\(\int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx\) [60]

Optimal result

Integrand size = 19, antiderivative size = 17 \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\log (1+\cos (c+d x))}{a d} \]

[Out]

-ln(1+cos(d*x+c))/a/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3964, 31} \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\log (\cos (c+d x)+1)}{a d} \]

[In]

Int[Tan[c + d*x]/(a + a*Sec[c + d*x]),x]

[Out]

-(Log[1 + Cos[c + d*x]]/(a*d))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3964

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{a+a x} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {\log (1+\cos (c+d x))}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \]

[In]

Integrate[Tan[c + d*x]/(a + a*Sec[c + d*x]),x]

[Out]

(-2*Log[Cos[(c + d*x)/2]])/(a*d)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\(\frac {-\ln \left (1+\sec \left (d x +c \right )\right )+\ln \left (\sec \left (d x +c \right )\right )}{d a}\)	\(27\)
default	\(\frac {-\ln \left (1+\sec \left (d x +c \right )\right )+\ln \left (\sec \left (d x +c \right )\right )}{d a}\)	\(27\)
risch	\(\frac {i x}{a}+\frac {2 i c}{d a}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}\)	\(39\)

[In]

int(tan(d*x+c)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-ln(1+sec(d*x+c))+ln(sec(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{a d} \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-log(1/2*cos(d*x + c) + 1/2)/(a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (14) = 28\).

Time = 2.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.41 \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=\begin {cases} \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a d} - \frac {\log {\left (\sec {\left (c + d x \right )} + 1 \right )}}{a d} & \text {for}\: d \neq 0 \\\frac {x \tan {\left (c \right )}}{a \sec {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c)),x)

[Out]

Piecewise((log(tan(c + d*x)**2 + 1)/(2*a*d) - log(sec(c + d*x) + 1)/(a*d), Ne(d, 0)), (x*tan(c)/(a*sec(c) + a)
, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a d} \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

-log(cos(d*x + c) + 1)/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.82 \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a d} \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/(a*d)

Mupad [B] (verification not implemented)

Time = 14.49 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \]

[In]

int(tan(c + d*x)/(a + a/cos(c + d*x)),x)

[Out]

log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d)