3.306 \(\int \frac{\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=354 \[ -\frac{4 i \tan ^{-1}\left (\frac{\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac{2 i \log (\tan (c+d x))}{3 a^{4/3} d}-\frac{2 i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}} \]

[Out]

x/(8*2^(1/3)*a^(4/3)) - ((4*I)*ArcTan[(a^(1/3) + 2*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*
a^(4/3)*d) - ((I/4)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))])/(2^(1/
3)*a^(4/3)*d) - ((I/8)*Log[Cos[c + d*x]])/(2^(1/3)*a^(4/3)*d) + (((2*I)/3)*Log[Tan[c + d*x]])/(a^(4/3)*d) - ((
2*I)*Log[a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(a^(4/3)*d) - (((3*I)/8)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[
c + d*x])^(1/3)])/(2^(1/3)*a^(4/3)*d) - ((11*I)/8)/(d*(a + I*a*Tan[c + d*x])^(4/3)) - Cot[c + d*x]/(d*(a + I*a
*Tan[c + d*x])^(4/3)) - ((19*I)/4)/(a*d*(a + I*a*Tan[c + d*x])^(1/3))

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Rubi [A]  time = 0.710554, antiderivative size = 354, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {3561, 3596, 3600, 3481, 55, 617, 204, 31, 3599} \[ -\frac{4 i \tan ^{-1}\left (\frac{\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac{2 i \log (\tan (c+d x))}{3 a^{4/3} d}-\frac{2 i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

x/(8*2^(1/3)*a^(4/3)) - ((4*I)*ArcTan[(a^(1/3) + 2*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*
a^(4/3)*d) - ((I/4)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))])/(2^(1/
3)*a^(4/3)*d) - ((I/8)*Log[Cos[c + d*x]])/(2^(1/3)*a^(4/3)*d) + (((2*I)/3)*Log[Tan[c + d*x]])/(a^(4/3)*d) - ((
2*I)*Log[a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(a^(4/3)*d) - (((3*I)/8)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[
c + d*x])^(1/3)])/(2^(1/3)*a^(4/3)*d) - ((11*I)/8)/(d*(a + I*a*Tan[c + d*x])^(4/3)) - Cot[c + d*x]/(d*(a + I*a
*Tan[c + d*x])^(4/3)) - ((19*I)/4)/(a*d*(a + I*a*Tan[c + d*x])^(1/3))

Rule 3561

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(c^2 + d^2)*
(n + 1)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*T
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^
2 + d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || 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 3600

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

Rule 3481

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3599

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

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx &=-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}+\frac{\int \frac{\cot (c+d x) \left (-\frac{4 i a}{3}-\frac{7}{3} a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{4/3}} \, dx}{a}\\ &=-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}+\frac{3 \int \frac{\cot (c+d x) \left (-\frac{32 i a^2}{9}-\frac{44}{9} a^2 \tan (c+d x)\right )}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{8 a^3}\\ &=-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{9 \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} \left (-\frac{64 i a^3}{27}-\frac{76}{27} a^3 \tan (c+d x)\right ) \, dx}{16 a^5}\\ &=-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{(4 i) \int \cot (c+d x) (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{2/3} \, dx}{3 a^3}-\frac{\int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2}\\ &=-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{i \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{4 a d}-\frac{(4 i) \operatorname{Subst}\left (\int \frac{1}{x \sqrt [3]{a+i a x}} \, dx,x,\tan (c+d x)\right )}{3 a d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{2 i \log (\tan (c+d x))}{3 a^{4/3} d}-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{a^{4/3} d}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 a d}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{a d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{2 i \log (\tan (c+d x))}{3 a^{4/3} d}-\frac{2 i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{(4 i) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{a^{4/3} d}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}\\ &=\frac{x}{8 \sqrt [3]{2} a^{4/3}}-\frac{4 i \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} a^{4/3} d}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac{2 i \log (\tan (c+d x))}{3 a^{4/3} d}-\frac{2 i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{a^{4/3} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{11 i}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac{\cot (c+d x)}{d (a+i a \tan (c+d x))^{4/3}}-\frac{19 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.87601, size = 233, normalized size = 0.66 \[ \frac{6 e^{4 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )+64 e^{4 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{2 e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )+63 e^{2 i (c+d x)}-35 e^{4 i (c+d x)}-95 e^{6 i (c+d x)}+3}{8 a d \left (-1+e^{2 i (c+d x)}\right ) \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

