3.296 \(\int \frac {\cot ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=327 \[ -\frac {i \tan ^{-1}\left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} d}-\frac {i \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {i \log (\tan (c+d x))}{6 \sqrt [3]{a} d}-\frac {i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

[Out]

1/8*x*2^(2/3)/a^(1/3)-1/8*I*ln(cos(d*x+c))*2^(2/3)/a^(1/3)/d+1/6*I*ln(tan(d*x+c))/a^(1/3)/d-1/2*I*ln(a^(1/3)-(
a+I*a*tan(d*x+c))^(1/3))/a^(1/3)/d-3/8*I*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(2/3)/a^(1/3)/d-1/3*I*
arctan(1/3*(a^(1/3)+2*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))/a^(1/3)/d*3^(1/2)-1/4*I*arctan(1/3*(a^(1/3)+2
^(2/3)*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)*2^(2/3)/a^(1/3)/d-5/2*I/d/(a+I*a*tan(d*x+c))^(1/3)-c
ot(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/3)

________________________________________________________________________________________

Rubi [A]  time = 0.58, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ -\frac {i \tan ^{-1}\left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} d}-\frac {i \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {i \log (\tan (c+d x))}{6 \sqrt [3]{a} d}-\frac {i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

x/(4*2^(1/3)*a^(1/3)) - (I*ArcTan[(a^(1/3) + 2*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(1
/3)*d) - ((I/2)*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
^(1/3)*d) - ((I/4)*Log[Cos[c + d*x]])/(2^(1/3)*a^(1/3)*d) + ((I/6)*Log[Tan[c + d*x]])/(a^(1/3)*d) - ((I/2)*Log
[a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(a^(1/3)*d) - (((3*I)/4)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x]
)^(1/3)])/(2^(1/3)*a^(1/3)*d) - ((5*I)/2)/(d*(a + I*a*Tan[c + d*x])^(1/3)) - Cot[c + d*x]/(d*(a + I*a*Tan[c +
d*x])^(1/3))

Rule 31

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

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 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 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 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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {\int \frac {\cot (c+d x) \left (-\frac {i a}{3}-\frac {4}{3} a \tan (c+d x)\right )}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{a}\\ &=-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} \left (-\frac {2 i a^2}{9}-\frac {5}{9} a^2 \tan (c+d x)\right ) \, dx}{2 a^3}\\ &=-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {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^2}-\frac {\int (a+i a \tan (c+d x))^{2/3} \, dx}{2 a}\\ &=-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{a+i a x}} \, dx,x,\tan (c+d x)\right )}{3 d}+\frac {i \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\tan (c+d x))}{6 \sqrt [3]{a} d}-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {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 )}{2 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 )}{4 d}+\frac {i \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a} 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 )}{4 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\tan (c+d x))}{6 \sqrt [3]{a} d}-\frac {i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {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 )}{\sqrt [3]{a} 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 )}{2 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {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} \sqrt [3]{a} 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 )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\tan (c+d x))}{6 \sqrt [3]{a} d}-\frac {i \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {5 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.45, size = 179, normalized size = 0.55 \[ \frac {\csc (c+d x) \sec (c+d x) \left (3 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (i \sin (2 (c+d x))+\cos (2 (c+d x))-1)+4 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {2 e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (i \sin (2 (c+d x))+\cos (2 (c+d x))-1)-20 i \sin (2 (c+d x))-8 \cos (2 (c+d x))-8\right )}{16 d \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

(Csc[c + d*x]*Sec[c + d*x]*(-8 - 8*Cos[2*(c + d*x)] + 3*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1
+ E^((2*I)*(c + d*x)))]*(-1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + 4*Hypergeometric2F1[2/3, 1, 5/3, (2*E^(
(2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - (20*I)*Sin[2*(c +
d*x)]))/(16*d*(a + I*a*Tan[c + d*x])^(1/3))

________________________________________________________________________________________

fricas [B]  time = 0.48, size = 717, normalized size = 2.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

1/4*(2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-7*I*e^(4*I*d*x + 4*I*c) - 4*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^
(4/3*I*d*x + 4/3*I*c) + 4*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*(1/16*I/(a*d^3))^(1/3)*log(8*a*d
^2*(1/16*I/(a*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)) + 4*(a*d*e^(4
*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*(1/27*I/(a*d^3))^(1/3)*log(9*a*d^2*(1/27*I/(a*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)) - 2*((-I*sqrt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c)
+ (I*sqrt(3)*a*d - a*d)*e^(2*I*d*x + 2*I*c))*(1/16*I/(a*d^3))^(1/3)*log(-4*(I*sqrt(3)*a*d^2 + a*d^2)*(1/16*I/(
a*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)) - 2*((I*sqrt(3)*a*d + a*d
)*e^(4*I*d*x + 4*I*c) + (-I*sqrt(3)*a*d - a*d)*e^(2*I*d*x + 2*I*c))*(1/16*I/(a*d^3))^(1/3)*log(-4*(-I*sqrt(3)*
a*d^2 + a*d^2)*(1/16*I/(a*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)) -
2*((-I*sqrt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c) + (I*sqrt(3)*a*d - a*d)*e^(2*I*d*x + 2*I*c))*(1/27*I/(a*d^3))^(
1/3)*log(-9/2*(I*sqrt(3)*a*d^2 + a*d^2)*(1/27*I/(a*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)) - 2*((I*sqrt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c) + (-I*sqrt(3)*a*d - a*d)*e^(2*I*d*x + 2
*I*c))*(1/27*I/(a*d^3))^(1/3)*log(-9/2*(-I*sqrt(3)*a*d^2 + a*d^2)*(1/27*I/(a*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*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.79, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{2}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 286, normalized size = 0.87 \[ -\frac {i \, a {\left (\frac {6 \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {4}{3}}} - \frac {3 \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {4}{3}}} + \frac {6 \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {4}{3}}} + \frac {8 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {4}{3}}} - \frac {12 \, {\left (-5 i \, a \tan \left (d x + c\right ) - 2 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{2}} - \frac {4 \, \log \left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {4}{3}}} + \frac {8 \, \log \left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{a^{\frac {4}{3}}}\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

