3.2.9 \(\int \tan (a+b x) \, dx\) [109]

Optimal. Leaf size=12 \[ -\frac {\log (\cos (a+b x))}{b} \]

[Out]

-ln(cos(b*x+a))/b

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Rubi [A]
time = 0.00, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3556} \begin {gather*} -\frac {\log (\cos (a+b x))}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[a + b*x],x]

[Out]

-(Log[Cos[a + b*x]]/b)

Rule 3556

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} -\frac {\log (\cos (a+b x))}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + b*x],x]

[Out]

-(Log[Cos[a + b*x]]/b)

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Maple [A]
time = 0.01, size = 17, normalized size = 1.42

method	result	size
derivativedivides	\(\frac {\ln \left (1+\tan ^{2}\left (b x +a \right )\right )}{2 b}\)	\(17\)
default	\(\frac {\ln \left (1+\tan ^{2}\left (b x +a \right )\right )}{2 b}\)	\(17\)
norman	\(\frac {\ln \left (1+\tan ^{2}\left (b x +a \right )\right )}{2 b}\)	\(17\)
risch	\(i x +\frac {2 i a}{b}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 1.57, size = 11, normalized size = 0.92 \begin {gather*} \frac {\log \left (\sec \left (b x + a\right )\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(b*x+a),x, algorithm="maxima")

[Out]

log(sec(b*x + a))/b

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Fricas [A]
time = 0.87, size = 18, normalized size = 1.50 \begin {gather*} -\frac {\log \left (\frac {1}{\tan \left (b x + a\right )^{2} + 1}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(b*x+a),x, algorithm="fricas")

[Out]

-1/2*log(1/(tan(b*x + a)^2 + 1))/b

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Sympy [A]
time = 0.05, size = 19, normalized size = 1.58 \begin {gather*} \begin {cases} \frac {\log {\left (\tan ^{2}{\left (a + b x \right )} + 1 \right )}}{2 b} & \text {for}\: b \neq 0 \\x \tan {\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(b*x+a),x)

[Out]

Piecewise((log(tan(a + b*x)**2 + 1)/(2*b), Ne(b, 0)), (x*tan(a), True))

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Giac [A]
time = 1.37, size = 13, normalized size = 1.08 \begin {gather*} -\frac {\log \left ({\left | \cos \left (b x + a\right ) \right |}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(b*x+a),x, algorithm="giac")

[Out]

-log(abs(cos(b*x + a)))/b

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Mupad [B]
time = 0.23, size = 16, normalized size = 1.33 \begin {gather*} \frac {\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a + b*x),x)

[Out]

log(tan(a + b*x)^2 + 1)/(2*b)

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