\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [19]

Optimal result

Integrand size = 24, antiderivative size = 38 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-2 a^2 x-\frac {a^2 \cot (c+d x)}{d}+\frac {2 i a^2 \log (\sin (c+d x))}{d} \]

[Out]

-2*a^2*x-a^2*cot(d*x+c)/d+2*I*a^2*ln(sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3623, 3612, 3556} \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \cot (c+d x)}{d}+\frac {2 i a^2 \log (\sin (c+d x))}{d}-2 a^2 x \]

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-2*a^2*x - (a^2*Cot[c + d*x])/d + ((2*I)*a^2*Log[Sin[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 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3623

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=a^2 \left (-\frac {\cot (c+d x)}{d}+\frac {2 i \log (\tan (c+d x))}{d}-\frac {2 i \log (i+\tan (c+d x))}{d}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.13

method	result	size
parallelrisch	\(\frac {a^{2} \left (2 i \ln \left (\tan \left (d x +c \right )\right )-i \ln \left (\sec ^{2}\left (d x +c \right )\right )-2 d x -\cot \left (d x +c \right )\right )}{d}\)	\(43\)
derivativedivides	\(\frac {-a^{2} \left (d x +c \right )+2 i a^{2} \ln \left (\sin \left (d x +c \right )\right )+a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(49\)
default	\(\frac {-a^{2} \left (d x +c \right )+2 i a^{2} \ln \left (\sin \left (d x +c \right )\right )+a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(49\)
risch	\(\frac {4 a^{2} c}{d}-\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(54\)
norman	\(\frac {-\frac {a^{2}}{d}-2 a^{2} x \tan \left (d x +c \right )}{\tan \left (d x +c \right )}+\frac {2 i a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (i \, a^{2} + {\left (-i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \]

[In]

integrate(cot(d*x+c)^2*(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 = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.34 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=- \frac {2 i a^{2}}{d e^{2 i c} e^{2 i d x} - d} + \frac {2 i a^{2} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} \]

[In]

integrate(cot(d*x+c)**2*(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 = 0.31 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.47 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (d x + c\right )} a^{2} + i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 i \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac {a^{2}}{\tan \left (d x + c\right )}}{d} \]

[In]

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

[Out]

-(2*(d*x + c)*a^2 + I*a^2*log(tan(d*x + c)^2 + 1) - 2*I*a^2*log(tan(d*x + c)) + a^2/tan(d*x + c))/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. 85 vs. \(2 (36) = 72\).

Time = 0.56 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.24 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {8 i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 4 i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {-4 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

[In]

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

[Out]

-1/2*(8*I*a^2*log(tan(1/2*d*x + 1/2*c) + I) - 4*I*a^2*log(tan(1/2*d*x + 1/2*c)) - a^2*tan(1/2*d*x + 1/2*c) - (
-4*I*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c))/d

Mupad [B] (verification not implemented)

Time = 4.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2\,\left (\mathrm {cot}\left (c+d\,x\right )+4\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\right )}{d} \]

[In]

int(cot(c + d*x)^2*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

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