### 3.21 $$\int \frac{\text{csch}^{-1}(\sqrt{x})}{x^4} \, dx$$

Optimal. Leaf size=115 $-\frac{5 \sqrt{-x-1}}{72 \sqrt{-x} x^{3/2}}+\frac{\sqrt{-x-1}}{18 \sqrt{-x} x^{5/2}}-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{5 \sqrt{-x-1}}{48 \sqrt{-x} \sqrt{x}}-\frac{5 \sqrt{x} \tan ^{-1}\left (\sqrt{-x-1}\right )}{48 \sqrt{-x}}$

[Out]

Sqrt[-1 - x]/(18*Sqrt[-x]*x^(5/2)) - (5*Sqrt[-1 - x])/(72*Sqrt[-x]*x^(3/2)) + (5*Sqrt[-1 - x])/(48*Sqrt[-x]*Sq
rt[x]) - ArcCsch[Sqrt[x]]/(3*x^3) - (5*Sqrt[x]*ArcTan[Sqrt[-1 - x]])/(48*Sqrt[-x])

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Rubi [A]  time = 0.0343055, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {6346, 12, 51, 63, 204} $-\frac{5 \sqrt{-x-1}}{72 \sqrt{-x} x^{3/2}}+\frac{\sqrt{-x-1}}{18 \sqrt{-x} x^{5/2}}-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{5 \sqrt{-x-1}}{48 \sqrt{-x} \sqrt{x}}-\frac{5 \sqrt{x} \tan ^{-1}\left (\sqrt{-x-1}\right )}{48 \sqrt{-x}}$

Antiderivative was successfully veriﬁed.

[In]

Int[ArcCsch[Sqrt[x]]/x^4,x]

[Out]

Sqrt[-1 - x]/(18*Sqrt[-x]*x^(5/2)) - (5*Sqrt[-1 - x])/(72*Sqrt[-x]*x^(3/2)) + (5*Sqrt[-1 - x])/(48*Sqrt[-x]*Sq
rt[x]) - ArcCsch[Sqrt[x]]/(3*x^3) - (5*Sqrt[x]*ArcTan[Sqrt[-1 - x]])/(48*Sqrt[-x])

Rule 6346

Int[((a_.) + ArcCsch[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcCsc
h[u]))/(d*(m + 1)), x] - Dist[(b*u)/(d*(m + 1)*Sqrt[-u^2]), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/
(u*Sqrt[-1 - u^2]), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !F
unctionOfQ[(c + d*x)^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{x^4} \, dx &=-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{\sqrt{x} \int \frac{1}{2 \sqrt{-1-x} x^4} \, dx}{3 \sqrt{-x}}\\ &=-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{\sqrt{x} \int \frac{1}{\sqrt{-1-x} x^4} \, dx}{6 \sqrt{-x}}\\ &=\frac{\sqrt{-1-x}}{18 \sqrt{-x} x^{5/2}}-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}-\frac{\left (5 \sqrt{x}\right ) \int \frac{1}{\sqrt{-1-x} x^3} \, dx}{36 \sqrt{-x}}\\ &=\frac{\sqrt{-1-x}}{18 \sqrt{-x} x^{5/2}}-\frac{5 \sqrt{-1-x}}{72 \sqrt{-x} x^{3/2}}-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{\left (5 \sqrt{x}\right ) \int \frac{1}{\sqrt{-1-x} x^2} \, dx}{48 \sqrt{-x}}\\ &=\frac{\sqrt{-1-x}}{18 \sqrt{-x} x^{5/2}}-\frac{5 \sqrt{-1-x}}{72 \sqrt{-x} x^{3/2}}+\frac{5 \sqrt{-1-x}}{48 \sqrt{-x} \sqrt{x}}-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}-\frac{\left (5 \sqrt{x}\right ) \int \frac{1}{\sqrt{-1-x} x} \, dx}{96 \sqrt{-x}}\\ &=\frac{\sqrt{-1-x}}{18 \sqrt{-x} x^{5/2}}-\frac{5 \sqrt{-1-x}}{72 \sqrt{-x} x^{3/2}}+\frac{5 \sqrt{-1-x}}{48 \sqrt{-x} \sqrt{x}}-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{\left (5 \sqrt{x}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{-1-x}\right )}{48 \sqrt{-x}}\\ &=\frac{\sqrt{-1-x}}{18 \sqrt{-x} x^{5/2}}-\frac{5 \sqrt{-1-x}}{72 \sqrt{-x} x^{3/2}}+\frac{5 \sqrt{-1-x}}{48 \sqrt{-x} \sqrt{x}}-\frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{3 x^3}-\frac{5 \sqrt{x} \tan ^{-1}\left (\sqrt{-1-x}\right )}{48 \sqrt{-x}}\\ \end{align*}

