Integrand size = 14, antiderivative size = 70 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=-\frac {1-a}{3 (1+a) x^3}+\frac {b}{(1+a)^2 x^2}-\frac {2 b^2}{(1+a)^3 x}-\frac {2 b^3 \log (x)}{(1+a)^4}+\frac {2 b^3 \log (1+a+b x)}{(1+a)^4} \] Output:
-1/3*(1-a)/(1+a)/x^3+b/(1+a)^2/x^2-2*b^2/(1+a)^3/x-2*b^3*ln(x)/(1+a)^4+2*b ^3*ln(b*x+a+1)/(1+a)^4
Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=-\frac {1-a}{3 (1+a) x^3}+\frac {b}{(1+a)^2 x^2}-\frac {2 b^2}{(1+a)^3 x}-\frac {2 b^3 \log (x)}{(1+a)^4}+\frac {2 b^3 \log (1+a+b x)}{(1+a)^4} \] Input:
Integrate[1/(E^(2*ArcTanh[a + b*x])*x^4),x]
Output:
-1/3*(1 - a)/((1 + a)*x^3) + b/((1 + a)^2*x^2) - (2*b^2)/((1 + a)^3*x) - ( 2*b^3*Log[x])/(1 + a)^4 + (2*b^3*Log[1 + a + b*x])/(1 + a)^4
Time = 0.29 (sec) , antiderivative size = 70, 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.
\(\displaystyle \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 6713 |
\(\displaystyle \int \frac {-a-b x+1}{x^4 (a+b x+1)}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {2 b^4}{(a+1)^4 (a+b x+1)}-\frac {2 b^3}{(a+1)^4 x}+\frac {2 b^2}{(a+1)^3 x^2}-\frac {2 b}{(a+1)^2 x^3}+\frac {1-a}{(a+1) x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 b^3 \log (x)}{(a+1)^4}+\frac {2 b^3 \log (a+b x+1)}{(a+1)^4}-\frac {2 b^2}{(a+1)^3 x}+\frac {b}{(a+1)^2 x^2}-\frac {1-a}{3 (a+1) x^3}\) |
Input:
Int[1/(E^(2*ArcTanh[a + b*x])*x^4),x]
Output:
-1/3*(1 - a)/((1 + a)*x^3) + b/((1 + a)^2*x^2) - (2*b^2)/((1 + a)^3*x) - ( 2*b^3*Log[x])/(1 + a)^4 + (2*b^3*Log[1 + a + b*x])/(1 + a)^4
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 = 69, normalized size of antiderivative = 0.99
method | result | size |
default | \(-\frac {1-a}{3 \left (a +1\right ) x^{3}}+\frac {b}{\left (a +1\right )^{2} x^{2}}-\frac {2 b^{2}}{\left (a +1\right )^{3} x}-\frac {2 b^{3} \ln \left (x \right )}{\left (a +1\right )^{4}}+\frac {2 b^{3} \ln \left (b x +a +1\right )}{\left (a +1\right )^{4}}\) | \(69\) |
parallelrisch | \(-\frac {1+6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a +1\right ) x^{3}+6 a \,b^{2} x^{2}+6 b^{2} x^{2}-3 a^{2} b x -a^{4}-6 a b x -2 a^{3}-3 b x +2 a}{3 \left (a^{4}+4 a^{3}+6 a^{2}+4 a +1\right ) x^{3}}\) | \(99\) |
risch | \(\frac {-\frac {2 b^{2} x^{2}}{a^{3}+3 a^{2}+3 a +1}+\frac {b x}{a^{2}+2 a +1}+\frac {-1+a}{3 a +3}}{x^{3}}-\frac {2 b^{3} \ln \left (x \right )}{a^{4}+4 a^{3}+6 a^{2}+4 a +1}+\frac {2 b^{3} \ln \left (-b x -a -1\right )}{a^{4}+4 a^{3}+6 a^{2}+4 a +1}\) | \(115\) |
norman | \(\frac {-\frac {2 b^{3} x^{3}}{a^{3}+3 a^{2}+3 a +1}-\frac {b^{2} x^{2}}{a^{2}+2 a +1}+\frac {b \left (2+a \right ) x}{3 a +3}-\frac {1}{3}+\frac {a}{3}}{\left (b x +a +1\right ) x^{3}}-\frac {2 b^{3} \ln \left (x \right )}{a^{4}+4 a^{3}+6 a^{2}+4 a +1}+\frac {2 b^{3} \ln \left (b x +a +1\right )}{a^{4}+4 a^{3}+6 a^{2}+4 a +1}\) | \(131\) |
Input:
int(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*(1-a)/(a+1)/x^3+b/(a+1)^2/x^2-2*b^2/(a+1)^3/x-2*b^3*ln(x)/(a+1)^4+2*b ^3*ln(b*x+a+1)/(a+1)^4
