3.1.0 ∫ e−sech−1(ax) ___________________

Integrand size = 12, antiderivative size = 200 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=-\frac {a^3}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {a^3}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {a^3}{2 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^4}+\frac {a^3}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}-\frac {a^3}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {a^3}{2 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {1}{4} a^3 \text {arctanh}\left (\sqrt {\frac {1-a x}{1+a x}}\right ) \]

-1/4*a^3*arctanh(((-a*x+1)/(a*x+1))^(1/2))-1/4*a^3/(1-((-a*x+1)/(a*x+1))^(1/2))^2+1/4*a^3/(1-((-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))^(

Rubi [A] (veriﬁed)

Time = 0.36 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6472, 1626, 213} \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=-\frac {1}{4} a^3 \text {arctanh}\left (\sqrt {\frac {1-a x}{a x+1}}\right )+\frac {a^3}{4 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}+\frac {a^3}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}-\frac {a^3}{4 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}-\frac {a^3}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}+\frac {a^3}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}-\frac {a^3}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4} \]

-1/4*a^3/(1 - Sqrt[(1 - a*x)/(1 + a*x)])^2 + a^3/(4*(1 - Sqrt[(1 - a*x)/(1 + a*x)])) - a^3/(2*(1 + Sqrt[(1 - a

*x)/(1 + a*x)])^4) + a^3/(1 + Sqrt[(1 - a*x)/(1 + a*x)])^3 - a^3/(1 + Sqrt[(1 - a*x)/(1 + a*x)])^2 + a^3/(2*(1

+ Sqrt[(1 - a*x)/(1 + a*x)])) - (a^3*ArcTanh[Sqrt[(1 - a*x)/(1 + a*x)]])/4

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E

xpandIntegrand[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] && Poly

Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1/u)*Sqrt[(1 - u)/(

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^4 \left (\frac {1}{a x}+\sqrt {\frac {1-a x}{1+a x}}+\frac {\sqrt {\frac {1-a x}{1+a x}}}{a x}\right )} \, dx \\ & = (4 a) \text {Subst}\left (\int \frac {x \left (a+a x^2\right )^2}{(-1+x)^3 (1+x)^5} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = (4 a) \text {Subst}\left (\int \left (\frac {a^2}{8 (-1+x)^3}+\frac {a^2}{16 (-1+x)^2}+\frac {a^2}{2 (1+x)^5}-\frac {3 a^2}{4 (1+x)^4}+\frac {a^2}{2 (1+x)^3}-\frac {a^2}{8 (1+x)^2}+\frac {a^2}{16 \left (-1+x^2\right )}\right ) \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = -\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 {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right ) \\ & = -\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 ) \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=-\frac {2+\sqrt {\frac {1-a x}{1+a x}} \left (-2-2 a x+a^2 x^2+a^3 x^3\right )-a^4 x^4 \log (x)+a^4 x^4 \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )}{8 a x^4} \]

-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[(

Maple [A] (veriﬁed)

Time = 0.93 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.58


method	result	size

default	\(a \left (-\frac {1}{4 a^{2} x^{4}}-\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) a^{4} x^{4}+a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}-2 \sqrt {-a^{2} x^{2}+1}\right )}{8 a \,x^{3} \sqrt {-a^{2} x^{2}+1}}\right )\)	\(115\)

int(1/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))/x^4,x,method=_RETURNVERBOSE)

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)*(arctanh(1/(-a^2*x^2+1)^(1/2))*a^4*x^4+a^2*

Fricas [A] (veriﬁcation not implemented)

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.69 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=-\frac {a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) + 2 \, {\left (a^{3} x^{3} - 2 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 4}{16 \, a x^{4}} \]

integrate(1/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))/x^4,x, algorithm="fricas")

-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)

Sympy [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=a \int \frac {1}{a x^{4} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + x^{3}}\, dx \]

a*Integral(1/(a*x**4*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + x**3), x)

Maxima [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=\int { \frac {1}{x^{4} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \]

integrate(1/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))/x^4,x, algorithm="maxima")

integrate(1/(x^4*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)

Giac [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=\int { \frac {1}{x^{4} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \]

integrate(1/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))/x^4,x, algorithm="giac")

integrate(1/(x^4*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)

Mupad [B] (veriﬁcation not implemented)

Time = 50.94 (sec) , antiderivative size = 1511, normalized size of antiderivative = 7.56 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx=\text {Too large to display} \]

((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)^1

0)/((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*a^3*((1/(a*x) - 1)^(1/2) - 1i))/((1/(a*x) + 1)^(1/2) - 1))/((6*

((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^(1/2) - 1)^4 - (4*((1/(a*x) - 1)^(1/2) - 1i)^2)/((1/(a*x) + 1)^(1

/2) - 1)^2 - (4*((1/(a*x) - 1)^(1/2) - 1i)^6)/((1/(a*x) + 1)^(1/2) - 1)^6 + ((1/(a*x) - 1)^(1/2) - 1i)^8/((1/(

a*x) + 1)^(1/2) - 1)^8 + 1) + ((23*a^3*((1/(a*x) - 1)^(1/2) - 1i)^3)/(2*((1/(a*x) + 1)^(1/2) - 1)^3) + (333*a^

3*((1/(a*x) - 1)^(1/2) - 1i)^5)/(2*((1/(a*x) + 1)^(1/2) - 1)^5) + (671*a^3*((1/(a*x) - 1)^(1/2) - 1i)^7)/(2*((

1/(a*x) + 1)^(1/2) - 1)^7) + (671*a^3*((1/(a*x) - 1)^(1/2) - 1i)^9)/(2*((1/(a*x) + 1)^(1/2) - 1)^9) + (333*a^3

*((1/(a*x) - 1)^(1/2) - 1i)^11)/(2*((1/(a*x) + 1)^(1/2) - 1)^11) + (23*a^3*((1/(a*x) - 1)^(1/2) - 1i)^13)/(2*(

(1/(a*x) + 1)^(1/2) - 1)^13) - (3*a^3*((1/(a*x) - 1)^(1/2) - 1i)^15)/(2*((1/(a*x) + 1)^(1/2) - 1)^15) - (3*a^3

*((1/(a*x) - 1)^(1/2) - 1i))/(2*((1/(a*x) + 1)^(1/2) - 1)))/((28*((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^

(1/2) - 1)^4 - (8*((1/(a*x) - 1)^(1/2) - 1i)^2)/((1/(a*x) + 1)^(1/2) - 1)^2 - (56*((1/(a*x) - 1)^(1/2) - 1i)^6

)/((1/(a*x) + 1)^(1/2) - 1)^6 + (70*((1/(a*x) - 1)^(1/2) - 1i)^8)/((1/(a*x) + 1)^(1/2) - 1)^8 - (56*((1/(a*x)

- 1)^(1/2) - 1i)^10)/((1/(a*x) + 1)^(1/2) - 1)^10 + (28*((1/(a*x) - 1)^(1/2) - 1i)^12)/((1/(a*x) + 1)^(1/2) -

1)^12 - (8*((1/(a*x) - 1)^(1/2) - 1i)^14)/((1/(a*x) + 1)^(1/2) - 1)^14 + ((1/(a*x) - 1)^(1/2) - 1i)^16/((1/(a*

\(\int \frac {e^{-\text {sech}^{-1}(a x)}}{x^4} \, dx\) [84]

Optimal result