3.4.5 \(\int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx\) [305]

Optimal. Leaf size=313 \[ -\frac {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{4/3} d}-\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {\log (\tan (c+d x))}{2 a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 a^{4/3} d}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \]

[Out]

-1/16*I*x*2^(2/3)/a^(4/3)-1/16*ln(cos(d*x+c))*2^(2/3)/a^(4/3)/d-1/2*ln(tan(d*x+c))/a^(4/3)/d+3/2*ln(a^(1/3)-(a
+I*a*tan(d*x+c))^(1/3))/a^(4/3)/d-3/16*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(2/3)/a^(4/3)/d+arctan(1
/3*(a^(1/3)+2*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)/a^(4/3)/d-1/8*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^(4/3)/d+3/8/d/(a+I*a*tan(d*x+c))^(4/3)+9/4/a/d/(a+I*
a*tan(d*x+c))^(1/3)

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Rubi [A]
time = 0.42, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3643, 3560, 3562, 57, 631, 210, 31, 3677, 3680} \begin {gather*} \frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{4/3} d}-\frac {\sqrt {3} \text {ArcTan}\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 {\log (\tan (c+d x))}{2 a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 a^{4/3} d}-\frac {3 \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 {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8*I)*x)/(2^(1/3)*a^(4/3)) + (Sqrt[3]*ArcTan[(a^(1/3) + 2*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))]
)/(a^(4/3)*d) - (Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))])/(4*2^(1/3
)*a^(4/3)*d) - Log[Cos[c + d*x]]/(8*2^(1/3)*a^(4/3)*d) - Log[Tan[c + d*x]]/(2*a^(4/3)*d) + (3*Log[a^(1/3) - (a
+ I*a*Tan[c + d*x])^(1/3)])/(2*a^(4/3)*d) - (3*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(8*2^(1/3
)*a^(4/3)*d) + 3/(8*d*(a + I*a*Tan[c + d*x])^(4/3)) + 9/(4*a*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 57

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 210

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

Rule 631

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 3560

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

Rule 3562

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 3643

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[d/(a*c - b*d), Int[(a + b*Tan[e + f*x])^m*((b + a*Tan[e
+ f*x])/(c + d*Tan[e + f*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]

Rule 3677

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 3680

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 (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx &=i \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx-\frac {i \int \frac {\cot (c+d x) (i a+a \tan (c+d x))}{(a+i a \tan (c+d x))^{4/3}} \, dx}{a}\\ &=\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac {(3 i) \int \frac {\cot (c+d x) \left (\frac {8 i a^2}{3}+\frac {8}{3} a^2 \tan (c+d x)\right )}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{8 a^3}+\frac {i \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{2 a}\\ &=\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {(9 i) \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} \left (\frac {16 i a^3}{9}+\frac {16}{9} a^3 \tan (c+d x)\right ) \, dx}{16 a^5}+\frac {i \int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2}\\ &=\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{4 a d}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt [3]{a+i a x}} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac {i x}{8 \sqrt [3]{2} a^{4/3}}-\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {\log (\tan (c+d x))}{2 a^{4/3} d}+\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 a^{4/3} d}+\frac {3 \text {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 \text {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 {3 \text {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 a d}\\ &=-\frac {i x}{8 \sqrt [3]{2} a^{4/3}}-\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {\log (\tan (c+d x))}{2 a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 a^{4/3} d}-\frac {3 \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 {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \text {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 \text {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 {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{4/3} d}-\frac {\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 {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {\log (\tan (c+d x))}{2 a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 a^{4/3} d}-\frac {3 \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 {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 1.69, size = 189, normalized size = 0.60 \begin {gather*} -\frac {3 i \sec ^2(c+d x) \left (7+7 \cos (2 (c+d x))+\, _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 ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-8 \, _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 ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+6 i \sin (2 (c+d x))\right )}{16 a d (-i+\tan (c+d x)) \sqrt [3]{a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/16)*Sec[c + d*x]^2*(7 + 7*Cos[2*(c + d*x)] + Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 +
E^((2*I)*(c + d*x)))]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - 8*Hypergeometric2F1[2/3, 1, 5/3, (2*E^((2*I)*(
c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + (6*I)*Sin[2*(c + d*x)]))/(a*d*
(-I + Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(1/3))

