3.1.80 \(\int \frac {1}{(1+\tan (x))^2} \, dx\) [80]

Optimal. Leaf size=21 \[ \frac {1}{2} \log (\cos (x)+\sin (x))-\frac {1}{2 (1+\tan (x))} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3564, 3611} \begin {gather*} \frac {1}{2} \log (\sin (x)+\cos (x))-\frac {1}{2 (\tan (x)+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + Tan[x])^(-2),x]

[Out]

Log[Cos[x] + Sin[x]]/2 - 1/(2*(1 + Tan[x]))

Rule 3564

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*
(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 27, normalized size = 1.29 \begin {gather*} \frac {\log (\cos (x)+\sin (x))+\tan (x)+\log (\cos (x)+\sin (x)) \tan (x)}{2+2 \tan (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tan[x])^(-2),x]

[Out]

(Log[Cos[x] + Sin[x]] + Tan[x] + Log[Cos[x] + Sin[x]]*Tan[x])/(2 + 2*Tan[x])

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Maple [A]
time = 0.03, size = 26, normalized size = 1.24

method	result	size
derivativedivides	\(-\frac {1}{2 \left (\tan \left (x \right )+1\right )}+\frac {\ln \left (\tan \left (x \right )+1\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (x \right )\right )}{4}\)	\(26\)
default	\(-\frac {1}{2 \left (\tan \left (x \right )+1\right )}+\frac {\ln \left (\tan \left (x \right )+1\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (x \right )\right )}{4}\)	\(26\)
norman	\(-\frac {1}{2 \left (\tan \left (x \right )+1\right )}+\frac {\ln \left (\tan \left (x \right )+1\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (x \right )\right )}{4}\)	\(26\)
risch	\(-\frac {i x}{2}-\frac {1}{2 \left ({\mathrm e}^{2 i x}+i\right )}+\frac {\ln \left ({\mathrm e}^{2 i x}+i\right )}{2}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.97, size = 25, normalized size = 1.19 \begin {gather*} -\frac {1}{2 \, {\left (\tan \left (x\right ) + 1\right )}} - \frac {1}{4} \, \log \left (\tan \left (x\right )^{2} + 1\right ) + \frac {1}{2} \, \log \left (\tan \left (x\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).
time = 1.01, size = 37, normalized size = 1.76 \begin {gather*} \frac {{\left (\tan \left (x\right ) + 1\right )} \log \left (\frac {\tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) + 1}{\tan \left (x\right )^{2} + 1}\right ) + \tan \left (x\right ) - 1}{4 \, {\left (\tan \left (x\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (17) = 34\).
time = 0.16, size = 75, normalized size = 3.57 \begin {gather*} \frac {2 \log {\left (\tan {\left (x \right )} + 1 \right )} \tan {\left (x \right )}}{4 \tan {\left (x \right )} + 4} + \frac {2 \log {\left (\tan {\left (x \right )} + 1 \right )}}{4 \tan {\left (x \right )} + 4} - \frac {\log {\left (\tan ^{2}{\left (x \right )} + 1 \right )} \tan {\left (x \right )}}{4 \tan {\left (x \right )} + 4} - \frac {\log {\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{4 \tan {\left (x \right )} + 4} - \frac {2}{4 \tan {\left (x \right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.64, size = 26, normalized size = 1.24 \begin {gather*} -\frac {1}{2 \, {\left (\tan \left (x\right ) + 1\right )}} - \frac {1}{4} \, \log \left (\tan \left (x\right )^{2} + 1\right ) + \frac {1}{2} \, \log \left ({\left | \tan \left (x\right ) + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.25, size = 27, normalized size = 1.29 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (x\right )+1\right )}{2}-\frac {\ln \left ({\mathrm {tan}\left (x\right )}^2+1\right )}{4}-\frac {1}{2\,\left (\mathrm {tan}\left (x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(x) + 1)^2,x)

[Out]

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

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