\(\int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\) [169]

Optimal result

Integrand size = 29, antiderivative size = 46 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {2 i \cos (c+d x)}{a^2 d}+\frac {2 \sin (c+d x)}{a^2 d} \]

[Out]

-arctanh(sin(d*x+c))/a^2/d+2*I*cos(d*x+c)/a^2/d+2*sin(d*x+c)/a^2/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3171, 3169, 2717, 2718, 2672, 327, 212} \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {2 \sin (c+d x)}{a^2 d}+\frac {2 i \cos (c+d x)}{a^2 d} \]

[In]

Int[Sec[c + d*x]/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

-(ArcTanh[Sin[c + d*x]]/(a^2*d)) + ((2*I)*Cos[c + d*x])/(a^2*d) + (2*Sin[c + d*x])/(a^2*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \sec (c+d x) (i a \cos (c+d x)+a \sin (c+d x))^2 \, dx}{a^4} \\ & = -\frac {\int \left (-a^2 \cos (c+d x)+2 i a^2 \sin (c+d x)+a^2 \sin (c+d x) \tan (c+d x)\right ) \, dx}{a^4} \\ & = -\frac {(2 i) \int \sin (c+d x) \, dx}{a^2}+\frac {\int \cos (c+d x) \, dx}{a^2}-\frac {\int \sin (c+d x) \tan (c+d x) \, dx}{a^2} \\ & = \frac {2 i \cos (c+d x)}{a^2 d}+\frac {\sin (c+d x)}{a^2 d}-\frac {\text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{a^2 d} \\ & = \frac {2 i \cos (c+d x)}{a^2 d}+\frac {2 \sin (c+d x)}{a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{a^2 d} \\ & = -\frac {\text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {2 i \cos (c+d x)}{a^2 d}+\frac {2 \sin (c+d x)}{a^2 d} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(184\) vs. \(2(46)=92\).

Time = 0.36 (sec) , antiderivative size = 184, normalized size of antiderivative = 4.00 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {\sec ^2(c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (2 i+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\left (2+i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {3}{2} (c+d x)\right )+i \sin \left (\frac {3}{2} (c+d x)\right )\right )}{a^2 d (-i+\tan (c+d x))^2} \]

[In]

Integrate[Sec[c + d*x]/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

-((Sec[c + d*x]^2*(Cos[(c + d*x)/2]*(2*I + Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]]) + (2 + I*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - I*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]])*Sin[(c + d*x)/2])*(Cos[(3*(c + d*x))/2] + I*Sin[(3*(c + d*x))/2]))/(a^2*d*(-I + Tan[c + d*x])^2))

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}}{a^{2} d}\)	\(54\)
default	\(\frac {-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}}{a^{2} d}\)	\(54\)
risch	\(\frac {2 i {\mathrm e}^{-i \left (d x +c \right )}}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{a^{2} d}\)	\(61\)
norman	\(\frac {\frac {4 i}{a d}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}\)	\(87\)

[In]

int(sec(d*x+c)/(cos(d*x+c)*a+I*a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {{\left (e^{\left (i \, d x + i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - e^{\left (i \, d x + i \, c\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 2 i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{- \sin ^{2}{\left (c + d x \right )} + 2 i \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} + \cos ^{2}{\left (c + d x \right )}}\, dx}{a^{2}} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)/(-sin(c + d*x)**2 + 2*I*sin(c + d*x)*cos(c + d*x) + cos(c + d*x)**2), x)/a**2

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. 117 vs. \(2 (44) = 88\).

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.54 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {-2 i \, \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) - 2 i \, \arctan \left (\cos \left (d x + c\right ), -\sin \left (d x + c\right ) + 1\right ) - 4 i \, \cos \left (d x + c\right ) + \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right ) - 4 \, \sin \left (d x + c\right )}{2 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(-2*I*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 2*I*arctan2(cos(d*x + c), -sin(d*x + c) + 1) - 4*I*cos(d*
x + c) + log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*s
in(d*x + c) + 1) - 4*sin(d*x + c))/(a^2*d)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.24 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {\frac {\log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2}} - \frac {\log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{2}} - \frac {4}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}}}{d} \]

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-(log(tan(1/2*d*x + 1/2*c) + 1)/a^2 - log(tan(1/2*d*x + 1/2*c) - 1)/a^2 - 4/(a^2*(tan(1/2*d*x + 1/2*c) - I)))/
d

Mupad [B] (verification not implemented)

Time = 22.95 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \frac {\sec (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {4{}\mathrm {i}}{a^2\,d\,\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )} \]

[In]

int(1/(cos(c + d*x)*(a*cos(c + d*x) + a*sin(c + d*x)*1i)^2),x)

[Out]

4i/(a^2*d*(tan(c/2 + (d*x)/2)*1i + 1)) - (2*atanh(tan(c/2 + (d*x)/2)))/(a^2*d)