Integrand size = 12, antiderivative size = 200 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=-\frac {a^3}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {a^3}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {a^3}{2 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^4}+\frac {a^3}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}-\frac {a^3}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {a^3}{2 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {1}{4} a^3 \text {arctanh}\left (\sqrt {\frac {1-a x}{1+a x}}\right ) \] Output:
-1/4*a^3/(1-((-a*x+1)/(a*x+1))^(1/2))^2+a^3/(4-4*((-a*x+1)/(a*x+1))^(1/2)) -1/2*a^3/(1+((-a*x+1)/(a*x+1))^(1/2))^4+a^3/(1+((-a*x+1)/(a*x+1))^(1/2))^3 -a^3/(1+((-a*x+1)/(a*x+1))^(1/2))^2+a^3/(2+2*((-a*x+1)/(a*x+1))^(1/2))-1/4 *a^3*arctanh(((-a*x+1)/(a*x+1))^(1/2))
Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=-\frac {2+\sqrt {\frac {1-a x}{1+a x}} \left (-2-2 a x+a^2 x^2+a^3 x^3\right )-a^4 x^4 \log (x)+a^4 x^4 \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a x^4} \] Input:
Integrate[1/(E^ArcSech[a*x]*x^4),x]
Output:
-1/8*(2 + Sqrt[(1 - a*x)/(1 + a*x)]*(-2 - 2*a*x + a^2*x^2 + a^3*x^3) - a^4 *x^4*Log[x] + a^4*x^4*Log[1 + Sqrt[(1 - a*x)/(1 + a*x)] + a*x*Sqrt[(1 - a* x)/(1 + a*x)]])/(a*x^4)
Time = 1.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6891, 7268, 25, 27, 2115, 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.
| \(\displaystyle \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx\) |
| \(\Big \downarrow \) 6891 |
| \(\displaystyle \int \frac {1}{x^4 \left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )}dx\) |
| \(\Big \downarrow \) 7268 |
| \(\displaystyle 4 a \int -\frac {a^2 \sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^2}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}d\sqrt {\frac {1-a x}{a x+1}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 a \int \frac {a^2 \sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^2}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}d\sqrt {\frac {1-a x}{a x+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -4 a^3 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^2}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}d\sqrt {\frac {1-a x}{a x+1}}\) |
| \(\Big \downarrow \) 2115 |
| \(\displaystyle -4 a^3 \int \left (-\frac {1}{16 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^2}+\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}-\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^3}-\frac {1}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}+\frac {3}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}-\frac {1}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}-\frac {1}{16 \left (\frac {1-a x}{a x+1}-1\right )}\right )d\sqrt {\frac {1-a x}{a x+1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 a^3 \left (\frac {1}{16} \text {arctanh}\left (\sqrt {\frac {1-a x}{a x+1}}\right )-\frac {1}{16 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}+\frac {1}{16 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}+\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}-\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}+\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}\right )\) |
Input:
Int[1/(E^ArcSech[a*x]*x^4),x]
Output:
-4*a^3*(1/(16*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^2) - 1/(16*(1 - Sqrt[(1 - a* x)/(1 + a*x)])) + 1/(8*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^4) - 1/(4*(1 + Sqrt [(1 - a*x)/(1 + a*x)])^3) + 1/(4*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^2) - 1/(8 *(1 + Sqrt[(1 - a*x)/(1 + a*x)])) + ArcTanh[Sqrt[(1 - a*x)/(1 + a*x)]]/16)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f _.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^ n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && IntegersQ[m, n]
Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1/u)*Sqrt[(1 - u)/(1 + u)])^n, x] /; FreeQ[m, x] && Integer Q[n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfQuotientOfLinears [u, x]}, Simp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/ls t[[2]])], x] /; !FalseQ[lst]]
Time = 0.47 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.58
| method | result | size |
| default | \(a \left (-\frac {1}{4 a^{2} x^{4}}-\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) a^{4} x^{4}+\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\right )}{8 a \,x^{3} \sqrt {-a^{2} x^{2}+1}}\right )\) | \(115\) |
Input:
int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^4,x,method=_RETURNVERBOSE )
Output:
a*(-1/4/a^2/x^4-1/8/a*(-(a*x-1)/a/x)^(1/2)/x^3*((a*x+1)/a/x)^(1/2)*(arctan h(1/(-a^2*x^2+1)^(1/2))*a^4*x^4+(-a^2*x^2+1)^(1/2)*a^2*x^2-2*(-a^2*x^2+1)^ (1/2))/(-a^2*x^2+1)^(1/2))
Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.69 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=-\frac {a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) + 2 \, {\left (a^{3} x^{3} - 2 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 4}{16 \, a x^{4}} \] Input:
integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^4,x, algorithm="fri cas")
Output:
-1/16*(a^4*x^4*log(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 1) - a^4*x^4*log(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 1) + 2*(a^ 3*x^3 - 2*a*x)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 4)/(a*x^4)
\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=a \int \frac {1}{a x^{4} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + x^{3}}\, dx \] Input:
integrate(1/(1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))/x**4,x)
Output:
a*Integral(1/(a*x**4*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + x**3), x)
\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=\int { \frac {1}{x^{4} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:
integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^4,x, algorithm="max ima")
Output:
integrate(1/(x^4*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=\int { \frac {1}{x^{4} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:
integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^4,x, algorithm="gia c")
Output:
integrate(1/(x^4*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
Time = 87.93 (sec) , antiderivative size = 1511, normalized size of antiderivative = 7.56 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=\text {Too large to display} \] Input:
int(1/(x^4*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))),x)
Output:
((a^3*((1/(a*x) - 1)^(1/2) - 1i)^4*192i)/((1/(a*x) + 1)^(1/2) - 1)^4 + (a^ 3*((1/(a*x) - 1)^(1/2) - 1i)^6*128i)/((1/(a*x) + 1)^(1/2) - 1)^6 + (a^3*(( 1/(a*x) - 1)^(1/2) - 1i)^8*192i)/((1/(a*x) + 1)^(1/2) - 1)^8)/(3*((15*((1/ (a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^(1/2) - 1)^4 - (6*((1/(a*x) - 1)^ (1/2) - 1i)^2)/((1/(a*x) + 1)^(1/2) - 1)^2 - (20*((1/(a*x) - 1)^(1/2) - 1i )^6)/((1/(a*x) + 1)^(1/2) - 1)^6 + (15*((1/(a*x) - 1)^(1/2) - 1i)^8)/((1/( a*x) + 1)^(1/2) - 1)^8 - (6*((1/(a*x) - 1)^(1/2) - 1i)^10)/((1/(a*x) + 1)^ (1/2) - 1)^10 + ((1/(a*x) - 1)^(1/2) - 1i)^12/((1/(a*x) + 1)^(1/2) - 1)^12 + 1)) - ((a^3*((1/(a*x) - 1)^(1/2) - 1i)^4*64i)/((1/(a*x) + 1)^(1/2) - 1) ^4 + (a^3*((1/(a*x) - 1)^(1/2) - 1i)^6*128i)/(3*((1/(a*x) + 1)^(1/2) - 1)^ 6) + (a^3*((1/(a*x) - 1)^(1/2) - 1i)^8*64i)/((1/(a*x) + 1)^(1/2) - 1)^8)/( (15*((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^(1/2) - 1)^4 - (6*((1/(a* x) - 1)^(1/2) - 1i)^2)/((1/(a*x) + 1)^(1/2) - 1)^2 - (20*((1/(a*x) - 1)^(1 /2) - 1i)^6)/((1/(a*x) + 1)^(1/2) - 1)^6 + (15*((1/(a*x) - 1)^(1/2) - 1i)^ 8)/((1/(a*x) + 1)^(1/2) - 1)^8 - (6*((1/(a*x) - 1)^(1/2) - 1i)^10)/((1/(a* x) + 1)^(1/2) - 1)^10 + ((1/(a*x) - 1)^(1/2) - 1i)^12/((1/(a*x) + 1)^(1/2) - 1)^12 + 1) - (a^3*atanh(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1)))/2 + ((14*a^3*((1/(a*x) - 1)^(1/2) - 1i)^3)/((1/(a*x) + 1)^(1/2) - 1)^3 + (14*a^3*((1/(a*x) - 1)^(1/2) - 1i)^5)/((1/(a*x) + 1)^(1/2) - 1)^5 + (2*a^3*((1/(a*x) - 1)^(1/2) - 1i)^7)/((1/(a*x) + 1)^(1/2) - 1)^7 + (2*...
Time = 0.16 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.78 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=\frac {a \left (-\sqrt {a x +1}\, \sqrt {-a x +1}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a^{2} x^{2}+\sqrt {a x +1}\, \sqrt {-a x +1}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a^{2} x^{2}-\sqrt {a x +1}\, \sqrt {-a x +1}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a^{2} x^{2}+\sqrt {a x +1}\, \sqrt {-a x +1}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a^{2} x^{2}+\sqrt {a x +1}\, \sqrt {-a x +1}\, a^{2} x^{2}-\sqrt {a x +1}\, \sqrt {-a x +1}-\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a^{2} x^{2}+\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a^{2} x^{2}-\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a^{2} x^{2}+\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a^{2} x^{2}+2 a^{2} x^{2}-3\right )}{8 x^{2} \left (\sqrt {a x +1}\, \sqrt {-a x +1}+1\right )} \] Input:
int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^4,x)
Output:
(a*( - sqrt(a*x + 1)*sqrt( - a*x + 1)*log( - sqrt(2) + tan(asin(sqrt( - a* x + 1)/sqrt(2))/2) - 1)*a**2*x**2 + sqrt(a*x + 1)*sqrt( - a*x + 1)*log( - sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) + 1)*a**2*x**2 - sqrt(a*x + 1)*sqrt( - a*x + 1)*log(sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) - 1)*a**2*x**2 + sqrt(a*x + 1)*sqrt( - a*x + 1)*log(sqrt(2) + tan(asin(sqr t( - a*x + 1)/sqrt(2))/2) + 1)*a**2*x**2 + sqrt(a*x + 1)*sqrt( - a*x + 1)* a**2*x**2 - sqrt(a*x + 1)*sqrt( - a*x + 1) - log( - sqrt(2) + tan(asin(sqr t( - a*x + 1)/sqrt(2))/2) - 1)*a**2*x**2 + log( - sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) + 1)*a**2*x**2 - log(sqrt(2) + tan(asin(sqrt( - a* x + 1)/sqrt(2))/2) - 1)*a**2*x**2 + log(sqrt(2) + tan(asin(sqrt( - a*x + 1 )/sqrt(2))/2) + 1)*a**2*x**2 + 2*a**2*x**2 - 3))/(8*x**2*(sqrt(a*x + 1)*sq rt( - a*x + 1) + 1))