\(\int \sec (c+d x) (a+b \tan (c+d x)) \, dx\) [512]

Optimal result

Integrand size = 17, antiderivative size = 24 \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \]

[Out]

a*arctanh(sin(d*x+c))/d+b*sec(d*x+c)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3567, 3855} \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \]

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (b*Sec[c + d*x])/d

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {b \sec (c+d x)}{d}+a \int \sec (c+d x) \, dx \\ & = \frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \]

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (b*Sec[c + d*x])/d

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {\frac {b}{\cos \left (d x +c \right )}+a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(32\)
default	\(\frac {\frac {b}{\cos \left (d x +c \right )}+a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(32\)
risch	\(\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(67\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.25 \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, b}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} \frac {a \log {\left (\tan {\left (c + d x \right )} + \sec {\left (c + d x \right )} \right )} + b \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right ) \sec {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac {b}{\cos \left (d x + c\right )}}{d} \]

[In]

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

[Out]

(a*log(sec(d*x + c) + tan(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. 54 vs. \(2 (24) = 48\).

Time = 0.37 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.25 \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, b}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.53 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \sec (c+d x) (a+b \tan (c+d x)) \, dx=\frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

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