\(\int \frac {1}{(a+i a \tan (c+d x))^2} \, dx\) [119]

Optimal result

Integrand size = 15, antiderivative size = 61 \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=\frac {x}{4 a^2}+\frac {i}{4 d (a+i a \tan (c+d x))^2}+\frac {i}{4 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3560, 8} \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=\frac {i}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {x}{4 a^2}+\frac {i}{4 d (a+i a \tan (c+d x))^2} \]

[In]

Int[(a + I*a*Tan[c + d*x])^(-2),x]

[Out]

x/(4*a^2) + (I/4)/(d*(a + I*a*Tan[c + d*x])^2) + (I/4)/(d*(a^2 + I*a^2*Tan[c + d*x]))

Rule 8

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

Rule 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=\frac {-2 i+\tan (c+d x)+\arctan (\tan (c+d x)) (-i+\tan (c+d x))^2}{4 a^2 d (-i+\tan (c+d x))^2} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^(-2),x]

[Out]

(-2*I + Tan[c + d*x] + ArcTan[Tan[c + d*x]]*(-I + Tan[c + d*x])^2)/(4*a^2*d*(-I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {x}{4 a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}\)	\(44\)
derivativedivides	\(\frac {\arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 a^{2} d \left (\tan \left (d x +c \right )-i\right )}\)	\(56\)
default	\(\frac {\arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 a^{2} d \left (\tan \left (d x +c \right )-i\right )}\)	\(56\)
norman	\(\frac {\frac {x}{4 a}+\frac {\tan ^{3}\left (d x +c \right )}{4 a d}+\frac {x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}+\frac {x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}+\frac {i}{2 a d}+\frac {3 \tan \left (d x +c \right )}{4 a d}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}\)	\(91\)

[In]

int(1/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x/a^2+1/4*I/a^2/d*exp(-2*I*(d*x+c))+1/16*I/a^2/d*exp(-4*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (4 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(4*d*x*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.92 \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 i a^{2} d e^{4 i c} e^{- 2 i d x} + 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} - \frac {1}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x}{4 a^{2}} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((16*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*d
**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((exp(4*I*c) + 2*exp(2*I*c) + 1)*exp(-4*I*c)/(4*a**2) - 1/(4*a**2)), Tr
ue)) + x/(4*a**2)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=-\frac {-\frac {2 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {2 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {-3 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right ) + 11 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(-2*I*log(tan(d*x + c) + I)/a^2 + 2*I*log(tan(d*x + c) - I)/a^2 + (-3*I*tan(d*x + c)^2 - 10*tan(d*x + c)
+ 11*I)/(a^2*(tan(d*x + c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 4.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.64 \[ \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx=\frac {x}{4\,a^2}-\frac {\frac {\mathrm {tan}\left (c+d\,x\right )}{4}-\frac {1}{2}{}\mathrm {i}}{a^2\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \]

[In]

int(1/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

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