3.197 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=270 \[ -\frac{\sqrt{3} a^{2/3} (B+i A) \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac{3 a^{2/3} (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac{a^{2/3} (B+i A) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac{a^{2/3} x (A-i B)}{2 \sqrt [3]{2}}-\frac{3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{9 B (a+i a \tan (c+d x))^{2/3}}{8 d} \]

[Out]

(a^(2/3)*(A - I*B)*x)/(2*2^(1/3)) - (Sqrt[3]*a^(2/3)*(I*A + B)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x]
)^(1/3))/(Sqrt[3]*a^(1/3))])/(2^(1/3)*d) - (a^(2/3)*(I*A + B)*Log[Cos[c + d*x]])/(2*2^(1/3)*d) - (3*a^(2/3)*(I
*A + B)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(2*2^(1/3)*d) - (9*B*(a + I*a*Tan[c + d*x])^(2/3)
)/(8*d) + (3*B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(2/3))/(8*d) - (3*((4*I)*A + B)*(a + I*a*Tan[c + d*x])^(5
/3))/(20*a*d)

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Rubi [A]  time = 0.444395, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3597, 3592, 3527, 3481, 55, 617, 204, 31} \[ -\frac{\sqrt{3} a^{2/3} (B+i A) \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac{3 a^{2/3} (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac{a^{2/3} (B+i A) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac{a^{2/3} x (A-i B)}{2 \sqrt [3]{2}}-\frac{3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{9 B (a+i a \tan (c+d x))^{2/3}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(2/3)*(A - I*B)*x)/(2*2^(1/3)) - (Sqrt[3]*a^(2/3)*(I*A + B)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x]
)^(1/3))/(Sqrt[3]*a^(1/3))])/(2^(1/3)*d) - (a^(2/3)*(I*A + B)*Log[Cos[c + d*x]])/(2*2^(1/3)*d) - (3*a^(2/3)*(I
*A + B)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(2*2^(1/3)*d) - (9*B*(a + I*a*Tan[c + d*x])^(2/3)
)/(8*d) + (3*B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(2/3))/(8*d) - (3*((4*I)*A + B)*(a + I*a*Tan[c + d*x])^(5
/3))/(20*a*d)

Rule 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*Tan[e + f*x], 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] && GtQ[n, 0]

Rule 3592

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

Rule 3527

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

Rubi steps

\begin{align*} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx &=\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac{3 \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} \left (-2 a B+\frac{2}{3} a (4 A-i B) \tan (c+d x)\right ) \, dx}{8 a}\\ &=\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac{3 \int (a+i a \tan (c+d x))^{2/3} \left (-\frac{2}{3} a (4 A-i B)-2 a B \tan (c+d x)\right ) \, dx}{8 a}\\ &=-\frac{9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+(-A+i B) \int (a+i a \tan (c+d x))^{2/3} \, dx\\ &=-\frac{9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac{(a (i A+B)) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac{a^{2/3} (A-i B) x}{2 \sqrt [3]{2}}-\frac{a^{2/3} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac{9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac{\left (3 a^{2/3} (i A+B)\right ) \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 )}{2 \sqrt [3]{2} d}-\frac{(3 a (i A+B)) \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 )}{2 d}\\ &=\frac{a^{2/3} (A-i B) x}{2 \sqrt [3]{2}}-\frac{a^{2/3} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac{3 a^{2/3} (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac{9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac{\left (3 a^{2/3} (i A+B)\right ) \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 )}{\sqrt [3]{2} d}\\ &=\frac{a^{2/3} (A-i B) x}{2 \sqrt [3]{2}}-\frac{\sqrt{3} a^{2/3} (i A+B) \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt [3]{2} d}-\frac{a^{2/3} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac{3 a^{2/3} (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac{9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac{3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac{3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}\\ \end{align*}

Mathematica [C]  time = 2.64225, size = 104, normalized size = 0.39 \[ \frac{3 (a+i a \tan (c+d x))^{2/3} \left (10 (B+i A) \text{Hypergeometric2F1}\left (\frac{2}{3},1,\frac{5}{3},\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )+(8 A-2 i B) \tan (c+d x)-8 i A+5 B \sec ^2(c+d x)-22 B\right )}{40 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + I*a*Tan[c + d*x])^(2/3)*((-8*I)*A - 22*B + 10*(I*A + B)*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d
*x))/(1 + E^((2*I)*(c + d*x)))] + 5*B*Sec[c + d*x]^2 + (8*A - (2*I)*B)*Tan[c + d*x]))/(40*d)

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Maple [A]  time = 0.027, size = 367, normalized size = 1.4 \begin{align*} -{\frac{3\,B}{8\,{a}^{2}d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{8}{3}}}}+{\frac{3\,B}{5\,ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}}}-{\frac{{\frac{3\,i}{5}}A}{ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}}}-{\frac{3\,B}{2\,d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}}-{\frac{{2}^{{\frac{2}{3}}}B}{2\,d}{a}^{{\frac{2}{3}}}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }-{\frac{{\frac{i}{2}}{2}^{{\frac{2}{3}}}A}{d}{a}^{{\frac{2}{3}}}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }+{\frac{{2}^{{\frac{2}{3}}}B}{4\,d}{a}^{{\frac{2}{3}}}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ) }+{\frac{{\frac{i}{4}}{2}^{{\frac{2}{3}}}A}{d}{a}^{{\frac{2}{3}}}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ) }-{\frac{\sqrt{3}{2}^{{\frac{2}{3}}}B}{2\,d}{a}^{{\frac{2}{3}}}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ) }-{\frac{{\frac{i}{2}}\sqrt{3}{2}^{{\frac{2}{3}}}A}{d}{a}^{{\frac{2}{3}}}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82168, size = 1762, normalized size = 6.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(2^(2/3)*((-12*I*A - 18*B)*e^(4*I*d*x + 4*I*c) + (-12*I*A - 18*B)*e^(2*I*d*x + 2*I*c) - 15*B)*(a/(e^(2*I*
d*x + 2*I*c) + 1))^(2/3)*e^(4/3*I*d*x + 4/3*I*c) + 10*(1/2)^(1/3)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*
I*c) + d)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - B^3)*a^2/d^3)^(1/3)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*
d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + 2*(1/2)^(2/3)*d^2*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - B^3)*a^2/
d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) - 5*(1/2)^(1/3)*((-I*sqrt(3)*d + d)*e^(4*I*d*x + 4*I*c) + 2*(-I*sqrt(3)
*d + d)*e^(2*I*d*x + 2*I*c) - I*sqrt(3)*d + d)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - B^3)*a^2/d^3)^(1/3)*log((2^(1/3
)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(I*sqrt(3)
*d^2 + d^2)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) - (1/2)^(1/3)*(5*(
I*sqrt(3)*d + d)*e^(4*I*d*x + 4*I*c) + 10*(I*sqrt(3)*d + d)*e^(2*I*d*x + 2*I*c) + 5*I*sqrt(3)*d + 5*d)*((I*A^3
+ 3*A^2*B - 3*I*A*B^2 - B^3)*a^2/d^3)^(1/3)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1)
)^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(-I*sqrt(3)*d^2 + d^2)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - B^3)*a^2/
d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{2}{3}} \left (A + B \tan{\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \tan \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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