### 3.14 $$\int x^3 \text{csch}^{-1}(\sqrt{x}) \, dx$$

Optimal. Leaf size=114 $\frac{1}{4} x^4 \text{csch}^{-1}\left (\sqrt{x}\right )-\frac{(-x-1)^{7/2} \sqrt{x}}{28 \sqrt{-x}}-\frac{3 (-x-1)^{5/2} \sqrt{x}}{20 \sqrt{-x}}-\frac{(-x-1)^{3/2} \sqrt{x}}{4 \sqrt{-x}}-\frac{\sqrt{-x-1} \sqrt{x}}{4 \sqrt{-x}}$

[Out]

-(Sqrt[-1 - x]*Sqrt[x])/(4*Sqrt[-x]) - ((-1 - x)^(3/2)*Sqrt[x])/(4*Sqrt[-x]) - (3*(-1 - x)^(5/2)*Sqrt[x])/(20*
Sqrt[-x]) - ((-1 - x)^(7/2)*Sqrt[x])/(28*Sqrt[-x]) + (x^4*ArcCsch[Sqrt[x]])/4

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Rubi [A]  time = 0.0323269, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {6346, 12, 43} $\frac{1}{4} x^4 \text{csch}^{-1}\left (\sqrt{x}\right )-\frac{(-x-1)^{7/2} \sqrt{x}}{28 \sqrt{-x}}-\frac{3 (-x-1)^{5/2} \sqrt{x}}{20 \sqrt{-x}}-\frac{(-x-1)^{3/2} \sqrt{x}}{4 \sqrt{-x}}-\frac{\sqrt{-x-1} \sqrt{x}}{4 \sqrt{-x}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*ArcCsch[Sqrt[x]],x]

[Out]

-(Sqrt[-1 - x]*Sqrt[x])/(4*Sqrt[-x]) - ((-1 - x)^(3/2)*Sqrt[x])/(4*Sqrt[-x]) - (3*(-1 - x)^(5/2)*Sqrt[x])/(20*
Sqrt[-x]) - ((-1 - x)^(7/2)*Sqrt[x])/(28*Sqrt[-x]) + (x^4*ArcCsch[Sqrt[x]])/4

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^3 \text{csch}^{-1}\left (\sqrt{x}\right ) \, dx &=\frac{1}{4} x^4 \text{csch}^{-1}\left (\sqrt{x}\right )-\frac{\sqrt{x} \int \frac{x^3}{2 \sqrt{-1-x}} \, dx}{4 \sqrt{-x}}\\ &=\frac{1}{4} x^4 \text{csch}^{-1}\left (\sqrt{x}\right )-\frac{\sqrt{x} \int \frac{x^3}{\sqrt{-1-x}} \, dx}{8 \sqrt{-x}}\\ &=\frac{1}{4} x^4 \text{csch}^{-1}\left (\sqrt{x}\right )-\frac{\sqrt{x} \int \left (-\frac{1}{\sqrt{-1-x}}-3 \sqrt{-1-x}-3 (-1-x)^{3/2}-(-1-x)^{5/2}\right ) \, dx}{8 \sqrt{-x}}\\ &=-\frac{\sqrt{-1-x} \sqrt{x}}{4 \sqrt{-x}}-\frac{(-1-x)^{3/2} \sqrt{x}}{4 \sqrt{-x}}-\frac{3 (-1-x)^{5/2} \sqrt{x}}{20 \sqrt{-x}}-\frac{(-1-x)^{7/2} \sqrt{x}}{28 \sqrt{-x}}+\frac{1}{4} x^4 \text{csch}^{-1}\left (\sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0327313, size = 47, normalized size = 0.41 $\frac{1}{140} \sqrt{\frac{1}{x}+1} \left (5 x^3-6 x^2+8 x-16\right ) \sqrt{x}+\frac{1}{4} x^4 \text{csch}^{-1}\left (\sqrt{x}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*ArcCsch[Sqrt[x]],x]

[Out]

(Sqrt[1 + x^(-1)]*Sqrt[x]*(-16 + 8*x - 6*x^2 + 5*x^3))/140 + (x^4*ArcCsch[Sqrt[x]])/4

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Maple [A]  time = 0.128, size = 43, normalized size = 0.4 \begin{align*}{\frac{{x}^{4}}{4}{\rm arccsch} \left (\sqrt{x}\right )}+{\frac{ \left ( 1+x \right ) \left ( 5\,{x}^{3}-6\,{x}^{2}+8\,x-16 \right ) }{140}{\frac{1}{\sqrt{{\frac{1+x}{x}}}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccsch(x^(1/2)),x)

[Out]

1/4*x^4*arccsch(x^(1/2))+1/140*(1+x)*(5*x^3-6*x^2+8*x-16)/((1+x)/x)^(1/2)/x^(1/2)

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Maxima [A]  time = 1.02526, size = 78, normalized size = 0.68 \begin{align*} \frac{1}{28} \, x^{\frac{7}{2}}{\left (\frac{1}{x} + 1\right )}^{\frac{7}{2}} - \frac{3}{20} \, x^{\frac{5}{2}}{\left (\frac{1}{x} + 1\right )}^{\frac{5}{2}} + \frac{1}{4} \, x^{4} \operatorname{arcsch}\left (\sqrt{x}\right ) + \frac{1}{4} \, x^{\frac{3}{2}}{\left (\frac{1}{x} + 1\right )}^{\frac{3}{2}} - \frac{1}{4} \, \sqrt{x} \sqrt{\frac{1}{x} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsch(x^(1/2)),x, algorithm="maxima")

[Out]

1/28*x^(7/2)*(1/x + 1)^(7/2) - 3/20*x^(5/2)*(1/x + 1)^(5/2) + 1/4*x^4*arccsch(sqrt(x)) + 1/4*x^(3/2)*(1/x + 1)
^(3/2) - 1/4*sqrt(x)*sqrt(1/x + 1)

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Fricas [A]  time = 2.58256, size = 142, normalized size = 1.25 \begin{align*} \frac{1}{4} \, x^{4} \log \left (\frac{x \sqrt{\frac{x + 1}{x}} + \sqrt{x}}{x}\right ) + \frac{1}{140} \,{\left (5 \, x^{3} - 6 \, x^{2} + 8 \, x - 16\right )} \sqrt{x} \sqrt{\frac{x + 1}{x}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsch(x^(1/2)),x, algorithm="fricas")

[Out]

1/4*x^4*log((x*sqrt((x + 1)/x) + sqrt(x))/x) + 1/140*(5*x^3 - 6*x^2 + 8*x - 16)*sqrt(x)*sqrt((x + 1)/x)

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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(x**3*acsch(x**(1/2)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arcsch}\left (\sqrt{x}\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsch(x^(1/2)),x, algorithm="giac")

[Out]

integrate(x^3*arccsch(sqrt(x)), x)