\(\int (a+i a \tan (c+d x))^2 \, dx\) [23]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int (a+i a \tan (c+d x))^2 \, dx=2 a^2 x-\frac {2 i a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \tan (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3558, 3556} \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x \]

[In]

Int[(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 3556

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

Rule 3558

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {i a (-2 a \log (i+\tan (c+d x))-i a \tan (c+d x))}{d} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^2,x]

[Out]

((-I)*a*(-2*a*Log[I + Tan[c + d*x]] - I*a*Tan[c + d*x]))/d

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\tan \left (d x +c \right )+i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(40\)
default	\(\frac {a^{2} \left (-\tan \left (d x +c \right )+i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(40\)
parallelrisch	\(\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 a^{2} d x -a^{2} \tan \left (d x +c \right )}{d}\)	\(41\)
norman	\(2 a^{2} x -\frac {a^{2} \tan \left (d x +c \right )}{d}+\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(42\)
parts	\(a^{2} x +\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(51\)
risch	\(-\frac {4 a^{2} c}{d}-\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(54\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.47 \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (i \, a^{2} + {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int (a+i a \tan (c+d x))^2 \, dx=- \frac {2 i a^{2}}{d e^{2 i c} e^{2 i d x} + d} - \frac {2 i a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int (a+i a \tan (c+d x))^2 \, dx=a^{2} x + \frac {{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2}}{d} + \frac {2 i \, a^{2} \log \left (\sec \left (d x + c\right )\right )}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.74 \[ \int (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + i \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + i \, a^{2}\right )}}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\left (-\mathrm {tan}\left (c+d\,x\right )+\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}\right )}{d} \]

[In]

int((a + a*tan(c + d*x)*1i)^2,x)

[Out]

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