3.383 \(\int \cos ^2(x) \sec (3 x) \, dx\)

Optimal. Leaf size=9 \[ \frac{1}{2} \tanh ^{-1}(2 \sin (x)) \]

[Out]

ArcTanh[2*Sin[x]]/2

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Rubi [A]  time = 0.0224052, antiderivative size = 9, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {206} \[ \frac{1}{2} \tanh ^{-1}(2 \sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2*Sec[3*x],x]

[Out]

ArcTanh[2*Sin[x]]/2

Rule 206

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

Rubi steps

\begin{align*} \int \cos ^2(x) \sec (3 x) \, dx &=\operatorname{Subst}\left (\int \frac{1}{1-4 x^2} \, dx,x,\sin (x)\right )\\ &=\frac{1}{2} \tanh ^{-1}(2 \sin (x))\\ \end{align*}

Mathematica [A]  time = 0.0066682, size = 9, normalized size = 1. \[ \frac{1}{2} \tanh ^{-1}(2 \sin (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2*Sec[3*x],x]

[Out]

ArcTanh[2*Sin[x]]/2

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Maple [B]  time = 0.04, size = 20, normalized size = 2.2 \begin{align*} -{\frac{\ln \left ( 2\,\sin \left ( x \right ) -1 \right ) }{4}}+{\frac{\ln \left ( 1+2\,\sin \left ( x \right ) \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/cos(3*x),x)

[Out]

-1/4*ln(2*sin(x)-1)+1/4*ln(1+2*sin(x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (x\right )^{2}}{\cos \left (3 \, x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(x)^2/cos(3*x), x)

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Fricas [B]  time = 2.19179, size = 65, normalized size = 7.22 \begin{align*} \frac{1}{4} \, \log \left (2 \, \sin \left (x\right ) + 1\right ) - \frac{1}{4} \, \log \left (-2 \, \sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(2*sin(x) + 1) - 1/4*log(-2*sin(x) + 1)

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Sympy [B]  time = 6.42523, size = 76, normalized size = 8.44 \begin{align*} - \frac{\log{\left (\sin{\left (3 x \right )} - 1 \right )}}{12} + \frac{\log{\left (\sin{\left (3 x \right )} + 1 \right )}}{12} - \frac{\log{\left (\tan{\left (\frac{x}{2} \right )} - 1 \right )}}{6} + \frac{\log{\left (\tan{\left (\frac{x}{2} \right )} + 1 \right )}}{6} - \frac{\log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} - 4 \tan{\left (\frac{x}{2} \right )} + 1 \right )}}{12} + \frac{\log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + 4 \tan{\left (\frac{x}{2} \right )} + 1 \right )}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**2/cos(3*x),x)

[Out]

-log(sin(3*x) - 1)/12 + log(sin(3*x) + 1)/12 - log(tan(x/2) - 1)/6 + log(tan(x/2) + 1)/6 - log(tan(x/2)**2 - 4
*tan(x/2) + 1)/12 + log(tan(x/2)**2 + 4*tan(x/2) + 1)/12

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Giac [B]  time = 1.10587, size = 28, normalized size = 3.11 \begin{align*} \frac{1}{4} \, \log \left ({\left | 2 \, \sin \left (x\right ) + 1 \right |}\right ) - \frac{1}{4} \, \log \left ({\left | 2 \, \sin \left (x\right ) - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(abs(2*sin(x) + 1)) - 1/4*log(abs(2*sin(x) - 1))