\(\int \sqrt {1+\sin (2 x)} \, dx\) [390]

Optimal result

Integrand size = 10, antiderivative size = 16 \[ \int \sqrt {1+\sin (2 x)} \, dx=-\frac {\cos (2 x)}{\sqrt {1+\sin (2 x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2725} \[ \int \sqrt {1+\sin (2 x)} \, dx=-\frac {\cos (2 x)}{\sqrt {\sin (2 x)+1}} \]

[In]

Int[Sqrt[1 + Sin[2*x]],x]

[Out]

-(Cos[2*x]/Sqrt[1 + Sin[2*x]])

Rule 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x
]])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (2 x)}{\sqrt {1+\sin (2 x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \sqrt {1+\sin (2 x)} \, dx=\frac {(-\cos (x)+\sin (x)) \sqrt {1+\sin (2 x)}}{\cos (x)+\sin (x)} \]

[In]

Integrate[Sqrt[1 + Sin[2*x]],x]

[Out]

((-Cos[x] + Sin[x])*Sqrt[1 + Sin[2*x]])/(Cos[x] + Sin[x])

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {\left (-1+\sin \left (2 x \right )\right ) \sqrt {1+\sin \left (2 x \right )}}{\cos \left (2 x \right )}\)	\(22\)

[In]

int((1+sin(2*x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.12 \[ \int \sqrt {1+\sin (2 x)} \, dx=-\frac {{\left (\cos \left (2 \, x\right ) - \sin \left (2 \, x\right ) + 1\right )} \sqrt {\sin \left (2 \, x\right ) + 1}}{\cos \left (2 \, x\right ) + \sin \left (2 \, x\right ) + 1} \]

[In]

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

[Out]

-(cos(2*x) - sin(2*x) + 1)*sqrt(sin(2*x) + 1)/(cos(2*x) + sin(2*x) + 1)

Sympy [F]

\[ \int \sqrt {1+\sin (2 x)} \, dx=\int \sqrt {\sin {\left (2 x \right )} + 1}\, dx \]

[In]

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

[Out]

Integral(sqrt(sin(2*x) + 1), x)

Maxima [F]

\[ \int \sqrt {1+\sin (2 x)} \, dx=\int { \sqrt {\sin \left (2 \, x\right ) + 1} \,d x } \]

[In]

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

[Out]

integrate(sqrt(sin(2*x) + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \sqrt {1+\sin (2 x)} \, dx=\sqrt {2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + x\right )\right ) \sin \left (-\frac {1}{4} \, \pi + x\right ) \]

[In]

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

[Out]

sqrt(2)*sgn(cos(-1/4*pi + x))*sin(-1/4*pi + x)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \sqrt {1+\sin (2 x)} \, dx=\frac {\left (\sin \left (2\,x\right )-1\right )\,\sqrt {\sin \left (2\,x\right )+1}}{\cos \left (2\,x\right )} \]

[In]

int((sin(2*x) + 1)^(1/2),x)

[Out]

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