Integrand size = 14, antiderivative size = 82 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=-\frac {1+a}{3 (1-a) x^3}-\frac {b}{(1-a)^2 x^2}-\frac {2 b^2}{(1-a)^3 x}+\frac {2 b^3 \log (x)}{(1-a)^4}-\frac {2 b^3 \log (1-a-b x)}{(1-a)^4} \] 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 = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {(-1+a) \left (1-a-a^2+a^3+3 b x-3 a b x+6 b^2 x^2\right )+6 b^3 x^3 \log (x)-6 b^3 x^3 \log (1-a-b x)}{3 (-1+a)^4 x^3} \] Input:
Integrate[E^(2*ArcTanh[a + b*x])/x^4,x]
Output:
((-1 + a)*(1 - a - a^2 + a^3 + 3*b*x - 3*a*b*x + 6*b^2*x^2) + 6*b^3*x^3*Lo g[x] - 6*b^3*x^3*Log[1 - a - b*x])/(3*(-1 + a)^4*x^3)
Time = 0.48 (sec) , antiderivative size = 82, 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 |
\(\displaystyle \int \left (-\frac {2 b^4}{(a-1)^4 (a+b x-1)}+\frac {2 b^3}{(a-1)^4 x}-\frac {2 b^2}{(a-1)^3 x^2}+\frac {2 b}{(a-1)^2 x^3}+\frac {-a-1}{(a-1) x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 b^3 \log (x)}{(1-a)^4}-\frac {2 b^3 \log (-a-b x+1)}{(1-a)^4}-\frac {2 b^2}{(1-a)^3 x}-\frac {b}{(1-a)^2 x^2}-\frac {a+1}{3 (1-a) x^3}\) |
Input:
Int[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 = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {-a -1}{3 x^{3} \left (-1+a \right )}-\frac {b}{\left (-1+a \right )^{2} x^{2}}+\frac {2 b^{2}}{\left (-1+a \right )^{3} x}+\frac {2 b^{3} \ln \left (x \right )}{\left (-1+a \right )^{4}}-\frac {2 b^{3} \ln \left (b x +a -1\right )}{\left (-1+a \right )^{4}}\) | \(70\) |
parallelrisch | \(\frac {-1+6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a -1\right ) x^{3}+6 a \,b^{2} x^{2}-6 b^{2} x^{2}-3 a^{2} b x +a^{4}+6 a b x -2 a^{3}-3 b x +2 a}{3 \left (a^{4}-4 a^{3}+6 a^{2}-4 a +1\right ) x^{3}}\) | \(97\) |
norman | \(\frac {\frac {a +1}{3 a -3}-\frac {b x}{a^{2}-2 a +1}+\frac {2 b^{2} x^{2}}{a^{3}-3 a^{2}+3 a -1}}{x^{3}}+\frac {2 b^{3} \ln \left (x \right )}{a^{4}-4 a^{3}+6 a^{2}-4 a +1}-\frac {2 b^{3} \ln \left (b x +a -1\right )}{a^{4}-4 a^{3}+6 a^{2}-4 a +1}\) | \(113\) |
risch | \(\frac {\frac {a +1}{3 a -3}-\frac {b x}{a^{2}-2 a +1}+\frac {2 b^{2} x^{2}}{a^{3}-3 a^{2}+3 a -1}}{x^{3}}+\frac {2 b^{3} \ln \left (-x \right )}{a^{4}-4 a^{3}+6 a^{2}-4 a +1}-\frac {2 b^{3} \ln \left (b x +a -1\right )}{a^{4}-4 a^{3}+6 a^{2}-4 a +1}\) | \(115\) |
Input:
int((b*x+a+1)^2/(1-(b*x+a)^2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*(-a-1)/x^3/(-1+a)-b/(-1+a)^2/x^2+2*b^2/(-1+a)^3/x+2*b^3/(-1+a)^4*ln(x )-2*b^3/(-1+a)^4*ln(b*x+a-1)
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=-\frac {6 \, b^{3} x^{3} \log \left (b x + a - 1\right ) - 6 \, b^{3} x^{3} \log \left (x\right ) - 6 \, {\left (a - 1\right )} b^{2} x^{2} - a^{4} + 2 \, a^{3} + 3 \, {\left (a^{2} - 2 \, a + 1\right )} b x - 2 \, a + 1}{3 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} x^{3}} \] Input:
integrate((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. 260 vs. \(2 (68) = 136\).