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Maple [F]
time = 0.46, size = 0, normalized size = 0.00 \[\int \frac {\cot \left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.60, size = 265, normalized size = 0.85 \begin {gather*} -\frac {\frac {2 \, \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 {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 {2 \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 {16 \, \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 {8 \, \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 {16 \, \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}}} - \frac {6 \, {\left (6 i \, a \tan \left (d x + c\right ) + 7 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*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) - 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) + 2*2^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3))/a^(4/3) - 16*sqrt(3)*arctan(
1/3*sqrt(3)*(2*(I*a*tan(d*x + c) + a)^(1/3) + a^(1/3))/a^(1/3))/a^(4/3) + 8*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) - 16*log((I*a*tan(d*x + c) + a)^(1/3) - a^(1/3))/a^(4
/3) - 6*(6*I*a*tan(d*x + c) + 7*a)/((I*a*tan(d*x + c) + a)^(4/3)*a))/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 633 vs. \(2 (227) = 454\).
time = 1.47, size = 633, normalized size = 2.02 \begin {gather*} \frac {{\left (8 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a^{2} d \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a^{3} d^{2} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 32 \, a^{2} d \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-a^{3} d^{2} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (13 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 16 \, {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {1}{2} \, {\left (i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 16 \, {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {1}{2} \, {\left (-i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*(1/2)^(1/3)*a^2*d*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-2*(1/2)^(2/3)*a^3*d^2*(-1/(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*(1/(a^4*d^3))^(1/3)*e^(
4*I*d*x + 4*I*c)*log(-a^3*d^2*(1/(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)) - 4*(1/2)^(1/3)*(I*sqrt(3)*a^2*d + a^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-(1/2)^(2/3)
*(I*sqrt(3)*a^3*d^2 - a^3*d^2)*(-1/(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)) - 4*(1/2)^(1/3)*(-I*sqrt(3)*a^2*d + a^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-(1/2)^(2
/3)*(-I*sqrt(3)*a^3*d^2 - a^3*d^2)*(-1/(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)) + 3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(13*e^(4*I*d*x + 4*I*c) + 14*e^(2*I*d*x + 2*I
*c) + 1)*e^(4/3*I*d*x + 4/3*I*c) - 16*(-I*sqrt(3)*a^2*d + a^2*d)*(1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(1
/2*(I*sqrt(3)*a^3*d^2 + a^3*d^2)*(1/(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 + a^2*d)*(1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(1/2*(-I*sqrt(3)*a^3*d
^2 + a^3*d^2)*(1/(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)))*e^(-
4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.98, size = 822, normalized size = 2.63 \begin {gather*} \ln \left (\left (\left (382205952\,a^{16}\,d^9\,{\left (\frac {1}{a^4\,d^3}\right )}^{2/3}-258785280\,a^{13}\,d^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{1/3}-125411328\,a^{12}\,d^6\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{2/3}+1990656\,a^9\,d^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{1/3}+\ln \left (\left (\left (382205952\,a^{16}\,d^9\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{2/3}-258785280\,a^{13}\,d^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{1/3}-125411328\,a^{12}\,d^6\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{2/3}+1990656\,a^9\,d^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{1/3}+\frac {\ln \left (1990656\,a^9\,d^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (125411328\,a^{12}\,d^6+\frac {\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (258785280\,a^{13}\,d^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-95551488\,a^{16}\,d^9\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (\frac {1}{a^4\,d^3}\right )}^{2/3}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{1/3}}{2}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{2/3}}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (1990656\,a^9\,d^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (125411328\,a^{12}\,d^6-\frac {\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (258785280\,a^{13}\,d^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-95551488\,a^{16}\,d^9\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (\frac {1}{a^4\,d^3}\right )}^{2/3}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{1/3}}{2}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{2/3}}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^4\,d^3}\right )}^{1/3}}{2}+\ln \left (1990656\,a^9\,d^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (125411328\,a^{12}\,d^6+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (258785280\,a^{13}\,d^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-382205952\,a^{16}\,d^9\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{1/3}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{2/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{1/3}-\ln \left (1990656\,a^9\,d^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (125411328\,a^{12}\,d^6-\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (258785280\,a^{13}\,d^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-382205952\,a^{16}\,d^9\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{1/3}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{2/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (-\frac {1}{128\,a^4\,d^3}\right )}^{1/3}+\frac {\frac {9\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{4\,a}+\frac {3}{8}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)/(a + a*tan(c + d*x)*1i)^(4/3),x)

