\(\int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx\) [49]

Optimal result

Integrand size = 19, antiderivative size = 38 \[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

arctanh(sin(d*x+c))/a/d-sin(d*x+c)/d/(a+a*cos(d*x+c))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2826, 3855, 2727} \[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{d (a \cos (c+d x)+a)} \]

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(a*d) - Sin[c + d*x]/(d*(a + a*Cos[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2826

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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 {\int \sec (c+d x) \, dx}{a}-\int \frac {1}{a+a \cos (c+d x)} \, dx \\ & = \frac {\text {arctanh}(\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{d (a+a \cos (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(103\) vs. \(2(38)=76\).

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.71 \[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )\right )}{a d (1+\cos (c+d x))} \]

[In]

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

[Out]

(-2*Cos[(c + d*x)/2]*(Cos[(c + d*x)/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]]) + Sec[c/2]*Sin[(d*x)/2]))/(a*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(46\)
default	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(46\)
parallelrisch	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\)	\(46\)
norman	\(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\)	\(58\)
risch	\(-\frac {2 i}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}\)	\(65\)

[In]

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

[Out]

1/d/a*(-tan(1/2*d*x+1/2*c)-ln(tan(1/2*d*x+1/2*c)-1)+ln(tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.71 \[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

Integral(sec(c + d*x)/(cos(c + d*x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.97 \[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

[In]

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

[Out]

(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - sin(d*x + c)/(a*(co
s(d*x + c) + 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.48 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {\sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \]

[In]

int(1/(cos(c + d*x)*(a + a*cos(c + d*x))),x)

[Out]

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