3.1.38 \(\int (1+\tan (2 x))^2 \, dx\) [38]

Optimal. Leaf size=16 \[ -\log (\cos (2 x))+\frac {1}{2} \tan (2 x) \]

[Out]

-ln(cos(2*x))+1/2*tan(2*x)

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3558, 3556} \begin {gather*} \frac {1}{2} \tan (2 x)-\log (\cos (2 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + Tan[2*x])^2,x]

[Out]

-Log[Cos[2*x]] + Tan[2*x]/2

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3558

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[b^2*(Tan[c + d*x]/d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

\begin {align*} \int (1+\tan (2 x))^2 \, dx &=\frac {1}{2} \tan (2 x)+2 \int \tan (2 x) \, dx\\ &=-\log (\cos (2 x))+\frac {1}{2} \tan (2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 26, normalized size = 1.62 \begin {gather*} x-\frac {1}{2} \tan ^{-1}(\tan (2 x))-\log (\cos (2 x))+\frac {1}{2} \tan (2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tan[2*x])^2,x]

[Out]

x - ArcTan[Tan[2*x]]/2 - Log[Cos[2*x]] + Tan[2*x]/2

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Maple [A]
time = 0.03, size = 19, normalized size = 1.19

method	result	size
derivativedivides	\(\frac {\tan \left (2 x \right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (2 x \right )\right )}{2}\)	\(19\)
default	\(\frac {\tan \left (2 x \right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (2 x \right )\right )}{2}\)	\(19\)
norman	\(\frac {\tan \left (2 x \right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (2 x \right )\right )}{2}\)	\(19\)
risch	\(2 i x +\frac {i}{{\mathrm e}^{4 i x}+1}-\ln \left ({\mathrm e}^{4 i x}+1\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+tan(2*x))^2,x,method=_RETURNVERBOSE)

[Out]

1/2*tan(2*x)+1/2*ln(1+tan(2*x)^2)

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Maxima [A]
time = 2.55, size = 12, normalized size = 0.75 \begin {gather*} \log \left (\sec \left (2 \, x\right )\right ) + \frac {1}{2} \, \tan \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(2*x))^2,x, algorithm="maxima")

[Out]

log(sec(2*x)) + 1/2*tan(2*x)

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Fricas [A]
time = 0.67, size = 20, normalized size = 1.25 \begin {gather*} -\frac {1}{2} \, \log \left (\frac {1}{\tan \left (2 \, x\right )^{2} + 1}\right ) + \frac {1}{2} \, \tan \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(2*x))^2,x, algorithm="fricas")

[Out]

-1/2*log(1/(tan(2*x)^2 + 1)) + 1/2*tan(2*x)

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Sympy [A]
time = 0.05, size = 17, normalized size = 1.06 \begin {gather*} \frac {\log {\left (\tan ^{2}{\left (2 x \right )} + 1 \right )}}{2} + \frac {\tan {\left (2 x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(2*x))**2,x)

[Out]

log(tan(2*x)**2 + 1)/2 + tan(2*x)/2

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Giac [A]
time = 0.85, size = 22, normalized size = 1.38 \begin {gather*} -\frac {1}{2} \, \log \left (\frac {4}{\tan \left (2 \, x\right )^{2} + 1}\right ) + \frac {1}{2} \, \tan \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(2*x))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.27, size = 18, normalized size = 1.12 \begin {gather*} \frac {\mathrm {tan}\left (2\,x\right )}{2}+\frac {\ln \left ({\mathrm {tan}\left (2\,x\right )}^2+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(2*x) + 1)^2,x)

[Out]

tan(2*x)/2 + log(tan(2*x)^2 + 1)/2

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