3.3.92 \(\int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [292]

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

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

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Rubi [A]
time = 0.12, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3624, 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 )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\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 {x}{4 \sqrt [3]{2} \sqrt [3]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3624

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\\ &=-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {\int (a+i a \tan (c+d x))^{2/3} \, dx}{2 a}\\ &=-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}+\frac {i \text {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 {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a 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 )}{4 d}+\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 )}{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 {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 {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}+\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 )}{2 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\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 {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 {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.71, size = 82, normalized size = 0.38 \begin {gather*} -\frac {3 \left (8 i-4 \tan (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 ) (-i+\tan (c+d x))\right )}{8 d \sqrt [3]{a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 172, normalized size = 0.81

method	result	size
derivativedivides	\(-\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+\frac {\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a}{2}+\frac {a}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d a}\)	\(172\)
default	\(-\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+\frac {\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a}{2}+\frac {a}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d a}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*I/d/a*(1/2*(a+I*a*tan(d*x+c))^(2/3)+1/2*(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+1/2*a/(a+I*a*tan(d
*x+c))^(1/3))

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Maxima [A]
time = 0.52, size = 172, normalized size = 0.81 \begin {gather*} -\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {8}{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 ) - 2^{\frac {2}{3}} a^{\frac {8}{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 ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {8}{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 ) + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{2} + \frac {12 \, a^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*I*(2*sqrt(3)*2^(2/3)*a^(8/3)*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)) - 2^(2/3)*a^(8/3)*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)) + 2*2^(2/3)*a^(8/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) + 12*(I*a*tan(d*x +
c) + a)^(2/3)*a^2 + 12*a^3/(I*a*tan(d*x + c) + a)^(1/3))/(a^3*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. 315 vs. \(2 (146) = 292\).
time = 1.32, size = 315, normalized size = 1.48 \begin {gather*} \frac {{\left (4 \, a d \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (8 \, a d^{2} \left (\frac {i}{16 \, a 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 (3 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 )} - 2 \, {\left (-i \, \sqrt {3} a d + a d\right )} \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (\frac {i}{16 \, a 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 ) - 2 \, {\left (i \, \sqrt {3} a d + a d\right )} \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (-i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (\frac {i}{16 \, a 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 (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*a*d*(1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*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)) - 3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(3*I*e^(2*I*
d*x + 2*I*c) + I)*e^(4/3*I*d*x + 4/3*I*c) - 2*(-I*sqrt(3)*a*d + a*d)*(1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c
)*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)*(1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*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)))*e^(
-2*I*d*x - 2*I*c)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 4.48, size = 210, normalized size = 0.99 \begin {gather*} -\frac {3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}\,3{}\mathrm {i}}{2\,a\,d}+\frac {{\left (\frac {1}{16}{}\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^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\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}}{4\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\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}}{4\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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