### 3.35 $$\int \frac{e^{\text{csch}^{-1}(a x)}}{x^4} \, dx$$

Optimal. Leaf size=65 $-\frac{a^2 \sqrt{\frac{1}{a^2 x^2}+1}}{8 x}-\frac{\sqrt{\frac{1}{a^2 x^2}+1}}{4 x^3}+\frac{1}{8} a^3 \text{csch}^{-1}(a x)-\frac{1}{4 a x^4}$

[Out]

-1/(4*a*x^4) - Sqrt[1 + 1/(a^2*x^2)]/(4*x^3) - (a^2*Sqrt[1 + 1/(a^2*x^2)])/(8*x) + (a^3*ArcCsch[a*x])/8

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Rubi [A]  time = 0.0434752, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {6336, 30, 335, 279, 321, 215} $-\frac{a^2 \sqrt{\frac{1}{a^2 x^2}+1}}{8 x}-\frac{\sqrt{\frac{1}{a^2 x^2}+1}}{4 x^3}+\frac{1}{8} a^3 \text{csch}^{-1}(a x)-\frac{1}{4 a x^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCsch[a*x]/x^4,x]

[Out]

-1/(4*a*x^4) - Sqrt[1 + 1/(a^2*x^2)]/(4*x^3) - (a^2*Sqrt[1 + 1/(a^2*x^2)])/(8*x) + (a^3*ArcCsch[a*x])/8

Rule 6336

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Dist[1/a, Int[x^(m - p), x], x] + Int[x^m*Sqrt[1 + 1/
(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0490127, size = 53, normalized size = 0.82 $\frac{-a x \sqrt{\frac{1}{a^2 x^2}+1} \left (a^2 x^2+2\right )+a^4 x^4 \sinh ^{-1}\left (\frac{1}{a x}\right )-2}{8 a x^4}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCsch[a*x]/x^4,x]

[Out]

(-2 - a*Sqrt[1 + 1/(a^2*x^2)]*x*(2 + a^2*x^2) + a^4*x^4*ArcSinh[1/(a*x)])/(8*a*x^4)

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Maple [B]  time = 0.183, size = 173, normalized size = 2.7 \begin{align*}{\frac{{a}^{2}}{8\,{x}^{3}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}} \left ( \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{{\frac{3}{2}}}\sqrt{{a}^{-2}}{x}^{2}{a}^{2}-\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}\sqrt{{a}^{-2}}{x}^{4}{a}^{2}+\ln \left ( 2\,{\frac{1}{{a}^{2}x} \left ( \sqrt{{a}^{-2}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}{a}^{2}+1 \right ) } \right ){x}^{4}-2\, \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{3/2}\sqrt{{a}^{-2}} \right ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}}}{\frac{1}{\sqrt{{a}^{-2}}}}}-{\frac{1}{4\,{x}^{4}a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1+1/a^2/x^2)^(1/2))/x^4,x)

[Out]

1/8*((a^2*x^2+1)/a^2/x^2)^(1/2)/x^3*a^2*(((a^2*x^2+1)/a^2)^(3/2)*(1/a^2)^(1/2)*x^2*a^2-((a^2*x^2+1)/a^2)^(1/2)
*(1/a^2)^(1/2)*x^4*a^2+ln(2*((1/a^2)^(1/2)*((a^2*x^2+1)/a^2)^(1/2)*a^2+1)/x/a^2)*x^4-2*((a^2*x^2+1)/a^2)^(3/2)
*(1/a^2)^(1/2))/((a^2*x^2+1)/a^2)^(1/2)/(1/a^2)^(1/2)-1/4/x^4/a

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Maxima [B]  time = 1.02206, size = 174, normalized size = 2.68 \begin{align*} \frac{1}{16} \, a^{3} \log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} + 1\right ) - \frac{1}{16} \, a^{3} \log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} - 1\right ) - \frac{a^{6} x^{3}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + a^{4} x \sqrt{\frac{1}{a^{2} x^{2}} + 1}}{8 \,{\left (a^{4} x^{4}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{2} - 2 \, a^{2} x^{2}{\left (\frac{1}{a^{2} x^{2}} + 1\right )} + 1\right )}} - \frac{1}{4 \, a x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a^3*log(a*x*sqrt(1/(a^2*x^2) + 1) + 1) - 1/16*a^3*log(a*x*sqrt(1/(a^2*x^2) + 1) - 1) - 1/8*(a^6*x^3*(1/(a
^2*x^2) + 1)^(3/2) + a^4*x*sqrt(1/(a^2*x^2) + 1))/(a^4*x^4*(1/(a^2*x^2) + 1)^2 - 2*a^2*x^2*(1/(a^2*x^2) + 1) +
1) - 1/4/(a*x^4)

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Fricas [B]  time = 2.58623, size = 250, normalized size = 3.85 \begin{align*} \frac{a^{4} x^{4} \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x - 1\right ) -{\left (a^{3} x^{3} + 2 \, a x\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - 2}{8 \, a x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^4*x^4*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - a*x + 1) - a^4*x^4*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2))
- a*x - 1) - (a^3*x^3 + 2*a*x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - 2)/(a*x^4)

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Sympy [A]  time = 5.32647, size = 83, normalized size = 1.28 \begin{align*} \frac{a^{3} \operatorname{asinh}{\left (\frac{1}{a x} \right )}}{8} - \frac{a^{2}}{8 x \sqrt{1 + \frac{1}{a^{2} x^{2}}}} - \frac{3}{8 x^{3} \sqrt{1 + \frac{1}{a^{2} x^{2}}}} - \frac{1}{4 a x^{4}} - \frac{1}{4 a^{2} x^{5} \sqrt{1 + \frac{1}{a^{2} x^{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a**2/x**2)**(1/2))/x**4,x)

[Out]

a**3*asinh(1/(a*x))/8 - a**2/(8*x*sqrt(1 + 1/(a**2*x**2))) - 3/(8*x**3*sqrt(1 + 1/(a**2*x**2))) - 1/(4*a*x**4)
- 1/(4*a**2*x**5*sqrt(1 + 1/(a**2*x**2)))

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Giac [A]  time = 1.15207, size = 139, normalized size = 2.14 \begin{align*} \frac{a^{6}{\left | a \right |} \log \left (\sqrt{a^{2} x^{2} + 1} + 1\right ) \mathrm{sgn}\left (x\right ) - a^{6}{\left | a \right |} \log \left (\sqrt{a^{2} x^{2} + 1} - 1\right ) \mathrm{sgn}\left (x\right ) - \frac{2 \,{\left ({\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{6}{\left | a \right |} \mathrm{sgn}\left (x\right ) + \sqrt{a^{2} x^{2} + 1} a^{6}{\left | a \right |} \mathrm{sgn}\left (x\right ) + 2 \, a^{7}\right )}}{a^{4} x^{4}}}{16 \, a^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(a^6*abs(a)*log(sqrt(a^2*x^2 + 1) + 1)*sgn(x) - a^6*abs(a)*log(sqrt(a^2*x^2 + 1) - 1)*sgn(x) - 2*((a^2*x^
2 + 1)^(3/2)*a^6*abs(a)*sgn(x) + sqrt(a^2*x^2 + 1)*a^6*abs(a)*sgn(x) + 2*a^7)/(a^4*x^4))/a^4