3.73 \(\int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=98 \[ \frac {7}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\log (\sin (c+d x))}{a^3 d}-\frac {7 i x}{8 a^3}+\frac {3}{8 a d (a+i a \tan (c+d x))^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3} \]

[Out]

-7/8*I*x/a^3+ln(sin(d*x+c))/a^3/d+1/6/d/(a+I*a*tan(d*x+c))^3+3/8/a/d/(a+I*a*tan(d*x+c))^2+7/8/d/(a^3+I*a^3*tan
(d*x+c))

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Rubi [A]  time = 0.23, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3559, 3596, 3531, 3475} \[ \frac {7}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\log (\sin (c+d x))}{a^3 d}-\frac {7 i x}{8 a^3}+\frac {3}{8 a d (a+i a \tan (c+d x))^2}+\frac {1}{6 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((-7*I)/8)*x)/a^3 + Log[Sin[c + d*x]]/(a^3*d) + 1/(6*d*(a + I*a*Tan[c + d*x])^3) + 3/(8*a*d*(a + I*a*Tan[c +
d*x])^2) + 7/(8*d*(a^3 + I*a^3*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 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 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]

Rubi steps

\begin {align*} \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot (c+d x) (6 a-3 i a \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {3}{8 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot (c+d x) \left (24 a^2-18 i a^2 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {3}{8 a d (a+i a \tan (c+d x))^2}+\frac {7}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (48 a^3-42 i a^3 \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=-\frac {7 i x}{8 a^3}+\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {3}{8 a d (a+i a \tan (c+d x))^2}+\frac {7}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \, dx}{a^3}\\ &=-\frac {7 i x}{8 a^3}+\frac {\log (\sin (c+d x))}{a^3 d}+\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {3}{8 a d (a+i a \tan (c+d x))^2}+\frac {7}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 118, normalized size = 1.20 \[ \frac {\sec ^3(c+d x) (-51 \sin (c+d x)+84 i d x \sin (3 (c+d x))+2 \sin (3 (c+d x))+81 i \cos (c+d x)-96 \sin (3 (c+d x)) \log (\sin (c+d x))+\cos (3 (c+d x)) (96 i \log (\sin (c+d x))+84 d x+2 i))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*((81*I)*Cos[c + d*x] + Cos[3*(c + d*x)]*(2*I + 84*d*x + (96*I)*Log[Sin[c + d*x]]) - 51*Sin[c +
d*x] + 2*Sin[3*(c + d*x)] + (84*I)*d*x*Sin[3*(c + d*x)] - 96*Log[Sin[c + d*x]]*Sin[3*(c + d*x)]))/(96*a^3*d*(
-I + Tan[c + d*x])^3)

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fricas [A]  time = 0.43, size = 77, normalized size = 0.79 \[ \frac {{\left (-180 i \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 \, e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 66 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(-180*I*d*x*e^(6*I*d*x + 6*I*c) + 96*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) + 66*e^(4*I*d*x + 4
*I*c) + 15*e^(2*I*d*x + 2*I*c) + 2)*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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giac [A]  time = 2.63, size = 93, normalized size = 0.95 \[ -\frac {\frac {90 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {6 \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} - \frac {96 \, \log \left (\tan \left (d x + c\right )\right )}{a^{3}} - \frac {165 \, \tan \left (d x + c\right )^{3} - 579 i \, \tan \left (d x + c\right )^{2} - 699 \, \tan \left (d x + c\right ) + 301 i}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(90*log(tan(d*x + c) - I)/a^3 + 6*log(I*tan(d*x + c) - 1)/a^3 - 96*log(tan(d*x + c))/a^3 - (165*tan(d*x
+ c)^3 - 579*I*tan(d*x + c)^2 - 699*tan(d*x + c) + 301*I)/(a^3*(tan(d*x + c) - I)^3))/d

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maple [A]  time = 0.50, size = 111, normalized size = 1.13 \[ -\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}+\frac {i}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {7 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {3}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {15 \ln \left (\tan \left (d x +c \right )-i\right )}{16 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a+I*a*tan(d*x+c))^3,x)

[Out]

-1/16/d/a^3*ln(tan(d*x+c)+I)+1/a^3/d*ln(tan(d*x+c))+1/6*I/d/a^3/(tan(d*x+c)-I)^3-7/8*I/d/a^3/(tan(d*x+c)-I)-3/
8/d/a^3/(tan(d*x+c)-I)^2-15/16/d/a^3*ln(tan(d*x+c)-I)

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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)/(a+I*a*tan(d*x+c))^3,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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mupad [B]  time = 3.87, size = 120, normalized size = 1.22 \[ \frac {\frac {17}{12\,a^3}-\frac {7\,{\mathrm {tan}\left (c+d\,x\right )}^2}{8\,a^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {15\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

((tan(c + d*x)*17i)/(8*a^3) + 17/(12*a^3) - (7*tan(c + d*x)^2)/(8*a^3))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2
- tan(c + d*x)^3*1i + 1)) - (15*log(tan(c + d*x) - 1i))/(16*a^3*d) - log(tan(c + d*x) + 1i)/(16*a^3*d) + log(
tan(c + d*x))/(a^3*d)

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sympy [A]  time = 0.61, size = 184, normalized size = 1.88 \[ \begin {cases} \frac {\left (16896 a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 3840 a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {i \left (15 e^{6 i c} + 11 e^{4 i c} + 5 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} + \frac {15 i}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {15 i x}{8 a^{3}} + \frac {\log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((16896*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 3840*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) + 512*a**6*d**
2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(24576*a**9*d**3*exp(12*I*c), 0)), (x*(-I*(15*ex
p(6*I*c) + 11*exp(4*I*c) + 5*exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) + 15*I/(8*a**3)), True)) - 15*I*x/(8*a**3) +
log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d)

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