3.1.15 \(\int x^2 \text {csch}^{-1}(\sqrt {x}) \, dx\) [15]

Optimal. Leaf size=89 \[ \frac {\sqrt {-1-x} \sqrt {x}}{3 \sqrt {-x}}+\frac {2 (-1-x)^{3/2} \sqrt {x}}{9 \sqrt {-x}}+\frac {(-1-x)^{5/2} \sqrt {x}}{15 \sqrt {-x}}+\frac {1}{3} x^3 \text {csch}^{-1}\left (\sqrt {x}\right ) \]

[Out]

1/3*x^3*arccsch(x^(1/2))+2/9*(-1-x)^(3/2)*x^(1/2)/(-x)^(1/2)+1/15*(-1-x)^(5/2)*x^(1/2)/(-x)^(1/2)+1/3*(-1-x)^(
1/2)*x^(1/2)/(-x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6481, 12, 45} \begin {gather*} \frac {1}{3} x^3 \text {csch}^{-1}\left (\sqrt {x}\right )+\frac {(-x-1)^{5/2} \sqrt {x}}{15 \sqrt {-x}}+\frac {2 (-x-1)^{3/2} \sqrt {x}}{9 \sqrt {-x}}+\frac {\sqrt {-x-1} \sqrt {x}}{3 \sqrt {-x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCsch[Sqrt[x]],x]

[Out]

(Sqrt[-1 - x]*Sqrt[x])/(3*Sqrt[-x]) + (2*(-1 - x)^(3/2)*Sqrt[x])/(9*Sqrt[-x]) + ((-1 - x)^(5/2)*Sqrt[x])/(15*S
qrt[-x]) + (x^3*ArcCsch[Sqrt[x]])/3

Rule 12

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

Rule 45

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])

Rule 6481

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]

Rubi steps

\begin {align*} \int x^2 \text {csch}^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{3} x^3 \text {csch}^{-1}\left (\sqrt {x}\right )-\frac {\sqrt {x} \int \frac {x^2}{2 \sqrt {-1-x}} \, dx}{3 \sqrt {-x}}\\ &=\frac {1}{3} x^3 \text {csch}^{-1}\left (\sqrt {x}\right )-\frac {\sqrt {x} \int \frac {x^2}{\sqrt {-1-x}} \, dx}{6 \sqrt {-x}}\\ &=\frac {1}{3} x^3 \text {csch}^{-1}\left (\sqrt {x}\right )-\frac {\sqrt {x} \int \left (\frac {1}{\sqrt {-1-x}}+2 \sqrt {-1-x}+(-1-x)^{3/2}\right ) \, dx}{6 \sqrt {-x}}\\ &=\frac {\sqrt {-1-x} \sqrt {x}}{3 \sqrt {-x}}+\frac {2 (-1-x)^{3/2} \sqrt {x}}{9 \sqrt {-x}}+\frac {(-1-x)^{5/2} \sqrt {x}}{15 \sqrt {-x}}+\frac {1}{3} x^3 \text {csch}^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 42, normalized size = 0.47 \begin {gather*} \frac {1}{45} \sqrt {1+\frac {1}{x}} \sqrt {x} \left (8-4 x+3 x^2\right )+\frac {1}{3} x^3 \text {csch}^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCsch[Sqrt[x]],x]

[Out]

(Sqrt[1 + x^(-1)]*Sqrt[x]*(8 - 4*x + 3*x^2))/45 + (x^3*ArcCsch[Sqrt[x]])/3

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Maple [A]
time = 0.15, size = 38, normalized size = 0.43

method	result	size
derivativedivides	\(\frac {x^{3} \mathrm {arccsch}\left (\sqrt {x}\right )}{3}+\frac {\left (1+x \right ) \left (3 x^{2}-4 x +8\right )}{45 \sqrt {\frac {1+x}{x}}\, \sqrt {x}}\)	\(38\)
default	\(\frac {x^{3} \mathrm {arccsch}\left (\sqrt {x}\right )}{3}+\frac {\left (1+x \right ) \left (3 x^{2}-4 x +8\right )}{45 \sqrt {\frac {1+x}{x}}\, \sqrt {x}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccsch(x^(1/2)),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3*arccsch(x^(1/2))+1/45*(1+x)*(3*x^2-4*x+8)/((1+x)/x)^(1/2)/x^(1/2)

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Maxima [A]
time = 0.27, size = 46, normalized size = 0.52 \begin {gather*} \frac {1}{15} \, x^{\frac {5}{2}} {\left (\frac {1}{x} + 1\right )}^{\frac {5}{2}} + \frac {1}{3} \, x^{3} \operatorname {arcsch}\left (\sqrt {x}\right ) - \frac {2}{9} \, x^{\frac {3}{2}} {\left (\frac {1}{x} + 1\right )}^{\frac {3}{2}} + \frac {1}{3} \, \sqrt {x} \sqrt {\frac {1}{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*x^(5/2)*(1/x + 1)^(5/2) + 1/3*x^3*arccsch(sqrt(x)) - 2/9*x^(3/2)*(1/x + 1)^(3/2) + 1/3*sqrt(x)*sqrt(1/x +
1)

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Fricas [A]
time = 0.37, size = 50, normalized size = 0.56 \begin {gather*} \frac {1}{3} \, x^{3} \log \left (\frac {x \sqrt {\frac {x + 1}{x}} + \sqrt {x}}{x}\right ) + \frac {1}{45} \, {\left (3 \, x^{2} - 4 \, x + 8\right )} \sqrt {x} \sqrt {\frac {x + 1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log((x*sqrt((x + 1)/x) + sqrt(x))/x) + 1/45*(3*x^2 - 4*x + 8)*sqrt(x)*sqrt((x + 1)/x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {acsch}{\left (\sqrt {x} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*acsch(sqrt(x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccsch(sqrt(x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\mathrm {asinh}\left (\frac {1}{\sqrt {x}}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*asinh(1/x^(1/2)),x)

[Out]

int(x^2*asinh(1/x^(1/2)), x)

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