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {6 \, b^{3} x^{3} \log \left (b x + a + 1\right ) - 6 \, b^{3} x^{3} \log \left (x\right ) - 6 \, {\left (a + 1\right )} b^{2} x^{2} + a^{4} + 2 \, a^{3} + 3 \, {\left (a^{2} + 2 \, a + 1\right )} b x - 2 \, a - 1}{3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} x^{3}} \] Input:
integrate(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x, algorithm="fricas")
Output:
1/3*(6*b^3*x^3*log(b*x + a + 1) - 6*b^3*x^3*log(x) - 6*(a + 1)*b^2*x^2 + a ^4 + 2*a^3 + 3*(a^2 + 2*a + 1)*b*x - 2*a - 1)/((a^4 + 4*a^3 + 6*a^2 + 4*a + 1)*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (66) = 132\).
Time = 0.42 (sec) , antiderivative size = 262, normalized size of antiderivative = 3.74 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=- \frac {2 b^{3} \log {\left (x + \frac {- \frac {2 a^{5} b^{3}}{\left (a + 1\right )^{4}} - \frac {10 a^{4} b^{3}}{\left (a + 1\right )^{4}} - \frac {20 a^{3} b^{3}}{\left (a + 1\right )^{4}} - \frac {20 a^{2} b^{3}}{\left (a + 1\right )^{4}} + 2 a b^{3} - \frac {10 a b^{3}}{\left (a + 1\right )^{4}} + 2 b^{3} - \frac {2 b^{3}}{\left (a + 1\right )^{4}}}{4 b^{4}} \right )}}{\left (a + 1\right )^{4}} + \frac {2 b^{3} \log {\left (x + \frac {\frac {2 a^{5} b^{3}}{\left (a + 1\right )^{4}} + \frac {10 a^{4} b^{3}}{\left (a + 1\right )^{4}} + \frac {20 a^{3} b^{3}}{\left (a + 1\right )^{4}} + \frac {20 a^{2} b^{3}}{\left (a + 1\right )^{4}} + 2 a b^{3} + \frac {10 a b^{3}}{\left (a + 1\right )^{4}} + 2 b^{3} + \frac {2 b^{3}}{\left (a + 1\right )^{4}}}{4 b^{4}} \right )}}{\left (a + 1\right )^{4}} - \frac {- a^{3} - a^{2} + a + 6 b^{2} x^{2} + x \left (- 3 a b - 3 b\right ) + 1}{x^{3} \cdot \left (3 a^{3} + 9 a^{2} + 9 a + 3\right )} \] Input:
integrate(1/(b*x+a+1)**2*(1-(b*x+a)**2)/x**4,x)
Output:
-2*b**3*log(x + (-2*a**5*b**3/(a + 1)**4 - 10*a**4*b**3/(a + 1)**4 - 20*a* *3*b**3/(a + 1)**4 - 20*a**2*b**3/(a + 1)**4 + 2*a*b**3 - 10*a*b**3/(a + 1 )**4 + 2*b**3 - 2*b**3/(a + 1)**4)/(4*b**4))/(a + 1)**4 + 2*b**3*log(x + ( 2*a**5*b**3/(a + 1)**4 + 10*a**4*b**3/(a + 1)**4 + 20*a**3*b**3/(a + 1)**4 + 20*a**2*b**3/(a + 1)**4 + 2*a*b**3 + 10*a*b**3/(a + 1)**4 + 2*b**3 + 2* b**3/(a + 1)**4)/(4*b**4))/(a + 1)**4 - (-a**3 - a**2 + a + 6*b**2*x**2 + x*(-3*a*b - 3*b) + 1)/(x**3*(3*a**3 + 9*a**2 + 9*a + 3))
Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {2 \, b^{3} \log \left (b x + a + 1\right )}{a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1} - \frac {2 \, b^{3} \log \left (x\right )}{a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1} - \frac {6 \, b^{2} x^{2} - a^{3} - 3 \, {\left (a + 1\right )} b x - a^{2} + a + 1}{3 \, {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} x^{3}} \] Input:
integrate(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x, algorithm="maxima")
Output:
2*b^3*log(b*x + a + 1)/(a^4 + 4*a^3 + 6*a^2 + 4*a + 1) - 2*b^3*log(x)/(a^4 + 4*a^3 + 6*a^2 + 4*a + 1) - 1/3*(6*b^2*x^2 - a^3 - 3*(a + 1)*b*x - a^2 + a + 1)/((a^3 + 3*a^2 + 3*a + 1)*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (66) = 132\).