(3 + 63*E^((2*I)*(c + d*x)) - 35*E^((4*I)*(c + d*x)) - 95*E^((6*I)*(c + d*x)) + 6*E^((4*I)*(c + d*x))*(-1 + E^
((2*I)*(c + d*x)))*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))] + 64*E^((4*I)
*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*Hypergeometric2F1[2/3, 1, 5/3, (2*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c
+ d*x)))])/(8*a*d*(-1 + E^((2*I)*(c + d*x)))*(1 + E^((2*I)*(c + d*x)))^2*(-I + Tan[c + d*x])*(a + I*a*Tan[c +
d*x])^(1/3))

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Maple [F]  time = 0.122, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cot \left ( dx+c \right ) \right ) ^{2} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x)

[Out]

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82143, size = 2327, normalized size = 6.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

1/32*(2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-95*I*e^(6*I*d*x + 6*I*c) - 35*I*e^(4*I*d*x + 4*I*c) + 63*I
*e^(2*I*d*x + 2*I*c) + 3*I)*e^(4/3*I*d*x + 4/3*I*c) + 32*(a^2*d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4*I*c
))*(64/27*I/(a^4*d^3))^(1/3)*log(9/16*a^3*d^2*(64/27*I/(a^4*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1)
)^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + 32*(a^2*d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4*I*c))*(1/128*I/(a^4*d^
3))^(1/3)*log(32*a^3*d^2*(1/128*I/(a^4*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x
+ 2/3*I*c)) + ((-16*I*sqrt(3)*a^2*d - 16*a^2*d)*e^(6*I*d*x + 6*I*c) + (16*I*sqrt(3)*a^2*d + 16*a^2*d)*e^(4*I*d
*x + 4*I*c))*(64/27*I/(a^4*d^3))^(1/3)*log(1/32*(9*I*sqrt(3)*a^3*d^2 - 9*a^3*d^2)*(64/27*I/(a^4*d^3))^(2/3) +
2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + ((16*I*sqrt(3)*a^2*d - 16*a^2*d)*e^(6*I
*d*x + 6*I*c) + (-16*I*sqrt(3)*a^2*d + 16*a^2*d)*e^(4*I*d*x + 4*I*c))*(64/27*I/(a^4*d^3))^(1/3)*log(1/32*(-9*I
*sqrt(3)*a^3*d^2 - 9*a^3*d^2)*(64/27*I/(a^4*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I
*d*x + 2/3*I*c)) + ((-16*I*sqrt(3)*a^2*d - 16*a^2*d)*e^(6*I*d*x + 6*I*c) + (16*I*sqrt(3)*a^2*d + 16*a^2*d)*e^(
4*I*d*x + 4*I*c))*(1/128*I/(a^4*d^3))^(1/3)*log((16*I*sqrt(3)*a^3*d^2 - 16*a^3*d^2)*(1/128*I/(a^4*d^3))^(2/3)
+ 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + ((16*I*sqrt(3)*a^2*d - 16*a^2*d)*e^(6
*I*d*x + 6*I*c) + (-16*I*sqrt(3)*a^2*d + 16*a^2*d)*e^(4*I*d*x + 4*I*c))*(1/128*I/(a^4*d^3))^(1/3)*log((-16*I*s
qrt(3)*a^3*d^2 - 16*a^3*d^2)*(1/128*I/(a^4*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*
d*x + 2/3*I*c)))/(a^2*d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4*I*c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{2}{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2/(a+I*a*tan(d*x+c))**(4/3),x)

[Out]

Integral(cot(c + d*x)**2/(a*(I*tan(c + d*x) + 1))**(4/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate(cot(d*x + c)^2/(I*a*tan(d*x + c) + a)^(4/3), x)