3.192 \(\int \sin (\sqrt {x}) \, dx\)

Optimal. Leaf size=22 \[ 2 \sin \left (\sqrt {x}\right )-2 \sqrt {x} \cos \left (\sqrt {x}\right ) \]

[Out]

2*sin(x^(1/2))-2*cos(x^(1/2))*x^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3361, 3296, 2637} \[ 2 \sin \left (\sqrt {x}\right )-2 \sqrt {x} \cos \left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[Sqrt[x]],x]

[Out]

-2*Sqrt[x]*Cos[Sqrt[x]] + 2*Sin[Sqrt[x]]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3361

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Sin[c + d*x])^p, x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && In
tegerQ[1/n]

Rubi steps

\begin {align*} \int \sin \left (\sqrt {x}\right ) \, dx &=2 \operatorname {Subst}\left (\int x \sin (x) \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {x} \cos \left (\sqrt {x}\right )+2 \operatorname {Subst}\left (\int \cos (x) \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {x} \cos \left (\sqrt {x}\right )+2 \sin \left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \[ 2 \sin \left (\sqrt {x}\right )-2 \sqrt {x} \cos \left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[Sqrt[x]],x]

[Out]

-2*Sqrt[x]*Cos[Sqrt[x]] + 2*Sin[Sqrt[x]]

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fricas [A]  time = 0.42, size = 16, normalized size = 0.73 \[ -2 \, \sqrt {x} \cos \left (\sqrt {x}\right ) + 2 \, \sin \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(x)*cos(sqrt(x)) + 2*sin(sqrt(x))

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giac [A]  time = 0.93, size = 16, normalized size = 0.73 \[ -2 \, \sqrt {x} \cos \left (\sqrt {x}\right ) + 2 \, \sin \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(x)*cos(sqrt(x)) + 2*sin(sqrt(x))

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maple [A]  time = 0.02, size = 17, normalized size = 0.77 \[ -2 \sqrt {x}\, \cos \left (\sqrt {x}\right )+2 \sin \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x^(1/2)),x)

[Out]

2*sin(x^(1/2))-2*cos(x^(1/2))*x^(1/2)

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maxima [A]  time = 0.45, size = 16, normalized size = 0.73 \[ -2 \, \sqrt {x} \cos \left (\sqrt {x}\right ) + 2 \, \sin \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(x)*cos(sqrt(x)) + 2*sin(sqrt(x))

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mupad [B]  time = 0.25, size = 16, normalized size = 0.73 \[ 2\,\sin \left (\sqrt {x}\right )-2\,\sqrt {x}\,\cos \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x^(1/2)),x)

[Out]

2*sin(x^(1/2)) - 2*x^(1/2)*cos(x^(1/2))

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sympy [A]  time = 0.31, size = 20, normalized size = 0.91 \[ - 2 \sqrt {x} \cos {\left (\sqrt {x} \right )} + 2 \sin {\left (\sqrt {x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(x)*cos(sqrt(x)) + 2*sin(sqrt(x))

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