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

Optimal. Leaf size=61 \[ \frac{\log \left (\sqrt{2} \cos (3 x+2)-\sin (3 x+2)\right )}{6 \sqrt{2}}-\frac{\log \left (\sin (3 x+2)+\sqrt{2} \cos (3 x+2)\right )}{6 \sqrt{2}} \]

[Out]

Log[Sqrt[2]*Cos[2 + 3*x] - Sin[2 + 3*x]]/(6*Sqrt[2]) - Log[Sqrt[2]*Cos[2 + 3*x] + Sin[2 + 3*x]]/(6*Sqrt[2])

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Rubi [A]  time = 0.0429416, antiderivative size = 61, 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, 207} \[ \frac{\log \left (\sqrt{2} \cos (3 x+2)-\sin (3 x+2)\right )}{6 \sqrt{2}}-\frac{\log \left (\sin (3 x+2)+\sqrt{2} \cos (3 x+2)\right )}{6 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Sqrt[2]*Cos[2 + 3*x] - Sin[2 + 3*x]]/(6*Sqrt[2]) - Log[Sqrt[2]*Cos[2 + 3*x] + Sin[2 + 3*x]]/(6*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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[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{\log \left (\sqrt{2} \cos (2+3 x)-\sin (2+3 x)\right )}{6 \sqrt{2}}-\frac{\log \left (\sqrt{2} \cos (2+3 x)+\sin (2+3 x)\right )}{6 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0210346, size = 22, normalized size = 0.36 \[ -\frac{\tanh ^{-1}\left (\frac{\tan (3 x+2)}{\sqrt{2}}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.059, size = 18, normalized size = 0.3 \begin{align*} -{\frac{\sqrt{2}}{6}{\it Artanh} \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)*arctanh(1/2*tan(2+3*x)*2^(1/2))

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Maxima [A]  time = 1.64842, size = 43, normalized size = 0.7 \begin{align*} \frac{1}{12} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - \tan \left (3 \, x + 2\right )}{\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]

1/12*sqrt(2)*log(-(sqrt(2) - tan(3*x + 2))/(sqrt(2) + tan(3*x + 2)))

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Fricas [A]  time = 1.72201, size = 230, normalized size = 3.77 \begin{align*} \frac{1}{24} \, \sqrt{2} \log \left (-\frac{7 \, \cos \left (3 \, x + 2\right )^{4} - 10 \, \cos \left (3 \, x + 2\right )^{2} + 4 \,{\left (\sqrt{2} \cos \left (3 \, x + 2\right )^{3} + \sqrt{2} \cos \left (3 \, x + 2\right )\right )} \sin \left (3 \, x + 2\right ) - 1}{9 \, \cos \left (3 \, x + 2\right )^{4} - 6 \, \cos \left (3 \, x + 2\right )^{2} + 1}\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]

1/24*sqrt(2)*log(-(7*cos(3*x + 2)^4 - 10*cos(3*x + 2)^2 + 4*(sqrt(2)*cos(3*x + 2)^3 + sqrt(2)*cos(3*x + 2))*si
n(3*x + 2) - 1)/(9*cos(3*x + 2)^4 - 6*cos(3*x + 2)^2 + 1))

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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 = 1.87334, size = 53, normalized size = 0.87 \begin{align*} \frac{1}{12} \, \sqrt{2} \log \left (\frac{{\left | -2 \, \sqrt{2} + 2 \, \tan \left (3 \, x + 2\right ) \right |}}{{\left | 2 \, \sqrt{2} + 2 \, \tan \left (3 \, x + 2\right ) \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="giac")

[Out]

1/12*sqrt(2)*log(abs(-2*sqrt(2) + 2*tan(3*x + 2))/abs(2*sqrt(2) + 2*tan(3*x + 2)))