3.4.4 \(\int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx\) [304]

Optimal. Leaf size=213 \[ -\frac {x}{8 \sqrt [3]{2} a^{4/3}}+\frac {i \sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3560, 3562, 57, 631, 210, 31} \begin {gather*} \frac {i \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 {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 {3 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-4/3),x]

[Out]

-1/8*x/(2^(1/3)*a^(4/3)) + ((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) + (((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) + ((3*I)/8)/(d*(a + I*a*Tan[c + d*x])^(4/3)) + ((3*I)/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]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx &=\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {\int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{2 a}\\ &=\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {\int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2}\\ &=\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \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 {x}{8 \sqrt [3]{2} a^{4/3}}+\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {(3 i) \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 i) \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 {x}{8 \sqrt [3]{2} a^{4/3}}+\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} 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 {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {(3 i) \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 {x}{8 \sqrt [3]{2} a^{4/3}}+\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 {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 {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 i}{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 = 0.46, size = 128, normalized size = 0.60 \begin {gather*} -\frac {3 \sec ^2(c+d x) \left (-3-3 \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)))-2 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[(a + I*a*Tan[c + d*x])^(-4/3),x]

[Out]

(-3*Sec[c + d*x]^2*(-3 - 3*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)]) - (2*I)*Sin[2*(c + d*x)]))/(16*a*d*(-I + Tan[c + d*x])*(
a + I*a*Tan[c + d*x])^(1/3))

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Maple [A]
time = 0.10, size = 177, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {3 i a \left (\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{4 a^{2}}+\frac {1}{4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {1}{8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\right )}{d}\)	\(177\)
default	\(\frac {3 i a \left (\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{4 a^{2}}+\frac {1}{4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {1}{8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\right )}{d}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(d*x+c))^(4/3),x,method=_RETURNVERBOSE)

[Out]

3*I/d*a*(1/4*(1/6*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/12*2^(2/3)/a^(1/3)*ln((a+I*a*
tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))+1/6*3^(1/2)*2^(2/3)/a^(1/3)*arctan
(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))/a^2+1/4/a^2/(a+I*a*tan(d*x+c))^(1/3)+1/8/a/(a+I*a*
tan(d*x+c))^(4/3))

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Maxima [A]
time = 0.52, size = 164, normalized size = 0.77 \begin {gather*} \frac {i \, {\left (\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 {1}{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 {1}{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 {1}{3}}} + \frac {6 \, {\left (2 i \, a \tan \left (d x + c\right ) + 3 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\right )}}{16 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

1/16*I*(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^(1/3) - 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^(1/3) + 2*2^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3))/a^(1/3) + 6*(2*I*a*tan(d*x
+ c) + 3*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. 346 vs. \(2 (146) = 292\).
time = 1.57, size = 346, normalized size = 1.62 \begin {gather*} \frac {{\left (32 \, a^{2} d \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (32 \, a^{3} d^{2} \left (-\frac {i}{128 \, 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 (-5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\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 {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-16 \, {\left (i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (-\frac {i}{128 \, 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 {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-16 \, {\left (-i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (-\frac {i}{128 \, 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(1/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

1/32*(32*a^2*d*(-1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*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)) - 3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/
3)*(-5*I*e^(4*I*d*x + 4*I*c) - 6*I*e^(2*I*d*x + 2*I*c) - I)*e^(4/3*I*d*x + 4/3*I*c) - 16*(-I*sqrt(3)*a^2*d + a
^2*d)*(-1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-16*(I*sqrt(3)*a^3*d^2 + 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 + a^2*d)*
(-1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-16*(-I*sqrt(3)*a^3*d^2 + 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)))*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 {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**(4/3),x)

[Out]

Integral((I*a*tan(c + d*x) + a)**(-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(1/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.42, size = 217, normalized size = 1.02 \begin {gather*} \frac {\frac {3{}\mathrm {i}}{8\,d}+\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{4\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}-\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (9\,{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{a^{4/3}\,d}+\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d}-\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3i/(8*d) + ((a + a*tan(c + d*x)*1i)*3i)/(4*a*d))/(a + a*tan(c + d*x)*1i)^(4/3) - ((1i/128)^(1/3)*log(9*(a*(ta
n(c + d*x)*1i + 1))^(1/3) + 9*(-1)^(1/3)*2^(1/3)*a^(1/3)))/(a^(4/3)*d) + ((1i/128)^(1/3)*log(- (9*(a + a*tan(c
+ d*x)*1i)^(1/3))/(16*a^2*d^2) - (9*(-1)^(1/3)*2^(1/3)*(3^(1/2)*1i - 1))/(32*a^(5/3)*d^2))*((3^(1/2)*1i)/2 +
1/2))/(a^(4/3)*d) - ((1i/128)^(1/3)*log((9*(-1)^(1/3)*2^(1/3)*(3^(1/2)*1i + 1))/(32*a^(5/3)*d^2) - (9*(a + a*t
an(c + d*x)*1i)^(1/3))/(16*a^2*d^2))*((3^(1/2)*1i)/2 - 1/2))/(a^(4/3)*d)

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