3.105 \(\int \cos (x) \tan (2 x) \, dx\)

Optimal. Leaf size=20 \[ \frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{\sqrt{2}}-\cos (x) \]

[Out]

ArcTanh[Sqrt[2]*Cos[x]]/Sqrt[2] - Cos[x]

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Rubi [A]  time = 0.0254347, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {12, 321, 207} \[ \frac{\tanh ^{-1}\left (\sqrt{2} \cos (x)\right )}{\sqrt{2}}-\cos (x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*Tan[2*x],x]

[Out]

ArcTanh[Sqrt[2]*Cos[x]]/Sqrt[2] - Cos[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

Mathematica [C]  time = 0.23959, size = 183, normalized size = 9.15 \[ \frac{-4 \sqrt{2} \cos (x)+4 \tanh ^{-1}\left (\tan \left (\frac{x}{2}\right )+\sqrt{2}\right )-\log \left (-\sqrt{2} \sin (x)-\sqrt{2} \cos (x)+2\right )+\log \left (-\sqrt{2} \sin (x)+\sqrt{2} \cos (x)+2\right )+2 i \tan ^{-1}\left (\frac{\cos \left (\frac{x}{2}\right )-\left (\sqrt{2}-1\right ) \sin \left (\frac{x}{2}\right )}{\left (1+\sqrt{2}\right ) \cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )}\right )-2 i \tan ^{-1}\left (\frac{\cos \left (\frac{x}{2}\right )-\left (1+\sqrt{2}\right ) \sin \left (\frac{x}{2}\right )}{\left (\sqrt{2}-1\right ) \cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )}\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Tan[2*x],x]

[Out]

((2*I)*ArcTan[(Cos[x/2] - (-1 + Sqrt[2])*Sin[x/2])/((1 + Sqrt[2])*Cos[x/2] - Sin[x/2])] - (2*I)*ArcTan[(Cos[x/
2] - (1 + Sqrt[2])*Sin[x/2])/((-1 + Sqrt[2])*Cos[x/2] - Sin[x/2])] + 4*ArcTanh[Sqrt[2] + Tan[x/2]] - 4*Sqrt[2]
*Cos[x] - Log[2 - Sqrt[2]*Cos[x] - Sqrt[2]*Sin[x]] + Log[2 + Sqrt[2]*Cos[x] - Sqrt[2]*Sin[x]])/(4*Sqrt[2])

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Maple [A]  time = 0.015, size = 18, normalized size = 0.9 \begin{align*} -\cos \left ( x \right ) +{\frac{{\it Artanh} \left ( \cos \left ( x \right ) \sqrt{2} \right ) \sqrt{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*tan(2*x),x)

[Out]

-cos(x)+1/2*arctanh(cos(x)*2^(1/2))*2^(1/2)

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Maxima [B]  time = 1.57688, size = 180, normalized size = 9. \begin{align*} \frac{1}{8} \, \sqrt{2} \log \left (2 \, \sqrt{2} \sin \left (2 \, x\right ) \sin \left (x\right ) + 2 \,{\left (\sqrt{2} \cos \left (x\right ) + 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} + 2 \, \sqrt{2} \cos \left (x\right ) + 1\right ) - \frac{1}{8} \, \sqrt{2} \log \left (-2 \, \sqrt{2} \sin \left (2 \, x\right ) \sin \left (x\right ) - 2 \,{\left (\sqrt{2} \cos \left (x\right ) - 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} - 2 \, \sqrt{2} \cos \left (x\right ) + 1\right ) - \cos \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*tan(2*x),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*log(2*sqrt(2)*sin(2*x)*sin(x) + 2*(sqrt(2)*cos(x) + 1)*cos(2*x) + cos(2*x)^2 + 2*cos(x)^2 + sin(2*
x)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) + 1) - 1/8*sqrt(2)*log(-2*sqrt(2)*sin(2*x)*sin(x) - 2*(sqrt(2)*cos(x) - 1
)*cos(2*x) + cos(2*x)^2 + 2*cos(x)^2 + sin(2*x)^2 + 2*sin(x)^2 - 2*sqrt(2)*cos(x) + 1) - cos(x)

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Fricas [B]  time = 2.82491, size = 109, normalized size = 5.45 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (-\frac{2 \, \cos \left (x\right )^{2} + 2 \, \sqrt{2} \cos \left (x\right ) + 1}{2 \, \cos \left (x\right )^{2} - 1}\right ) - \cos \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*tan(2*x),x, algorithm="fricas")

[Out]

1/4*sqrt(2)*log(-(2*cos(x)^2 + 2*sqrt(2)*cos(x) + 1)/(2*cos(x)^2 - 1)) - cos(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*tan(2*x),x)

[Out]

Integral(cos(x)*tan(2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*tan(2*x),x, algorithm="giac")

[Out]

integrate(cos(x)*tan(2*x), x)