3.405 \(\int (\cos (x)-\sin (x)) \sqrt{\sin (2 x)} \, dx\)

Optimal. Leaf size=47 \[ \frac{1}{2} \sin (x) \sqrt{\sin (2 x)}+\frac{1}{2} \sqrt{\sin (2 x)} \cos (x)-\frac{1}{2} \log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.09618, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4401, 4301, 4306, 4302, 4305} \[ \frac{1}{2} \sin (x) \sqrt{\sin (2 x)}+\frac{1}{2} \sqrt{\sin (2 x)} \cos (x)-\frac{1}{2} \log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Cos[x] - Sin[x])*Sqrt[Sin[2*x]],x]

[Out]

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

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 4301

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(2*Sin[a + b*x]*(g*Sin[c +
d*x])^p)/(d*(2*p + 1)), x] + Dist[(2*p*g)/(2*p + 1), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fre
eQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4306

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

Rule 4302

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-2*Cos[a + b*x]*(g*Sin[c
+ d*x])^p)/(d*(2*p + 1)), x] + Dist[(2*p*g)/(2*p + 1), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fr
eeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4305

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

Rubi steps

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

Mathematica [A]  time = 0.0618678, size = 43, normalized size = 0.91 \[ \frac{1}{2} \left (\sin (x) \sqrt{\sin (2 x)}+\sqrt{\sin (2 x)} \cos (x)-\log \left (\sin (x)+\sqrt{\sin (2 x)}+\cos (x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[x] - Sin[x])*Sqrt[Sin[2*x]],x]

[Out]

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

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Maple [C]  time = 0.08, size = 443, normalized size = 9.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^2-1)*(3*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1
/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(
1/2*x)^2-4*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2
*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^2+3*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+
2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2
*x)))^(1/2)-4*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),
1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)-4*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^4+2
*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^3-4*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^2-
2*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x))/(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)/(tan(1/2*x)
^3-tan(1/2*x))^(1/2)/(tan(1/2*x)^2+1)/(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\cos \left (x\right ) - \sin \left (x\right )\right )} \sqrt{\sin \left (2 \, x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((cos(x) - sin(x))*sqrt(sin(2*x)), x)

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Fricas [B]  time = 1.89208, size = 262, normalized size = 5.57 \begin{align*} \frac{1}{2} \, \sqrt{2} \sqrt{\cos \left (x\right ) \sin \left (x\right )}{\left (\cos \left (x\right ) + \sin \left (x\right )\right )} + \frac{1}{8} \, \log \left (-32 \, \cos \left (x\right )^{4} + 4 \, \sqrt{2}{\left (4 \, \cos \left (x\right )^{3} -{\left (4 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right ) - 5 \, \cos \left (x\right )\right )} \sqrt{\cos \left (x\right ) \sin \left (x\right )} + 32 \, \cos \left (x\right )^{2} + 16 \, \cos \left (x\right ) \sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*sqrt(cos(x)*sin(x))*(cos(x) + sin(x)) + 1/8*log(-32*cos(x)^4 + 4*sqrt(2)*(4*cos(x)^3 - (4*cos(x)^2
+ 1)*sin(x) - 5*cos(x))*sqrt(cos(x)*sin(x)) + 32*cos(x)^2 + 16*cos(x)*sin(x) + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\cos \left (x\right ) - \sin \left (x\right )\right )} \sqrt{\sin \left (2 \, x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((cos(x) - sin(x))*sqrt(sin(2*x)), x)