3.48 \(\int \frac {x \cos (\sqrt {1+x^2})}{\sqrt {1+x^2}} \, dx\)

Optimal. Leaf size=10 \[ \sin \left (\sqrt {x^2+1}\right ) \]

[Out]

sin((x^2+1)^(1/2))

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Rubi [A]  time = 0.14, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6715, 3432, 15, 2637} \[ \sin \left (\sqrt {x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x*Cos[Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]

[Out]

Sin[Sqrt[1 + x^2]]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 2637

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

Rule 3432

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Cos[c + d*x])^p, x^(1/n - 1)*(g - (e*h)/f + (h*x^(1/n))/f)^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {align*} \int \frac {x \cos \left (\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\cos \left (\sqrt {1+x}\right )}{\sqrt {1+x}} \, dx,x,x^2\right )\\ &=\operatorname {Subst}\left (\int \frac {x \cos (x)}{\sqrt {x^2}} \, dx,x,\sqrt {1+x^2}\right )\\ &=1 \operatorname {Subst}\left (\int \cos (x) \, dx,x,\sqrt {1+x^2}\right )\\ &=\sin \left (\sqrt {1+x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 10, normalized size = 1.00 \[ \sin \left (\sqrt {x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Cos[Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]

[Out]

Sin[Sqrt[1 + x^2]]

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fricas [A]  time = 1.63, size = 8, normalized size = 0.80 \[ \sin \left (\sqrt {x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(sqrt(x^2 + 1))

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giac [A]  time = 0.14, size = 8, normalized size = 0.80 \[ \sin \left (\sqrt {x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(sqrt(x^2 + 1))

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maple [A]  time = 0.04, size = 9, normalized size = 0.90 \[ \sin \left (\sqrt {x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos((x^2+1)^(1/2))/(x^2+1)^(1/2),x)

[Out]

sin((x^2+1)^(1/2))

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maxima [A]  time = 0.35, size = 8, normalized size = 0.80 \[ \sin \left (\sqrt {x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(sqrt(x^2 + 1))

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mupad [B]  time = 2.26, size = 8, normalized size = 0.80 \[ \sin \left (\sqrt {x^2+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*cos((x^2 + 1)^(1/2)))/(x^2 + 1)^(1/2),x)

[Out]

sin((x^2 + 1)^(1/2))

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sympy [A]  time = 0.43, size = 8, normalized size = 0.80 \[ \sin {\left (\sqrt {x^{2} + 1} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos((x**2+1)**(1/2))/(x**2+1)**(1/2),x)

[Out]

sin(sqrt(x**2 + 1))

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