\(\int (a+b \sec (c+d x)) \tan (c+d x) \, dx\) [259]

Optimal result

Integrand size = 17, antiderivative size = 25 \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=-\frac {a \log (\cos (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3969, 3556, 2686, 8} \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=\frac {b \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

-((a*Log[Cos[c + d*x]])/d) + (b*Sec[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3556

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

Rule 3969

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=-\frac {a \log (\cos (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \]

[In]

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

[Out]

-((a*Log[Cos[c + d*x]])/d) + (b*Sec[c + d*x])/d

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {b \sec \left (d x +c \right )+a \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(23\)
default	\(\frac {b \sec \left (d x +c \right )+a \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(23\)
parts	\(\frac {a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}+\frac {b \sec \left (d x +c \right )}{d}\)	\(30\)
risch	\(i a x +\frac {2 i a c}{d}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=-\frac {a \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - b}{d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=\begin {cases} \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\left (c \right )}\right ) \tan {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=-\frac {a \log \left (\cos \left (d x + c\right )\right ) - \frac {b}{\cos \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (25) = 50\).

Time = 0.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.28 \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=\frac {a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {a + 2 \, b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1}}{d} \]

[In]

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

[Out]

(a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1
)) + (a + 2*b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/d

Mupad [B] (verification not implemented)

Time = 14.90 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int (a+b \sec (c+d x)) \tan (c+d x) \, dx=\frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)^2))/d - (2*b)/(d*(tan(c/2 + (d*x)/2)^2 - 1))