-1/24*I*a*(6*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(I*a*tan(d*x + c) + a)^(1/3))/a^(
1/3))/a^(4/3) - 3*2^(2/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*tan(d*x +
c) + a)^(2/3))/a^(4/3) + 6*2^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3))/a^(4/3) + 8*sqrt(3)*ar
ctan(1/3*sqrt(3)*(2*(I*a*tan(d*x + c) + a)^(1/3) + a^(1/3))/a^(1/3))/a^(4/3) - 12*(-5*I*a*tan(d*x + c) - 2*a)/
((I*a*tan(d*x + c) + a)^(4/3)*a - (I*a*tan(d*x + c) + a)^(1/3)*a^2) - 4*log((I*a*tan(d*x + c) + a)^(2/3) + (I*
a*tan(d*x + c) + a)^(1/3)*a^(1/3) + a^(2/3))/a^(4/3) + 8*log((I*a*tan(d*x + c) + a)^(1/3) - a^(1/3))/a^(4/3))/
d

________________________________________________________________________________________

mupad [B]  time = 4.57, size = 887, normalized size = 2.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(1/3),x)

[Out]

log(a^5*d*(a + a*tan(c + d*x)*1i)^(1/3)*36i - ((5832*a^7*d^6*(1i/(16*a*d^3))^(2/3) + 675*a^6*d^4*(a + a*tan(c
+ d*x)*1i)^(1/3))*(1i/(16*a*d^3))^(1/3) - a^6*d^3*315i)*(1i/(16*a*d^3))^(2/3))*(1i/(16*a*d^3))^(1/3) + log(a^5
*d*(a + a*tan(c + d*x)*1i)^(1/3)*36i - ((5832*a^7*d^6*(1i/(27*a*d^3))^(2/3) + 675*a^6*d^4*(a + a*tan(c + d*x)*
1i)^(1/3))*(1i/(27*a*d^3))^(1/3) - a^6*d^3*315i)*(1i/(27*a*d^3))^(2/3))*(1i/(27*a*d^3))^(1/3) + (((a + a*tan(c
+ d*x)*1i)*5i)/(2*d) - (a*3i)/(2*d))/(a*(a + a*tan(c + d*x)*1i)^(1/3) - (a + a*tan(c + d*x)*1i)^(4/3)) + (log
(((3^(1/2)*1i - 1)^2*(a^6*d^3*315i - ((3^(1/2)*1i - 1)*(675*a^6*d^4*(a + a*tan(c + d*x)*1i)^(1/3) + 1458*a^7*d
^6*(3^(1/2)*1i - 1)^2*(1i/(16*a*d^3))^(2/3))*(1i/(16*a*d^3))^(1/3))/2)*(1i/(16*a*d^3))^(2/3))/4 + a^5*d*(a + a
*tan(c + d*x)*1i)^(1/3)*36i)*(3^(1/2)*1i - 1)*(1i/(16*a*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(a^6*d^3*315
i + ((3^(1/2)*1i + 1)*(675*a^6*d^4*(a + a*tan(c + d*x)*1i)^(1/3) + 1458*a^7*d^6*(3^(1/2)*1i + 1)^2*(1i/(16*a*d
^3))^(2/3))*(1i/(16*a*d^3))^(1/3))/2)*(1i/(16*a*d^3))^(2/3))/4 + a^5*d*(a + a*tan(c + d*x)*1i)^(1/3)*36i)*(3^(
1/2)*1i + 1)*(1i/(16*a*d^3))^(1/3))/2 + (log(((3^(1/2)*1i - 1)^2*(a^6*d^3*315i - ((3^(1/2)*1i - 1)*(675*a^6*d^
4*(a + a*tan(c + d*x)*1i)^(1/3) + 1458*a^7*d^6*(3^(1/2)*1i - 1)^2*(1i/(27*a*d^3))^(2/3))*(1i/(27*a*d^3))^(1/3)
)/2)*(1i/(27*a*d^3))^(2/3))/4 + a^5*d*(a + a*tan(c + d*x)*1i)^(1/3)*36i)*(3^(1/2)*1i - 1)*(1i/(27*a*d^3))^(1/3
))/2 - (log(((3^(1/2)*1i + 1)^2*(a^6*d^3*315i + ((3^(1/2)*1i + 1)*(675*a^6*d^4*(a + a*tan(c + d*x)*1i)^(1/3) +
1458*a^7*d^6*(3^(1/2)*1i + 1)^2*(1i/(27*a*d^3))^(2/3))*(1i/(27*a*d^3))^(1/3))/2)*(1i/(27*a*d^3))^(2/3))/4 + a
^5*d*(a + a*tan(c + d*x)*1i)^(1/3)*36i)*(3^(1/2)*1i + 1)*(1i/(27*a*d^3))^(1/3))/2

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{2}{\left (c + d x \right )}}{\sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________