3.233 \(\int x \cosh ^{-1}(\sqrt {x}) \, dx\)

Optimal. Leaf size=86 \[ -\frac {1}{8} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}+\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\frac {3}{16} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}-\frac {3}{16} \cosh ^{-1}\left (\sqrt {x}\right ) \]

[Out]

-3/16*arccosh(x^(1/2))+1/2*x^2*arccosh(x^(1/2))-1/8*x^(3/2)*(-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)-3/16*x^(1/2)*
(-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5903, 12, 323, 330, 52} \[ -\frac {1}{8} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}+\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\frac {3}{16} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}-\frac {3}{16} \cosh ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCosh[Sqrt[x]],x]

[Out]

(-3*Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x])/16 - (Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*x^(3/2))/8 - (3*A
rcCosh[Sqrt[x]])/16 + (x^2*ArcCosh[Sqrt[x]])/2

Rule 12

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

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 323

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(2
*n - 1)*(c*x)^(m - 2*n + 1)*(a1 + b1*x^n)^(p + 1)*(a2 + b2*x^n)^(p + 1))/(b1*b2*(m + 2*n*p + 1)), x] - Dist[(a
1*a2*c^(2*n)*(m - 2*n + 1))/(b1*b2*(m + 2*n*p + 1)), Int[(c*x)^(m - 2*n)*(a1 + b1*x^n)^p*(a2 + b2*x^n)^p, x],
x] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && GtQ[m, 2*n - 1] && NeQ[m +
2*n*p + 1, 0] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]

Rule 330

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k =
Denominator[m]}, Dist[k/c, Subst[Int[x^(k*(m + 1) - 1)*(a1 + (b1*x^(k*n))/c^n)^p*(a2 + (b2*x^(k*n))/c^n)^p, x]
, x, (c*x)^(1/k)], x]] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && Fractio
nQ[m] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]

Rule 5903

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

Rubi steps

\begin {align*} \int x \cosh ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\frac {1}{2} \int \frac {x^{3/2}}{2 \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}}} \, dx\\ &=\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\frac {1}{4} \int \frac {x^{3/2}}{\sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}}} \, dx\\ &=-\frac {1}{8} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} x^{3/2}+\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\frac {3}{16} \int \frac {\sqrt {x}}{\sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}}} \, dx\\ &=-\frac {3}{16} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} \sqrt {x}-\frac {1}{8} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} x^{3/2}+\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\frac {3}{32} \int \frac {1}{\sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} \sqrt {x}} \, dx\\ &=-\frac {3}{16} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} \sqrt {x}-\frac {1}{8} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} x^{3/2}+\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\frac {3}{16} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} \sqrt {1+x}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {3}{16} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} \sqrt {x}-\frac {1}{8} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} x^{3/2}-\frac {3}{16} \cosh ^{-1}\left (\sqrt {x}\right )+\frac {1}{2} x^2 \cosh ^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 74, normalized size = 0.86 \[ \frac {1}{16} \left (8 x^2 \cosh ^{-1}\left (\sqrt {x}\right )-\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} (2 x+3) \sqrt {x}-6 \tanh ^{-1}\left (\sqrt {\frac {\sqrt {x}-1}{\sqrt {x}+1}}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*ArcCosh[Sqrt[x]],x]

[Out]

(-(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x]*(3 + 2*x)) + 8*x^2*ArcCosh[Sqrt[x]] - 6*ArcTanh[Sqrt[(-1 + Sqr
t[x])/(1 + Sqrt[x])]])/16

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fricas [A]  time = 0.62, size = 35, normalized size = 0.41 \[ -\frac {1}{16} \, {\left (2 \, x + 3\right )} \sqrt {x - 1} \sqrt {x} + \frac {1}{16} \, {\left (8 \, x^{2} - 3\right )} \log \left (\sqrt {x - 1} + \sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.64, size = 55, normalized size = 0.64 \[ \frac {1}{2} \, x^{2} \log \left (\sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} + \sqrt {x}\right ) - \frac {1}{16} \, {\left (2 \, x + 3\right )} \sqrt {x - 1} \sqrt {x} + \frac {3}{16} \, \log \left (-\sqrt {x - 1} + \sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(sqrt(sqrt(x) + 1)*sqrt(sqrt(x) - 1) + sqrt(x)) - 1/16*(2*x + 3)*sqrt(x - 1)*sqrt(x) + 3/16*log(-sq
rt(x - 1) + sqrt(x))

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maple [A]  time = 0.00, size = 65, normalized size = 0.76 \[ \frac {x^{2} \mathrm {arccosh}\left (\sqrt {x}\right )}{2}-\frac {\sqrt {-1+\sqrt {x}}\, \sqrt {1+\sqrt {x}}\, \left (2 x^{\frac {3}{2}} \sqrt {-1+x}+3 \sqrt {x}\, \sqrt {-1+x}+3 \ln \left (\sqrt {x}+\sqrt {-1+x}\right )\right )}{16 \sqrt {-1+x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccosh(x^(1/2))-1/16*(-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)*(2*x^(3/2)*(-1+x)^(1/2)+3*x^(1/2)*(-1+x)^(1
/2)+3*ln(x^(1/2)+(-1+x)^(1/2)))/(-1+x)^(1/2)

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maxima [A]  time = 0.38, size = 46, normalized size = 0.53 \[ \frac {1}{2} \, x^{2} \operatorname {arcosh}\left (\sqrt {x}\right ) - \frac {1}{8} \, \sqrt {x - 1} x^{\frac {3}{2}} - \frac {3}{16} \, \sqrt {x - 1} \sqrt {x} - \frac {3}{16} \, \log \left (2 \, \sqrt {x - 1} + 2 \, \sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccosh(sqrt(x)) - 1/8*sqrt(x - 1)*x^(3/2) - 3/16*sqrt(x - 1)*sqrt(x) - 3/16*log(2*sqrt(x - 1) + 2*sqr
t(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {acosh}\left (\sqrt {x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*acosh(x^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {acosh}{\left (\sqrt {x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acosh(sqrt(x)), x)

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