\(\int \frac {(-1+2 x) \cos (\sqrt {6+3 (-1+2 x)^2})}{\sqrt {6+3 (-1+2 x)^2}} \, dx\) [50]

Optimal result

Integrand size = 37, antiderivative size = 24 \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {1}{6} \sin \left (\sqrt {3} \sqrt {2+(-1+2 x)^2}\right ) \]

[Out]

1/6*sin(3^(1/2)*(2+(-1+2*x)^2)^(1/2))

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6847, 3513, 15, 2717} \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {1}{6} \sin \left (\sqrt {3} \sqrt {(2 x-1)^2+2}\right ) \]

[In]

Int[((-1 + 2*x)*Cos[Sqrt[6 + 3*(-1 + 2*x)^2]])/Sqrt[6 + 3*(-1 + 2*x)^2],x]

[Out]

Sin[Sqrt[3]*Sqrt[2 + (-1 + 2*x)^2]]/6

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 2717

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

Rule 3513

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 6847

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x \cos \left (\sqrt {6+3 x^2}\right )}{\sqrt {6+3 x^2}} \, dx,x,-1+2 x\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {\cos \left (\sqrt {6+3 x}\right )}{\sqrt {6+3 x}} \, dx,x,(-1+2 x)^2\right ) \\ & = \frac {1}{6} \text {Subst}\left (\int \frac {x \cos (x)}{\sqrt {x^2}} \, dx,x,\sqrt {3} \sqrt {2+(-1+2 x)^2}\right ) \\ & = \frac {1}{6} \text {Subst}\left (\int \cos (x) \, dx,x,\sqrt {3} \sqrt {2+(-1+2 x)^2}\right ) \\ & = \frac {1}{6} \sin \left (\sqrt {3} \sqrt {2+(-1+2 x)^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {1}{6} \sin \left (\sqrt {6+3 (1-2 x)^2}\right ) \]

[In]

Integrate[((-1 + 2*x)*Cos[Sqrt[6 + 3*(-1 + 2*x)^2]])/Sqrt[6 + 3*(-1 + 2*x)^2],x]

[Out]

Sin[Sqrt[6 + 3*(1 - 2*x)^2]]/6

Maple [A] (verified)

Time = 4.67 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
default	\(\frac {\sin \left (\sqrt {12 x^{2}-12 x +9}\right )}{6}\)	\(16\)

[In]

int((-1+2*x)*cos((6+3*(-1+2*x)^2)^(1/2))/(6+3*(-1+2*x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*sin((12*x^2-12*x+9)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {1}{6} \, \sin \left (\sqrt {12 \, x^{2} - 12 \, x + 9}\right ) \]

[In]

integrate((-1+2*x)*cos((6+3*(-1+2*x)^2)^(1/2))/(6+3*(-1+2*x)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*sin(sqrt(12*x^2 - 12*x + 9))

Sympy [A] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {\sin {\left (\sqrt {3 \left (2 x - 1\right )^{2} + 6} \right )}}{6} \]

[In]

integrate((-1+2*x)*cos((6+3*(-1+2*x)**2)**(1/2))/(6+3*(-1+2*x)**2)**(1/2),x)

[Out]

sin(sqrt(3*(2*x - 1)**2 + 6))/6

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {1}{6} \, \sin \left (\sqrt {3 \, {\left (2 \, x - 1\right )}^{2} + 6}\right ) \]

[In]

integrate((-1+2*x)*cos((6+3*(-1+2*x)^2)^(1/2))/(6+3*(-1+2*x)^2)^(1/2),x, algorithm="maxima")

[Out]

1/6*sin(sqrt(3*(2*x - 1)^2 + 6))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {1}{6} \, \sin \left (\sqrt {3} \sqrt {4 \, x^{2} - 4 \, x + 3}\right ) \]

[In]

integrate((-1+2*x)*cos((6+3*(-1+2*x)^2)^(1/2))/(6+3*(-1+2*x)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*sin(sqrt(3)*sqrt(4*x^2 - 4*x + 3))

Mupad [B] (verification not implemented)

Time = 26.75 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {(-1+2 x) \cos \left (\sqrt {6+3 (-1+2 x)^2}\right )}{\sqrt {6+3 (-1+2 x)^2}} \, dx=\frac {\sin \left (\sqrt {3\,{\left (2\,x-1\right )}^2+6}\right )}{6} \]

[In]

int((cos((3*(2*x - 1)^2 + 6)^(1/2))*(2*x - 1))/(3*(2*x - 1)^2 + 6)^(1/2),x)

[Out]

sin((3*(2*x - 1)^2 + 6)^(1/2))/6