3.20 \(\int \frac{\sec ^2(2+3 x)}{2+\tan ^2(2+3 x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0422988, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3675, 203} \[ \frac{x}{\sqrt{2}}-\frac{\tan ^{-1}\left (\frac{\sin (3 x+2) \cos (3 x+2)}{\cos ^2(3 x+2)+\sqrt{2}+1}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[2 + 3*x]^2/(2 + Tan[2 + 3*x]^2),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 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

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

Mathematica [A]  time = 0.0184973, size = 22, normalized size = 0.46 \[ \frac{\tan ^{-1}\left (\frac{\tan (3 x+2)}{\sqrt{2}}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[2 + 3*x]^2/(2 + Tan[2 + 3*x]^2),x]

[Out]

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

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Maple [A]  time = 0.06, size = 18, normalized size = 0.4 \begin{align*}{\frac{\sqrt{2}}{6}\arctan \left ({\frac{\tan \left ( 2+3\,x \right ) \sqrt{2}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(2+3*x)^2/(2+tan(2+3*x)^2),x)

[Out]

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

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Maxima [A]  time = 1.60711, size = 23, normalized size = 0.48 \begin{align*} \frac{1}{6} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \tan \left (3 \, x + 2\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2+3*x)^2/(2+tan(2+3*x)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8668, size = 124, normalized size = 2.58 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2+3*x)^2/(2+tan(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)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (3 x + 2 \right )}}{\tan ^{2}{\left (3 x + 2 \right )} + 2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2+3*x)**2/(2+tan(2+3*x)**2),x)

[Out]

Integral(sec(3*x + 2)**2/(tan(3*x + 2)**2 + 2), x)

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Giac [A]  time = 1.81969, 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 ) - \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2+3*x)^2/(2+tan(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)