[Out]

log(((382205952*a^16*d^9*(1/(a^4*d^3))^(2/3) - 258785280*a^13*d^7*(a + a*tan(c + d*x)*1i)^(1/3))*(1/(a^4*d^3))
^(1/3) - 125411328*a^12*d^6)*(1/(a^4*d^3))^(2/3) + 1990656*a^9*d^4*(a + a*tan(c + d*x)*1i)^(1/3))*(1/(a^4*d^3)
)^(1/3) + log(((382205952*a^16*d^9*(-1/(128*a^4*d^3))^(2/3) - 258785280*a^13*d^7*(a + a*tan(c + d*x)*1i)^(1/3)
)*(-1/(128*a^4*d^3))^(1/3) - 125411328*a^12*d^6)*(-1/(128*a^4*d^3))^(2/3) + 1990656*a^9*d^4*(a + a*tan(c + d*x
)*1i)^(1/3))*(-1/(128*a^4*d^3))^(1/3) + (log(1990656*a^9*d^4*(a + a*tan(c + d*x)*1i)^(1/3) - ((3^(1/2)*1i - 1)
^2*(125411328*a^12*d^6 + ((3^(1/2)*1i - 1)*(258785280*a^13*d^7*(a + a*tan(c + d*x)*1i)^(1/3) - 95551488*a^16*d
^9*(3^(1/2)*1i - 1)^2*(1/(a^4*d^3))^(2/3))*(1/(a^4*d^3))^(1/3))/2)*(1/(a^4*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*(1
/(a^4*d^3))^(1/3))/2 - (log(1990656*a^9*d^4*(a + a*tan(c + d*x)*1i)^(1/3) - ((3^(1/2)*1i + 1)^2*(125411328*a^1
2*d^6 - ((3^(1/2)*1i + 1)*(258785280*a^13*d^7*(a + a*tan(c + d*x)*1i)^(1/3) - 95551488*a^16*d^9*(3^(1/2)*1i +
1)^2*(1/(a^4*d^3))^(2/3))*(1/(a^4*d^3))^(1/3))/2)*(1/(a^4*d^3))^(2/3))/4)*(3^(1/2)*1i + 1)*(1/(a^4*d^3))^(1/3)
)/2 + log(1990656*a^9*d^4*(a + a*tan(c + d*x)*1i)^(1/3) - ((3^(1/2)*1i)/2 - 1/2)^2*(125411328*a^12*d^6 + ((3^(
1/2)*1i)/2 - 1/2)*(258785280*a^13*d^7*(a + a*tan(c + d*x)*1i)^(1/3) - 382205952*a^16*d^9*((3^(1/2)*1i)/2 - 1/2
)^2*(-1/(128*a^4*d^3))^(2/3))*(-1/(128*a^4*d^3))^(1/3))*(-1/(128*a^4*d^3))^(2/3))*((3^(1/2)*1i)/2 - 1/2)*(-1/(
128*a^4*d^3))^(1/3) - log(1990656*a^9*d^4*(a + a*tan(c + d*x)*1i)^(1/3) - ((3^(1/2)*1i)/2 + 1/2)^2*(125411328*
a^12*d^6 - ((3^(1/2)*1i)/2 + 1/2)*(258785280*a^13*d^7*(a + a*tan(c + d*x)*1i)^(1/3) - 382205952*a^16*d^9*((3^(
1/2)*1i)/2 + 1/2)^2*(-1/(128*a^4*d^3))^(2/3))*(-1/(128*a^4*d^3))^(1/3))*(-1/(128*a^4*d^3))^(2/3))*((3^(1/2)*1i
)/2 + 1/2)*(-1/(128*a^4*d^3))^(1/3) + ((9*(a + a*tan(c + d*x)*1i))/(4*a) + 3/8)/(d*(a + a*tan(c + d*x)*1i)^(4/
3))

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