3.293 \(\int x \sinh ^{-1}(\sqrt {x}) \, dx\)

Optimal. Leaf size=56 \[ -\frac {1}{8} \sqrt {x+1} x^{3/2}+\frac {1}{2} x^2 \sinh ^{-1}\left (\sqrt {x}\right )+\frac {3}{16} \sqrt {x+1} \sqrt {x}-\frac {3}{16} \sinh ^{-1}\left (\sqrt {x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 56, 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 = {5902, 12, 50, 54, 215} \[ -\frac {1}{8} \sqrt {x+1} x^{3/2}+\frac {1}{2} x^2 \sinh ^{-1}\left (\sqrt {x}\right )+\frac {3}{16} \sqrt {x+1} \sqrt {x}-\frac {3}{16} \sinh ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSinh[Sqrt[x]],x]

[Out]

(3*Sqrt[x]*Sqrt[1 + x])/16 - (x^(3/2)*Sqrt[1 + x])/8 - (3*ArcSinh[Sqrt[x]])/16 + (x^2*ArcSinh[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 50

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

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 215

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

Rule 5902

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.66 \[ \frac {1}{16} \left (\left (8 x^2-3\right ) \sinh ^{-1}\left (\sqrt {x}\right )+\sqrt {x} \sqrt {x+1} (3-2 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSinh[Sqrt[x]],x]

[Out]

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

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fricas [A]  time = 0.72, size = 35, normalized size = 0.62 \[ -\frac {1}{16} \, {\left (2 \, x - 3\right )} \sqrt {x + 1} \sqrt {x} + \frac {1}{16} \, {\left (8 \, x^{2} - 3\right )} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arcsinh(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 = 0.48, size = 48, normalized size = 0.86 \[ \frac {1}{2} \, x^{2} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) - \frac {1}{16} \, \sqrt {x^{2} + x} {\left (2 \, x - 3\right )} + \frac {3}{32} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} + x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(sqrt(x + 1) + sqrt(x)) - 1/16*sqrt(x^2 + x)*(2*x - 3) + 3/32*log(abs(-2*x + 2*sqrt(x^2 + x) - 1))

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maple [A]  time = 0.00, size = 37, normalized size = 0.66 \[ -\frac {3 \arcsinh \left (\sqrt {x}\right )}{16}+\frac {x^{2} \arcsinh \left (\sqrt {x}\right )}{2}-\frac {x^{\frac {3}{2}} \sqrt {1+x}}{8}+\frac {3 \sqrt {x}\, \sqrt {1+x}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.79, size = 36, normalized size = 0.64 \[ \frac {1}{2} \, x^{2} \operatorname {arsinh}\left (\sqrt {x}\right ) - \frac {1}{8} \, \sqrt {x + 1} x^{\frac {3}{2}} + \frac {3}{16} \, \sqrt {x + 1} \sqrt {x} - \frac {3}{16} \, \operatorname {arsinh}\left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\mathrm {asinh}\left (\sqrt {x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*asinh(x^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asinh(sqrt(x)), x)

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