3.1.14 \(\int x^3 \text {csch}^{-1}(\sqrt {x}) \, dx\) [14]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 114, 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}{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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[-1 - x]*Sqrt[x])/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 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^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*}

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Mathematica [A]
time = 0.03, size = 47, normalized size = 0.41 \begin {gather*} \frac {1}{140} \sqrt {1+\frac {1}{x}} \sqrt {x} \left (-16+8 x-6 x^2+5 x^3\right )+\frac {1}{4} x^4 \text {csch}^{-1}\left (\sqrt {x}\right ) \end {gather*}

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 = 0.15, size = 43, normalized size = 0.38

method	result	size
derivativedivides	\(\frac {x^{4} \mathrm {arccsch}\left (\sqrt {x}\right )}{4}+\frac {\left (1+x \right ) \left (5 x^{3}-6 x^{2}+8 x -16\right )}{140 \sqrt {\frac {1+x}{x}}\, \sqrt {x}}\)	\(43\)
default	\(\frac {x^{4} \mathrm {arccsch}\left (\sqrt {x}\right )}{4}+\frac {\left (1+x \right ) \left (5 x^{3}-6 x^{2}+8 x -16\right )}{140 \sqrt {\frac {1+x}{x}}\, \sqrt {x}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.26, size = 58, normalized size = 0.51 \begin {gather*} \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 {gather*}

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 = 0.37, size = 55, normalized size = 0.48 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \operatorname {acsch}{\left (\sqrt {x} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^3*arccsch(sqrt(x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\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^3*asinh(1/x^(1/2)),x)

[Out]

int(x^3*asinh(1/x^(1/2)), x)

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