\(\int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx\) [19]

Optimal result

Integrand size = 21, antiderivative size = 48 \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=\frac {x}{\sqrt {2}}-\frac {\arctan \left (\frac {\cos (2+3 x) \sin (2+3 x)}{1+\sqrt {2}+\cos ^2(2+3 x)}\right )}{3 \sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {209} \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=\frac {x}{\sqrt {2}}-\frac {\arctan \left (\frac {\sin (3 x+2) \cos (3 x+2)}{\cos ^2(3 x+2)+\sqrt {2}+1}\right )}{3 \sqrt {2}} \]

[In]

Int[(2*Cos[2 + 3*x]^2 + Sin[2 + 3*x]^2)^(-1),x]

[Out]

x/Sqrt[2] - ArcTan[(Cos[2 + 3*x]*Sin[2 + 3*x])/(1 + Sqrt[2] + Cos[2 + 3*x]^2)]/(3*Sqrt[2])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.46 \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=\frac {\arctan \left (\frac {\tan (2+3 x)}{\sqrt {2}}\right )}{3 \sqrt {2}} \]

[In]

Integrate[(2*Cos[2 + 3*x]^2 + Sin[2 + 3*x]^2)^(-1),x]

[Out]

ArcTan[Tan[2 + 3*x]/Sqrt[2]]/(3*Sqrt[2])

Maple [A] (verified)

Time = 1.43 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.38

method	result	size
derivativedivides	\(\frac {\sqrt {2}\, \arctan \left (\frac {\tan \left (2+3 x \right ) \sqrt {2}}{2}\right )}{6}\)	\(18\)
default	\(\frac {\sqrt {2}\, \arctan \left (\frac {\tan \left (2+3 x \right ) \sqrt {2}}{2}\right )}{6}\)	\(18\)
risch	\(\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (2+3 x \right )}+3+2 \sqrt {2}\right )}{12}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (2+3 x \right )}+3-2 \sqrt {2}\right )}{12}\)	\(48\)

[In]

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

[Out]

1/6*2^(1/2)*arctan(1/2*tan(2+3*x)*2^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=-\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (3 \, x + 2\right )^{2} - \sqrt {2}}{4 \, \cos \left (3 \, x + 2\right ) \sin \left (3 \, x + 2\right )}\right ) \]

[In]

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

[Out]

-1/12*sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(3*x + 2)^2 - sqrt(2))/(cos(3*x + 2)*sin(3*x + 2)))

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.58 \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=\frac {\sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {3 x}{2} + 1 \right )} - 1 \right )} + \pi \left \lfloor {\frac {\frac {3 x}{2} - \frac {\pi }{2} + 1}{\pi }}\right \rfloor \right )}{6} + \frac {\sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (\frac {3 x}{2} + 1 \right )} + 1 \right )} + \pi \left \lfloor {\frac {\frac {3 x}{2} - \frac {\pi }{2} + 1}{\pi }}\right \rfloor \right )}{6} \]

[In]

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

[Out]

sqrt(2)*(atan(sqrt(2)*tan(3*x/2 + 1) - 1) + pi*floor((3*x/2 - pi/2 + 1)/pi))/6 + sqrt(2)*(atan(sqrt(2)*tan(3*x
/2 + 1) + 1) + pi*floor((3*x/2 - pi/2 + 1)/pi))/6

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.35 \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=\frac {1}{6} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (3 \, x + 2\right )\right ) \]

[In]

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

[Out]

1/6*sqrt(2)*arctan(1/2*sqrt(2)*tan(3*x + 2))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=\frac {1}{6} \, \sqrt {2} {\left (3 \, x + \arctan \left (-\frac {\sqrt {2} \sin \left (6 \, x + 4\right ) - \sin \left (6 \, x + 4\right )}{\sqrt {2} \cos \left (6 \, x + 4\right ) + \sqrt {2} - \cos \left (6 \, x + 4\right ) + 1}\right ) + 2\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 26.48 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75 \[ \int \frac {1}{2 \cos ^2(2+3 x)+\sin ^2(2+3 x)} \, dx=\frac {\sqrt {2}\,\left (3\,x-\mathrm {atan}\left (\mathrm {tan}\left (3\,x+2\right )\right )\right )}{6}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (3\,x+2\right )}{2}\right )}{6} \]

[In]

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

[Out]

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