3.53 \(\int \frac {\cot ^2(c+d x)}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=70 \[ -\frac {3 \cot (c+d x)}{2 a d}-\frac {i \log (\sin (c+d x))}{a d}+\frac {\cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {3 x}{2 a} \]

[Out]

-3/2*x/a-3/2*cot(d*x+c)/a/d-I*ln(sin(d*x+c))/a/d+1/2*cot(d*x+c)/d/(a+I*a*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3552, 3529, 3531, 3475} \[ -\frac {3 \cot (c+d x)}{2 a d}-\frac {i \log (\sin (c+d x))}{a d}+\frac {\cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {3 x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(2*a) - (3*Cot[c + d*x])/(2*a*d) - (I*Log[Sin[c + d*x]])/(a*d) + Cot[c + d*x]/(2*d*(a + I*a*Tan[c + d*x
]))

Rule 3475

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

Rule 3529

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

Rule 3531

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 3552

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

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x)}{a+i a \tan (c+d x)} \, dx &=\frac {\cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \cot ^2(c+d x) (-3 a+2 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac {3 \cot (c+d x)}{2 a d}+\frac {\cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \cot (c+d x) (2 i a+3 a \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac {3 x}{2 a}-\frac {3 \cot (c+d x)}{2 a d}+\frac {\cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \int \cot (c+d x) \, dx}{a}\\ &=-\frac {3 x}{2 a}-\frac {3 \cot (c+d x)}{2 a d}-\frac {i \log (\sin (c+d x))}{a d}+\frac {\cot (c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.66, size = 286, normalized size = 4.09 \[ \frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \csc (c+d x) \sec (c+d x) \left (-4 d x \sin (c)-2 d x \sin (c+2 d x)-7 i \sin (c+2 d x)+2 d x \sin (3 c+2 d x)-i \sin (3 c+2 d x)+2 i d x \cos (c+2 d x)-9 \cos (c+2 d x)-2 i d x \cos (3 c+2 d x)+\cos (3 c+2 d x)-4 i \sin (c) \log \left (\sin ^2(c+d x)\right )-2 i \sin (c+2 d x) \log \left (\sin ^2(c+d x)\right )+2 i \sin (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )-2 \cos (c+2 d x) \log \left (\sin ^2(c+d x)\right )+2 \cos (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )+16 i \sin (c) \tan ^{-1}(\tan (d x)) \sin (c+d x) (\cos (c+d x)+i \sin (c+d x))+10 i \sin (c)+8 \cos (c)\right )}{32 a d (\tan (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c/2]*Csc[c + d*x]*Sec[c/2]*Sec[c + d*x]*(8*Cos[c] - 9*Cos[c + 2*d*x] + (2*I)*d*x*Cos[c + 2*d*x] + Cos[3*c
+ 2*d*x] - (2*I)*d*x*Cos[3*c + 2*d*x] - 2*Cos[c + 2*d*x]*Log[Sin[c + d*x]^2] + 2*Cos[3*c + 2*d*x]*Log[Sin[c +
d*x]^2] + (10*I)*Sin[c] - 4*d*x*Sin[c] - (4*I)*Log[Sin[c + d*x]^2]*Sin[c] + (16*I)*ArcTan[Tan[d*x]]*Sin[c]*(C
os[c + d*x] + I*Sin[c + d*x])*Sin[c + d*x] - (7*I)*Sin[c + 2*d*x] - 2*d*x*Sin[c + 2*d*x] - (2*I)*Log[Sin[c + d
*x]^2]*Sin[c + 2*d*x] - I*Sin[3*c + 2*d*x] + 2*d*x*Sin[3*c + 2*d*x] + (2*I)*Log[Sin[c + d*x]^2]*Sin[3*c + 2*d*
x]))/(32*a*d*(-I + Tan[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 99, normalized size = 1.41 \[ -\frac {10 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (10 \, d x - 9 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (-4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - i}{4 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(10*d*x*e^(4*I*d*x + 4*I*c) - (10*d*x - 9*I)*e^(2*I*d*x + 2*I*c) - (-4*I*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I
*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - I)/(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

________________________________________________________________________________________

giac [A]  time = 1.83, size = 91, normalized size = 1.30 \[ -\frac {-\frac {10 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {2 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac {8 i \, \log \left (\tan \left (d x + c\right )\right )}{a} + \frac {\tan \left (d x + c\right )^{2} - 13 i \, \tan \left (d x + c\right ) - 8}{{\left (-i \, \tan \left (d x + c\right )^{2} - \tan \left (d x + c\right )\right )} a}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-10*I*log(tan(d*x + c) - I)/a + 2*I*log(-I*tan(d*x + c) + 1)/a + 8*I*log(tan(d*x + c))/a + (tan(d*x + c)
^2 - 13*I*tan(d*x + c) - 8)/((-I*tan(d*x + c)^2 - tan(d*x + c))*a))/d

________________________________________________________________________________________

maple [A]  time = 0.40, size = 91, normalized size = 1.30 \[ -\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {1}{d a \tan \left (d x +c \right )}-\frac {i \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{4 d a}-\frac {1}{2 d a \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c)),x)

[Out]

-1/4*I/d/a*ln(tan(d*x+c)+I)-1/d/a/tan(d*x+c)-I/a/d*ln(tan(d*x+c))+5/4*I/a/d*ln(tan(d*x+c)-I)-1/2/d/a/(tan(d*x+
c)-I)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

mupad [B]  time = 3.99, size = 96, normalized size = 1.37 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,5{}\mathrm {i}}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a\,d}-\frac {\frac {1}{a}+\frac {\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}}{2\,a}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,1{}\mathrm {i}}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c + d*x) - 1i)*5i)/(4*a*d) - (log(tan(c + d*x) + 1i)*1i)/(4*a*d) - ((tan(c + d*x)*3i)/(2*a) + 1/a)/(d
*(tan(c + d*x) + tan(c + d*x)^2*1i)) - (log(tan(c + d*x))*1i)/(a*d)

________________________________________________________________________________________

sympy [A]  time = 0.37, size = 117, normalized size = 1.67 \[ \begin {cases} - \frac {i e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 4 a d e^{2 i c} \neq 0 \\x \left (\frac {\left (- 5 e^{2 i c} - 1\right ) e^{- 2 i c}}{2 a} + \frac {5}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {2 i}{- a d e^{2 i c} e^{2 i d x} + a d} - \frac {5 x}{2 a} - \frac {i \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(4*a*d*exp(2*I*c), 0)), (x*((-5*exp(2*I*c) - 1)*exp(-2*I*c)
/(2*a) + 5/(2*a)), True)) + 2*I/(-a*d*exp(2*I*c)*exp(2*I*d*x) + a*d) - 5*x/(2*a) - I*log(exp(2*I*d*x) - exp(-2
*I*c))/(a*d)

________________________________________________________________________________________