\(\int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx\) [27]

Optimal result

Integrand size = 21, antiderivative size = 61 \[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=\frac {\log \left (\sqrt {2} \cos (2+3 x)-\sin (2+3 x)\right )}{6 \sqrt {2}}-\frac {\log \left (\sqrt {2} \cos (2+3 x)+\sin (2+3 x)\right )}{6 \sqrt {2}} \]

[Out]

1/12*ln(-sin(2+3*x)+cos(2+3*x)*2^(1/2))*2^(1/2)-1/12*ln(sin(2+3*x)+cos(2+3*x)*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3756, 213} \[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=\frac {\log \left (\sqrt {2} \cos (3 x+2)-\sin (3 x+2)\right )}{6 \sqrt {2}}-\frac {\log \left (\sin (3 x+2)+\sqrt {2} \cos (3 x+2)\right )}{6 \sqrt {2}} \]

[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 213

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

Rule 3756

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])

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 {\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] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.36 \[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=-\frac {\text {arctanh}\left (\frac {\tan (2+3 x)}{\sqrt {2}}\right )}{3 \sqrt {2}} \]

[In]

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

[Out]

-1/3*ArcTanh[Tan[2 + 3*x]/Sqrt[2]]/Sqrt[2]

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.30

method	result	size
derivativedivides	\(-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\tan \left (2+3 x \right ) \sqrt {2}}{2}\right )}{6}\)	\(18\)
default	\(-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\tan \left (2+3 x \right ) \sqrt {2}}{2}\right )}{6}\)	\(18\)
risch	\(-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (2+3 x \right )}+\frac {1}{3}+\frac {2 i \sqrt {2}}{3}\right )}{12}+\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (2+3 x \right )}+\frac {1}{3}-\frac {2 i \sqrt {2}}{3}\right )}{12}\)	\(48\)

[In]

int(sec(2+3*x)^2/(-2+tan(2+3*x)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=\frac {1}{24} \, \sqrt {2} \log \left (-\frac {7 \, \cos \left (3 \, x + 2\right )^{4} - 10 \, \cos \left (3 \, x + 2\right )^{2} + 4 \, {\left (\sqrt {2} \cos \left (3 \, x + 2\right )^{3} + \sqrt {2} \cos \left (3 \, x + 2\right )\right )} \sin \left (3 \, x + 2\right ) - 1}{9 \, \cos \left (3 \, x + 2\right )^{4} - 6 \, \cos \left (3 \, x + 2\right )^{2} + 1}\right ) \]

[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))

Sympy [F]

\[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=\int \frac {\sec ^{2}{\left (3 x + 2 \right )}}{\tan ^{2}{\left (3 x + 2 \right )} - 2}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.52 \[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=\frac {1}{12} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \tan \left (3 \, x + 2\right )}{\sqrt {2} + \tan \left (3 \, x + 2\right )}\right ) \]

[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)))

Giac [A] (verification not implemented)

none

Time = 0.79 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=-\frac {1}{12} \, \sqrt {2} \log \left ({\left | \sqrt {2} + \tan \left (3 \, x + 2\right ) \right |}\right ) + \frac {1}{12} \, \sqrt {2} \log \left ({\left | -\sqrt {2} + \tan \left (3 \, x + 2\right ) \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 28.69 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.28 \[ \int \frac {\sec ^2(2+3 x)}{-2+\tan ^2(2+3 x)} \, dx=-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (3\,x+2\right )}{2}\right )}{6} \]

[In]

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

[Out]

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