 \, b^{4}} + \frac {2 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} \] Input:
integrate(x^4/(b*x+a+1)^2*(1-(b*x+a)^2),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
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (65) = 130\).
Time = 0.14 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.85 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=\frac {{\left (b x + a + 1\right )}^{5} {\left (\frac {15 \, {\left (2 \, a b + 3 \, b\right )}}{{\left (b x + a + 1\right )} b} - \frac {20 \, {\left (3 \, a^{2} b^{2} + 10 \, a b^{2} + 7 \, b^{2}\right )}}{{\left (b x + a + 1\right )}^{2} b^{2}} + \frac {60 \, {\left (a^{3} b^{3} + 6 \, a^{2} b^{3} + 9 \, a b^{3} + 4 \, b^{3}\right )}}{{\left (b x + a + 1\right )}^{3} b^{3}} - \frac {30 \, {\left (a^{4} b^{4} + 12 \, a^{3} b^{4} + 30 \, a^{2} b^{4} + 28 \, a b^{4} + 9 \, b^{4}\right )}}{{\left (b x + a + 1\right )}^{4} b^{4}} - 6\right )}}{30 \, b^{5}} - \frac {2 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2} {\left | b \right |}}\right )}{b^{5}} \] Input:
integrate(x^4/(b*x+a+1)^2*(1-(b*x+a)^2),x, algorithm="giac")
Output:
1/30*(b*x + a + 1)^5*(15*(2*a*b + 3*b)/((b*x + a + 1)*b) - 20*(3*a^2*b^2 + 10*a*b^2 + 7*b^2)/((b*x + a + 1)^2*b^2) + 60*(a^3*b^3 + 6*a^2*b^3 + 9*a*b ^3 + 4*b^3)/((b*x + a + 1)^3*b^3) - 30*(a^4*b^4 + 12*a^3*b^4 + 30*a^2*b^4 + 28*a*b^4 + 9*b^4)/((b*x + a + 1)^4*b^4) - 6)/b^5 - 2*(a^4 + 4*a^3 + 6*a^ 2 + 4*a + 1)*log(abs(b*x + a + 1)/((b*x + a + 1)^2*abs(b)))/b^5
Time = 0.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.99 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=\frac {\ln \left (a+b\,x+1\right )\,\left (2\,a^4+8\,a^3+12\,a^2+8\,a+2\right )}{b^5}-\frac {x^5}{5}-x^4\,\left (\frac {a-1}{4\,b}-\frac {a+1}{4\,b}\right )-\frac {x^2\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,{\left (a+1\right )}^2}{2\,b^2}+\frac {x^3\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,\left (a+1\right )}{3\,b}+\frac {x\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,{\left (a+1\right )}^3}{b^3} \] Input:
int(-(x^4*((a + b*x)^2 - 1))/(a + b*x + 1)^2,x)
Output:
(log(a + b*x + 1)*(8*a + 12*a^2 + 8*a^3 + 2*a^4 + 2))/b^5 - x^5/5 - x^4*(( a - 1)/(4*b) - (a + 1)/(4*b)) - (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 = 0.17 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.04 \[ \int e^{-2 \text {arctanh}(a+b x)} x^4 \, dx=\frac {60 \,\mathrm {log}\left (b x +a +1\right ) a^{4}+240 \,\mathrm {log}\left (b x +a +1\right ) a^{3}+360 \,\mathrm {log}\left (b x +a +1\right ) a^{2}+240 \,\mathrm {log}\left (b x +a +1\right ) a +60 \,\mathrm {log}\left (b x +a +1\right )-60 a^{3} b x +30 a^{2} b^{2} x^{2}-180 a^{2} b x -20 a \,b^{3} x^{3}+60 a \,b^{2} x^{2}-180 a b x -6 b^{5} x^{5}+15 b^{4} x^{4}-20 b^{3} x^{3}+30 b^{2} x^{2}-60 b x}{30 b^{5}} \] Input:
int(x^4/(b*x+a+1)^2*(1-(b*x+a)^2),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)