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

Optimal. Leaf size=17 \[ -\frac {\log (\cos (c+d x)+1)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3879, 31} \[ -\frac {\log (\cos (c+d x)+1)}{a d} \]

Antiderivative was successfully verified.

[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 3879

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*} \int \frac {\tan (c+d x)}{a+a \sec (c+d x)} \, dx &=-\frac {\operatorname {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*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.12 \[ -\frac {2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Log[Cos[(c + d*x)/2]])/(a*d)

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fricas [A]  time = 0.76, size = 19, normalized size = 1.12 \[ -\frac {\log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.28, size = 31, normalized size = 1.82 \[ \frac {\log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.12, size = 33, normalized size = 1.94 \[ \frac {\ln \left (\sec \left (d x +c \right )\right )}{a d}-\frac {\ln \left (1+\sec \left (d x +c \right )\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 17, normalized size = 1.00 \[ -\frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(cos(d*x + c) + 1)/(a*d)

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mupad [B]  time = 1.21, size = 21, normalized size = 1.24 \[ \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 3.80, size = 41, normalized size = 2.41 \[ \begin {cases} \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a d} - \frac {\log {\left (\sec {\left (c + d x \right )} + 1 \right )}}{a d} & \text {for}\: d \neq 0 \\\frac {x \tan {\relax (c )}}{a \sec {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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