3.82 \(\int \frac{\cot (c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=120 \[ \frac{15}{16 a^4 d (1+i \tan (c+d x))}+\frac{7}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{\log (\sin (c+d x))}{a^4 d}-\frac{15 i x}{16 a^4}+\frac{1}{4 a d (a+i a \tan (c+d x))^3}+\frac{1}{8 d (a+i a \tan (c+d x))^4} \]

[Out]

(((-15*I)/16)*x)/a^4 + Log[Sin[c + d*x]]/(a^4*d) + 7/(16*a^4*d*(1 + I*Tan[c + d*x])^2) + 15/(16*a^4*d*(1 + I*T
an[c + d*x])) + 1/(8*d*(a + I*a*Tan[c + d*x])^4) + 1/(4*a*d*(a + I*a*Tan[c + d*x])^3)

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Rubi [A]  time = 0.320424, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3559, 3596, 3531, 3475} \[ \frac{15}{16 a^4 d (1+i \tan (c+d x))}+\frac{7}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{\log (\sin (c+d x))}{a^4 d}-\frac{15 i x}{16 a^4}+\frac{1}{4 a d (a+i a \tan (c+d x))^3}+\frac{1}{8 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(((-15*I)/16)*x)/a^4 + Log[Sin[c + d*x]]/(a^4*d) + 7/(16*a^4*d*(1 + I*Tan[c + d*x])^2) + 15/(16*a^4*d*(1 + I*T
an[c + d*x])) + 1/(8*d*(a + I*a*Tan[c + d*x])^4) + 1/(4*a*d*(a + I*a*Tan[c + d*x])^3)

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*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] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3596

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

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 3475

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

Rubi steps

\begin{align*} \int \frac{\cot (c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=\frac{1}{8 d (a+i a \tan (c+d x))^4}+\frac{\int \frac{\cot (c+d x) (8 a-4 i a \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2}\\ &=\frac{1}{8 d (a+i a \tan (c+d x))^4}+\frac{1}{4 a d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\cot (c+d x) \left (48 a^2-36 i a^2 \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4}\\ &=\frac{7}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{1}{8 d (a+i a \tan (c+d x))^4}+\frac{1}{4 a d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\cot (c+d x) \left (192 a^3-168 i a^3 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6}\\ &=\frac{7}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{1}{8 d (a+i a \tan (c+d x))^4}+\frac{1}{4 a d (a+i a \tan (c+d x))^3}+\frac{15}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\int \cot (c+d x) \left (384 a^4-360 i a^4 \tan (c+d x)\right ) \, dx}{384 a^8}\\ &=-\frac{15 i x}{16 a^4}+\frac{7}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{1}{8 d (a+i a \tan (c+d x))^4}+\frac{1}{4 a d (a+i a \tan (c+d x))^3}+\frac{15}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\int \cot (c+d x) \, dx}{a^4}\\ &=-\frac{15 i x}{16 a^4}+\frac{\log (\sin (c+d x))}{a^4 d}+\frac{7}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{1}{8 d (a+i a \tan (c+d x))^4}+\frac{1}{4 a d (a+i a \tan (c+d x))^3}+\frac{15}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.397964, size = 123, normalized size = 1.02 \[ \frac{\sec ^4(c+d x) (96 i \sin (2 (c+d x))+120 d x \sin (4 (c+d x))-i \sin (4 (c+d x))+112 \cos (2 (c+d x))+128 i \sin (4 (c+d x)) \log (\sin (c+d x))+\cos (4 (c+d x)) (128 \log (\sin (c+d x))-120 i d x+1)+32)}{128 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^4*(32 + 112*Cos[2*(c + d*x)] + Cos[4*(c + d*x)]*(1 - (120*I)*d*x + 128*Log[Sin[c + d*x]]) + (96*
I)*Sin[2*(c + d*x)] - I*Sin[4*(c + d*x)] + 120*d*x*Sin[4*(c + d*x)] + (128*I)*Log[Sin[c + d*x]]*Sin[4*(c + d*x
)]))/(128*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.089, size = 130, normalized size = 1.1 \begin{align*}{\frac{{\frac{i}{4}}}{d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{15\,i}{16}}}{d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{1}{8\,d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}-{\frac{7}{16\,d{a}^{4} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{31\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{32\,d{a}^{4}}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{32\,d{a}^{4}}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) }{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a+I*a*tan(d*x+c))^4,x)

[Out]

1/4*I/d/a^4/(tan(d*x+c)-I)^3-15/16*I/d/a^4/(tan(d*x+c)-I)+1/8/d/a^4/(tan(d*x+c)-I)^4-7/16/d/a^4/(tan(d*x+c)-I)
^2-31/32/d/a^4*ln(tan(d*x+c)-I)-1/32/d/a^4*ln(tan(d*x+c)+I)+1/d/a^4*ln(tan(d*x+c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.32243, size = 274, normalized size = 2.28 \begin{align*} \frac{{\left (-248 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 128 \, e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 104 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 32 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{128 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/128*(-248*I*d*x*e^(8*I*d*x + 8*I*c) + 128*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) + 104*e^(6*I*d*x
+ 6*I*c) + 32*e^(4*I*d*x + 4*I*c) + 8*e^(2*I*d*x + 2*I*c) + 1)*e^(-8*I*d*x - 8*I*c)/(a^4*d)

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Sympy [A]  time = 4.90161, size = 221, normalized size = 1.84 \begin{align*} \begin{cases} \frac{\left (106496 a^{12} d^{3} e^{18 i c} e^{- 2 i d x} + 32768 a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 8192 a^{12} d^{3} e^{14 i c} e^{- 6 i d x} + 1024 a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{131072 a^{16} d^{4}} & \text{for}\: 131072 a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac{\left (31 i e^{8 i c} + 26 i e^{6 i c} + 16 i e^{4 i c} + 6 i e^{2 i c} + i\right ) e^{- 8 i c}}{16 a^{4}} + \frac{31 i}{16 a^{4}}\right ) & \text{otherwise} \end{cases} - \frac{31 i x}{16 a^{4}} + \frac{\log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((106496*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) + 32768*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + 8192*a*
*12*d**3*exp(14*I*c)*exp(-6*I*d*x) + 1024*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(131072*a**16*d**
4), Ne(131072*a**16*d**4*exp(20*I*c), 0)), (x*(-(31*I*exp(8*I*c) + 26*I*exp(6*I*c) + 16*I*exp(4*I*c) + 6*I*exp
(2*I*c) + I)*exp(-8*I*c)/(16*a**4) + 31*I/(16*a**4)), True)) - 31*I*x/(16*a**4) + log(exp(2*I*d*x) - exp(-2*I*
c))/(a**4*d)

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Giac [A]  time = 1.28059, size = 138, normalized size = 1.15 \begin{align*} -\frac{\frac{12 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} + \frac{372 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac{384 \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac{775 \, \tan \left (d x + c\right )^{4} - 3460 i \, \tan \left (d x + c\right )^{3} - 5898 \, \tan \left (d x + c\right )^{2} + 4612 i \, \tan \left (d x + c\right ) + 1447}{a^{4}{\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/384*(12*log(tan(d*x + c) + I)/a^4 + 372*log(tan(d*x + c) - I)/a^4 - 384*log(abs(tan(d*x + c)))/a^4 - (775*t
an(d*x + c)^4 - 3460*I*tan(d*x + c)^3 - 5898*tan(d*x + c)^2 + 4612*I*tan(d*x + c) + 1447)/(a^4*(tan(d*x + c) -
I)^4))/d