\(\int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\) [156]

Optimal result

Integrand size = 29, antiderivative size = 23 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {x}{a}+\frac {i \log (\cos (c+d x))}{a d} \]

[Out]

x/a+I*ln(cos(d*x+c))/a/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3171, 3169, 3556} \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {x}{a}+\frac {i \log (\cos (c+d x))}{a d} \]

[In]

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

[Out]

x/a + (I*Log[Cos[c + d*x]])/(a*d)

Rule 3169

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3171

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 0]

Rule 3556

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

Rubi steps \begin{align*} \text {integral}& = -\frac {i \int \sec (c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2} \\ & = -\frac {i \int (i a+a \tan (c+d x)) \, dx}{a^2} \\ & = \frac {x}{a}-\frac {i \int \tan (c+d x) \, dx}{a} \\ & = \frac {x}{a}+\frac {i \log (\cos (c+d x))}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {i \log (i-\tan (c+d x))}{a d} \]

[In]

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

[Out]

((-I)*Log[I - Tan[c + d*x]])/(a*d)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(-\frac {i \ln \left (i \tan \left (d x +c \right )+1\right )}{d a}\)	\(22\)
default	\(-\frac {i \ln \left (i \tan \left (d x +c \right )+1\right )}{d a}\)	\(22\)
risch	\(\frac {2 x}{a}+\frac {2 c}{a d}+\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a d}\)	\(38\)
norman	\(\frac {x}{a}+\frac {i \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}+\frac {i \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {i \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a d}\)	\(72\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {2 \, d x + i \, \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{a d} \]

[In]

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

[Out]

(2*d*x + I*log(e^(2*I*d*x + 2*I*c) + 1))/(a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (21) = 42\).

Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 4.39 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {-\frac {i \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {i \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {i \, \log \left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}{a}}{d} \]

[In]

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

[Out]

-(-I*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - I*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + I*log(-2*I*si
n(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/a)/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.48 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {-\frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a} + \frac {2 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a} - \frac {i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.63 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {\sec (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {\left (2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\right )\,1{}\mathrm {i}}{a\,d} \]

[In]

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

[Out]

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