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

Optimal. Leaf size=48 \[ \frac{x}{\sqrt{2}}+\frac{\tan ^{-1}\left (\frac{\sin (3 x+2) \cos (3 x+2)}{\sin ^2(3 x+2)+\sqrt{2}+1}\right )}{3 \sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0255043, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {203} \[ \frac{x}{\sqrt{2}}+\frac{\tan ^{-1}\left (\frac{\sin (3 x+2) \cos (3 x+2)}{\sin ^2(3 x+2)+\sqrt{2}+1}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.024852, size = 22, normalized size = 0.46 \[ \frac{\tan ^{-1}\left (\sqrt{2} \tan (3 x+2)\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 17, normalized size = 0.4 \begin{align*}{\frac{\sqrt{2}\arctan \left ( \tan \left ( 2+3\,x \right ) \sqrt{2} \right ) }{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.67127, size = 22, normalized size = 0.46 \begin{align*} \frac{1}{6} \, \sqrt{2} \arctan \left (\sqrt{2} \tan \left (3 \, x + 2\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.46917, size = 127, normalized size = 2.65 \begin{align*} -\frac{1}{12} \, \sqrt{2} \arctan \left (\frac{3 \, \sqrt{2} \cos \left (3 \, x + 2\right )^{2} - 2 \, \sqrt{2}}{4 \, \cos \left (3 \, x + 2\right ) \sin \left (3 \, x + 2\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cos(2+3*x)^2+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 - 2*sqrt(2))/(cos(3*x + 2)*sin(3*x + 2)))

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Sympy [B]  time = 9.4536, size = 343, normalized size = 7.15 \begin{align*} \frac{2 \sqrt{2} \left (\operatorname{atan}{\left (\frac{\tan{\left (\frac{3 x}{2} + 1 \right )}}{\sqrt{3 - 2 \sqrt{2}}} \right )} + \pi \left \lfloor{\frac{\frac{3 x}{2} - \frac{\pi }{2} + 1}{\pi }}\right \rfloor \right )}{21 \sqrt{2} \sqrt{3 - 2 \sqrt{2}} + 30 \sqrt{3 - 2 \sqrt{2}}} + \frac{3 \left (\operatorname{atan}{\left (\frac{\tan{\left (\frac{3 x}{2} + 1 \right )}}{\sqrt{3 - 2 \sqrt{2}}} \right )} + \pi \left \lfloor{\frac{\frac{3 x}{2} - \frac{\pi }{2} + 1}{\pi }}\right \rfloor \right )}{21 \sqrt{2} \sqrt{3 - 2 \sqrt{2}} + 30 \sqrt{3 - 2 \sqrt{2}}} + \frac{2 \sqrt{2} \sqrt{3 - 2 \sqrt{2}} \sqrt{2 \sqrt{2} + 3} \left (\operatorname{atan}{\left (\frac{\tan{\left (\frac{3 x}{2} + 1 \right )}}{\sqrt{2 \sqrt{2} + 3}} \right )} + \pi \left \lfloor{\frac{\frac{3 x}{2} - \frac{\pi }{2} + 1}{\pi }}\right \rfloor \right )}{21 \sqrt{2} \sqrt{3 - 2 \sqrt{2}} + 30 \sqrt{3 - 2 \sqrt{2}}} + \frac{3 \sqrt{3 - 2 \sqrt{2}} \sqrt{2 \sqrt{2} + 3} \left (\operatorname{atan}{\left (\frac{\tan{\left (\frac{3 x}{2} + 1 \right )}}{\sqrt{2 \sqrt{2} + 3}} \right )} + \pi \left \lfloor{\frac{\frac{3 x}{2} - \frac{\pi }{2} + 1}{\pi }}\right \rfloor \right )}{21 \sqrt{2} \sqrt{3 - 2 \sqrt{2}} + 30 \sqrt{3 - 2 \sqrt{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*(atan(tan(3*x/2 + 1)/sqrt(3 - 2*sqrt(2))) + pi*floor((3*x/2 - pi/2 + 1)/pi))/(21*sqrt(2)*sqrt(3 - 2*
sqrt(2)) + 30*sqrt(3 - 2*sqrt(2))) + 3*(atan(tan(3*x/2 + 1)/sqrt(3 - 2*sqrt(2))) + pi*floor((3*x/2 - pi/2 + 1)
/pi))/(21*sqrt(2)*sqrt(3 - 2*sqrt(2)) + 30*sqrt(3 - 2*sqrt(2))) + 2*sqrt(2)*sqrt(3 - 2*sqrt(2))*sqrt(2*sqrt(2)
+ 3)*(atan(tan(3*x/2 + 1)/sqrt(2*sqrt(2) + 3)) + pi*floor((3*x/2 - pi/2 + 1)/pi))/(21*sqrt(2)*sqrt(3 - 2*sqrt
(2)) + 30*sqrt(3 - 2*sqrt(2))) + 3*sqrt(3 - 2*sqrt(2))*sqrt(2*sqrt(2) + 3)*(atan(tan(3*x/2 + 1)/sqrt(2*sqrt(2)
+ 3)) + pi*floor((3*x/2 - pi/2 + 1)/pi))/(21*sqrt(2)*sqrt(3 - 2*sqrt(2)) + 30*sqrt(3 - 2*sqrt(2)))

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Giac [A]  time = 1.09728, size = 77, normalized size = 1.6 \begin{align*} \frac{1}{6} \, \sqrt{2}{\left (3 \, x + \arctan \left (-\frac{\sqrt{2} \sin \left (6 \, x + 4\right ) - 2 \, \sin \left (6 \, x + 4\right )}{\sqrt{2} \cos \left (6 \, x + 4\right ) + \sqrt{2} - 2 \, \cos \left (6 \, x + 4\right ) + 2}\right ) + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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