Mathematica [A]  time = 0.0446572, size = 52, normalized size = 0.45 $\frac{\sqrt{\frac{1}{x}+1} \left (15 x^2-10 x+8\right ) \sqrt{x}-15 x^3 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )-48 \text{csch}^{-1}\left (\sqrt{x}\right )}{144 x^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[ArcCsch[Sqrt[x]]/x^4,x]

[Out]

(Sqrt[1 + x^(-1)]*Sqrt[x]*(8 - 10*x + 15*x^2) - 48*ArcCsch[Sqrt[x]] - 15*x^3*ArcSinh[1/Sqrt[x]])/(144*x^3)

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Maple [A]  time = 0.117, size = 67, normalized size = 0.6 \begin{align*} -{\frac{1}{3\,{x}^{3}}{\rm arccsch} \left (\sqrt{x}\right )}-{\frac{1}{144}\sqrt{1+x} \left ( 15\,{\it Artanh} \left ({\frac{1}{\sqrt{1+x}}} \right ){x}^{3}-15\,{x}^{2}\sqrt{1+x}+10\,x\sqrt{1+x}-8\,\sqrt{1+x} \right ){\frac{1}{\sqrt{{\frac{1+x}{x}}}}}{x}^{-{\frac{7}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(arccsch(x^(1/2))/x^4,x)

[Out]

-1/3*arccsch(x^(1/2))/x^3-1/144*(1+x)^(1/2)*(15*arctanh(1/(1+x)^(1/2))*x^3-15*x^2*(1+x)^(1/2)+10*x*(1+x)^(1/2)
-8*(1+x)^(1/2))/((1+x)/x)^(1/2)/x^(7/2)

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Maxima [A]  time = 0.991373, size = 157, normalized size = 1.37 \begin{align*} \frac{15 \, x^{\frac{5}{2}}{\left (\frac{1}{x} + 1\right )}^{\frac{5}{2}} - 40 \, x^{\frac{3}{2}}{\left (\frac{1}{x} + 1\right )}^{\frac{3}{2}} + 33 \, \sqrt{x} \sqrt{\frac{1}{x} + 1}}{144 \,{\left (x^{3}{\left (\frac{1}{x} + 1\right )}^{3} - 3 \, x^{2}{\left (\frac{1}{x} + 1\right )}^{2} + 3 \, x{\left (\frac{1}{x} + 1\right )} - 1\right )}} - \frac{\operatorname{arcsch}\left (\sqrt{x}\right )}{3 \, x^{3}} - \frac{5}{96} \, \log \left (\sqrt{x} \sqrt{\frac{1}{x} + 1} + 1\right ) + \frac{5}{96} \, \log \left (\sqrt{x} \sqrt{\frac{1}{x} + 1} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(15*x^(5/2)*(1/x + 1)^(5/2) - 40*x^(3/2)*(1/x + 1)^(3/2) + 33*sqrt(x)*sqrt(1/x + 1))/(x^3*(1/x + 1)^3 -
3*x^2*(1/x + 1)^2 + 3*x*(1/x + 1) - 1) - 1/3*arccsch(sqrt(x))/x^3 - 5/96*log(sqrt(x)*sqrt(1/x + 1) + 1) + 5/96
*log(sqrt(x)*sqrt(1/x + 1) - 1)

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Fricas [A]  time = 2.58923, size = 150, normalized size = 1.3 \begin{align*} \frac{{\left (15 \, x^{2} - 10 \, x + 8\right )} \sqrt{x} \sqrt{\frac{x + 1}{x}} - 3 \,{\left (5 \, x^{3} + 16\right )} \log \left (\frac{x \sqrt{\frac{x + 1}{x}} + \sqrt{x}}{x}\right )}{144 \, x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*((15*x^2 - 10*x + 8)*sqrt(x)*sqrt((x + 1)/x) - 3*(5*x^3 + 16)*log((x*sqrt((x + 1)/x) + sqrt(x))/x))/x^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(x**(1/2))/x**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsch}\left (\sqrt{x}\right )}{x^{4}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsch(sqrt(x))/x^4, x)