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

Optimal. Leaf size=313 \[ \frac{\sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{4/3} d}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{\log (\tan (c+d x))}{2 a^{4/3} d}+\frac{3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 a^{4/3} d}-\frac{3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i x}{8 \sqrt [3]{2} a^{4/3}}+\frac{9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{3}{8 d (a+i a \tan (c+d x))^{4/3}} \]

[Out]

((-I/8)*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))

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Rubi [A]  time = 0.596277, antiderivative size = 313, normalized size of antiderivative = 1., 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 = {3562, 3479, 3481, 55, 617, 204, 31, 3596, 3599} \[ \frac{\sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{4/3} d}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac{\log (\tan (c+d x))}{2 a^{4/3} d}+\frac{3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 a^{4/3} d}-\frac{3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac{\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac{i x}{8 \sqrt [3]{2} a^{4/3}}+\frac{9}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac{3}{8 d (a+i a \tan (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*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 3562

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 3479

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 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 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]

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{\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{\operatorname{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 \operatorname{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 \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 \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{3 \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 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 \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 \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{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*}

Mathematica [C]  time = 1.43467, size = 189, normalized size = 0.6 \[ -\frac{3 i \sec ^2(c+d x) \left (\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-8 \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{2 e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+6 i \sin (2 (c+d x))+7 \cos (2 (c+d x))+7\right )}{16 a d (\tan (c+d x)-i) \sqrt [3]{a+i a \tan (c+d x)}} \]

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 = 0.116, size = 0, normalized size = 0. \begin{align*} \int{\cot \left ( dx+c \right ) \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)/(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 [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)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.07452, size = 1874, normalized size = 5.99 \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)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

1/128*(32*(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)) + 128*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)) + (1/2)^(1/3)*(16*I*sqrt(3)*a^2*d - 16*a^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(1/16*
(1/2)^(2/3)*(16*I*sqrt(3)*a^3*d^2 + 16*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)) + (1/2)^(1/3)*(-16*I*sqrt(3)*a^2*d - 16*a^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x +
4*I*c)*log(1/16*(1/2)^(2/3)*(-16*I*sqrt(3)*a^3*d^2 + 16*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)) + 12*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) + 4*(16*I*sqrt(3)*a^2*d - 16*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)) + 4*(-16*I*sqrt(3)*a^2*d - 16*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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\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)/(a+I*a*tan(d*x+c))**(4/3),x)

[Out]

Integral(cot(c + d*x)/(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 )}{{\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)/(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)