3.1.53 \(\int \frac {\cot ^2(c+d x)}{a+i a \tan (c+d x)} \, dx\) [53]

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

[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))

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Rubi [A]
time = 0.07, 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 = {3633, 3610, 3612, 3556} \begin {gather*} -\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} \end {gather*}

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 3556

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

Rule 3610

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 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 3633

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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(286\) vs. \(2(70)=140\).
time = 0.71, size = 286, normalized size = 4.09 \begin {gather*} \frac {\csc \left (\frac {c}{2}\right ) \csc (c+d x) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (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 \left (\sin ^2(c+d x)\right )+2 \cos (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )+10 i \sin (c)-4 d x \sin (c)-4 i \log \left (\sin ^2(c+d x)\right ) \sin (c)+16 i \text {ArcTan}(\tan (d x)) \sin (c) (\cos (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 \left (\sin ^2(c+d x)\right ) \sin (c+2 d x)-i \sin (3 c+2 d x)+2 d x \sin (3 c+2 d x)+2 i \log \left (\sin ^2(c+d x)\right ) \sin (3 c+2 d x)\right )}{32 a d (-i+\tan (c+d x))} \end {gather*}

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]))

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Maple [A]
time = 0.24, size = 68, normalized size = 0.97

method	result	size
derivativedivides	\(\frac {\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{4}-\frac {1}{2 \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{\tan \left (d x +c \right )}-i \ln \left (\tan \left (d x +c \right )\right )-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{4}}{d a}\)	\(68\)
default	\(\frac {\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{4}-\frac {1}{2 \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{\tan \left (d x +c \right )}-i \ln \left (\tan \left (d x +c \right )\right )-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{4}}{d a}\)	\(68\)
risch	\(-\frac {5 x}{2 a}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d a}-\frac {2 c}{d a}-\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\)	\(78\)
norman	\(\frac {-\frac {1}{d a}-\frac {3 x \tan \left (d x +c \right )}{2 a}-\frac {3 x \left (\tan ^{3}\left (d x +c \right )\right )}{2 a}-\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i \tan \left (d x +c \right )}{2 d a}}{\tan \left (d x +c \right ) \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {i \ln \left (\tan \left (d x +c \right )\right )}{d a}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}\)	\(125\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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.

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Fricas [A]
time = 0.38, size = 99, normalized size = 1.41 \begin {gather*} -\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 )} + 4 \, {\left (i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 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 )}} \end {gather*}

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) - 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))

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Sympy [A]
time = 0.19, size = 116, normalized size = 1.66 \begin {gather*} \begin {cases} - \frac {i e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 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} \end {gather*}

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(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)

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Giac [A]
time = 0.58, size = 91, normalized size = 1.30 \begin {gather*} -\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} \end {gather*}

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

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Mupad [B]
time = 3.99, size = 96, normalized size = 1.37 \begin {gather*} \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} \end {gather*}

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)

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