Time = 0.41 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.17 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {2 b^{3} \log {\left (x + \frac {- \frac {2 a^{5} b^{3}}{\left (a - 1\right )^{4}} + \frac {10 a^{4} b^{3}}{\left (a - 1\right )^{4}} - \frac {20 a^{3} b^{3}}{\left (a - 1\right )^{4}} + \frac {20 a^{2} b^{3}}{\left (a - 1\right )^{4}} + 2 a b^{3} - \frac {10 a b^{3}}{\left (a - 1\right )^{4}} - 2 b^{3} + \frac {2 b^{3}}{\left (a - 1\right )^{4}}}{4 b^{4}} \right )}}{\left (a - 1\right )^{4}} - \frac {2 b^{3} \log {\left (x + \frac {\frac {2 a^{5} b^{3}}{\left (a - 1\right )^{4}} - \frac {10 a^{4} b^{3}}{\left (a - 1\right )^{4}} + \frac {20 a^{3} b^{3}}{\left (a - 1\right )^{4}} - \frac {20 a^{2} b^{3}}{\left (a - 1\right )^{4}} + 2 a b^{3} + \frac {10 a b^{3}}{\left (a - 1\right )^{4}} - 2 b^{3} - \frac {2 b^{3}}{\left (a - 1\right )^{4}}}{4 b^{4}} \right )}}{\left (a - 1\right )^{4}} - \frac {- a^{3} + a^{2} + a - 6 b^{2} x^{2} + x \left (3 a b - 3 b\right ) - 1}{x^{3} \cdot \left (3 a^{3} - 9 a^{2} + 9 a - 3\right )} \] Input:
integrate((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 = 1.32 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=-\frac {2 \, b^{3} \log \left (b x + a - 1\right )}{a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1} + \frac {2 \, b^{3} \log \left (x\right )}{a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1} + \frac {6 \, b^{2} x^{2} + a^{3} - 3 \, {\left (a - 1\right )} b x - a^{2} - a + 1}{3 \, {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} x^{3}} \] Input:
integrate((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)
Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.46 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=-\frac {2 \, b^{4} \log \left ({\left | b x + a - 1 \right |}\right )}{a^{4} b - 4 \, a^{3} b + 6 \, a^{2} b - 4 \, a b + b} + \frac {2 \, b^{3} \log \left ({\left | x \right |}\right )}{a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1} + \frac {a^{4} - 2 \, a^{3} + 6 \, {\left (a b^{2} - b^{2}\right )} x^{2} - 3 \, {\left (a^{2} b - 2 \, a b + b\right )} x + 2 \, a - 1}{3 \, {\left (a - 1\right )}^{4} x^{3}} \] Input:
integrate((b*x+a+1)^2/(1-(b*x+a)^2)/x^4,x, algorithm="giac")
Output:
-2*b^4*log(abs(b*x + a - 1))/(a^4*b - 4*a^3*b + 6*a^2*b - 4*a*b + b) + 2*b ^3*log(abs(x))/(a^4 - 4*a^3 + 6*a^2 - 4*a + 1) + 1/3*(a^4 - 2*a^3 + 6*(a*b ^2 - b^2)*x^2 - 3*(a^2*b - 2*a*b + b)*x + 2*a - 1)/((a - 1)^4*x^3)
Time = 23.46 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {\frac {a+1}{3\,\left (a-1\right )}+\frac {2\,b^2\,x^2}{{\left (a-1\right )}^3}-\frac {b\,x}{{\left (a-1\right )}^2}}{x^3}-\frac {4\,b^3\,\mathrm {atanh}\left (\frac {a^4-4\,a^3+6\,a^2-4\,a+1}{{\left (a-1\right )}^4}+\frac {2\,b\,x}{a-1}\right )}{{\left (a-1\right )}^4} \] Input:
int(-(a + b*x + 1)^2/(x^4*((a + b*x)^2 - 1)),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((6*a^2 - 4*a - 4*a^3 + a^4 + 1)/(a - 1)^4 + (2*b*x)/(a - 1)))/(a - 1)^4
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 \text {arctanh}(a+b x)}}{x^4} \, dx=\frac {-6 \,\mathrm {log}\left (b x +a -1\right ) b^{3} x^{3}+6 \,\mathrm {log}\left (x \right ) b^{3} x^{3}+a^{4}-2 a^{3}-3 a^{2} b x +6 a \,b^{2} x^{2}+6 a b x +2 a -6 b^{2} x^{2}-3 b x -1}{3 x^{3} \left (a^{4}-4 a^{3}+6 a^{2}-4 a +1\right )} \] Input:
int((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))