Time = 0.14 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.27 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=-\frac {2 \, b^{4} \log \left ({\left | -\frac {a}{b x + a + 1} - \frac {1}{b x + a + 1} + 1 \right |}\right )}{a^{4} b + 4 \, a^{3} b + 6 \, a^{2} b + 4 \, a b + b} - \frac {\frac {a b^{3} - 10 \, b^{3}}{a + 1} - \frac {3 \, {\left (a b^{4} - 8 \, b^{4}\right )}}{{\left (b x + a + 1\right )} b} + \frac {3 \, {\left (a^{2} b^{5} - 4 \, a b^{5} - 5 \, b^{5}\right )}}{{\left (b x + a + 1\right )}^{2} b^{2}}}{3 \, {\left (a + 1\right )}^{3} {\left (\frac {a}{b x + a + 1} + \frac {1}{b x + a + 1} - 1\right )}^{3}} \] Input:
integrate(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x, algorithm="giac")
Output:
-2*b^4*log(abs(-a/(b*x + a + 1) - 1/(b*x + a + 1) + 1))/(a^4*b + 4*a^3*b + 6*a^2*b + 4*a*b + b) - 1/3*((a*b^3 - 10*b^3)/(a + 1) - 3*(a*b^4 - 8*b^4)/ ((b*x + a + 1)*b) + 3*(a^2*b^5 - 4*a*b^5 - 5*b^5)/((b*x + a + 1)^2*b^2))/( (a + 1)^3*(a/(b*x + a + 1) + 1/(b*x + a + 1) - 1)^3)
Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {\frac {a-1}{3\,\left (a+1\right )}-\frac {2\,b^2\,x^2}{{\left (a+1\right )}^3}+\frac {b\,x}{{\left (a+1\right )}^2}}{x^3}+\frac {4\,b^3\,\mathrm {atanh}\left (\frac {a^4+4\,a^3+6\,a^2+4\,a+1}{{\left (a+1\right )}^4}+\frac {2\,b\,x}{a+1}\right )}{{\left (a+1\right )}^4} \] Input:
int(-((a + b*x)^2 - 1)/(x^4*(a + b*x + 1)^2),x)
Output:
((a - 1)/(3*(a + 1)) - (2*b^2*x^2)/(a + 1)^3 + (b*x)/(a + 1)^2)/x^3 + (4*b ^3*atanh((4*a + 6*a^2 + 4*a^3 + a^4 + 1)/(a + 1)^4 + (2*b*x)/(a + 1)))/(a + 1)^4
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {6 \,\mathrm {log}\left (b x +a +1\right ) b^{3} x^{3}-6 \,\mathrm {log}\left (x \right ) b^{3} x^{3}+a^{4}+2 a^{3}+3 a^{2} b x -6 a \,b^{2} x^{2}+6 a b x -2 a -6 b^{2} x^{2}+3 b x -1}{3 x^{3} \left (a^{4}+4 a^{3}+6 a^{2}+4 a +1\right )} \] Input:
int(1/(b*x+a+1)^2*(1-(b*x+a)^2)/x^4,x)
Output:
(6*log(a + b*x + 1)*b**3*x**3 - 6*log(x)*b**3*x**3 + a**4 + 2*a**3 + 3*a** 2*b*x - 6*a*b**2*x**2 + 6*a*b*x - 2*a - 6*b**2*x**2 + 3*b*x - 1)/(3*x**3*( a**4 + 4*a**3 + 6*a**2 + 4*a + 1))