Integrand size = 36, antiderivative size = 270 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {a^{2/3} (A-i B) x}{2 \sqrt [3]{2}}-\frac {\sqrt {3} a^{2/3} (i A+B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac {a^{2/3} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {3 a^{2/3} (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d} \]
1/4*a^(2/3)*(A-I*B)*x*2^(2/3)-1/4*a^(2/3)*(I*A+B)*ln(cos(d*x+c))*2^(2/3)/d -3/4*a^(2/3)*(I*A+B)*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(2/3)/ d-1/2*a^(2/3)*(I*A+B)*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)/d-9/8*B*(a+I*a*tan(d*x+c))^(2/3)/d+3/8* B*tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(2/3)/d-3/20*(4*I*A+B)*(a+I*a*tan(d*x+c) )^(5/3)/a/d
Time = 5.25 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.73 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {10\ 2^{2/3} a^{2/3} (i A+B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )\right )+45 B (a+i a \tan (c+d x))^{2/3}-15 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}+\frac {6 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{a}}{40 d} \]
-1/40*(10*2^(2/3)*a^(2/3)*(I*A + B)*(2*Sqrt[3]*ArcTan[(1 + (2^(2/3)*(a + I *a*Tan[c + d*x])^(1/3))/a^(1/3))/Sqrt[3]] - Log[I + Tan[c + d*x]] + 3*Log[ 2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)]) + 45*B*(a + I*a*Tan[c + d *x])^(2/3) - 15*B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(2/3) + (6*((4*I)* A + B)*(a + I*a*Tan[c + d*x])^(5/3))/a)/d
Time = 0.84 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.80, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 4080, 27, 3042, 4075, 3042, 4010, 3042, 3962, 67, 16, 1082, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^2 (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {3 \int -\frac {2}{3} \tan (c+d x) (i \tan (c+d x) a+a)^{2/3} (3 a B-a (4 A-i B) \tan (c+d x))dx}{8 a}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {\int \tan (c+d x) (i \tan (c+d x) a+a)^{2/3} (3 a B-a (4 A-i B) \tan (c+d x))dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {\int \tan (c+d x) (i \tan (c+d x) a+a)^{2/3} (3 a B-a (4 A-i B) \tan (c+d x))dx}{4 a}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {\int (i \tan (c+d x) a+a)^{2/3} (a (4 A-i B)+3 a B \tan (c+d x))dx+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {\int (i \tan (c+d x) a+a)^{2/3} (a (4 A-i B)+3 a B \tan (c+d x))dx+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}}{4 a}\) |
| \(\Big \downarrow \) 4010 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {4 a (A-i B) \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}+\frac {9 a B (a+i a \tan (c+d x))^{2/3}}{2 d}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {4 a (A-i B) \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}+\frac {9 a B (a+i a \tan (c+d x))^{2/3}}{2 d}}{4 a}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {-\frac {4 i a^2 (A-i B) \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}+\frac {9 a B (a+i a \tan (c+d x))^{2/3}}{2 d}}{4 a}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {-\frac {4 i a^2 (A-i B) \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}+\frac {9 a B (a+i a \tan (c+d x))^{2/3}}{2 d}}{4 a}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {-\frac {4 i a^2 (A-i B) \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}+\frac {9 a B (a+i a \tan (c+d x))^{2/3}}{2 d}}{4 a}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {-\frac {4 i a^2 (A-i B) \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}+\frac {9 a B (a+i a \tan (c+d x))^{2/3}}{2 d}}{4 a}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {-\frac {4 i a^2 (A-i B) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{5 d}+\frac {9 a B (a+i a \tan (c+d x))^{2/3}}{2 d}}{4 a}\) |
(3*B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(2/3))/(8*d) - (((-4*I)*a^2*(A - I*B)*(((-I)*Sqrt[3]*ArcTanh[(a*Tan[c + d*x])/Sqrt[3]])/(2^(1/3)*a^(1/3)) - (3*Log[2^(1/3)*a^(1/3) - I*a*Tan[c + d*x]])/(2*2^(1/3)*a^(1/3)) + Log[a - I*a*Tan[c + d*x]]/(2*2^(1/3)*a^(1/3))))/d + (9*a*B*(a + I*a*Tan[c + d*x ])^(2/3))/(2*d) + (3*((4*I)*A + B)*(a + I*a*Tan[c + d*x])^(5/3))/(5*d))/(4 *a)
3.2.97.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[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]
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])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-b/d S ubst[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]
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] + Simp [(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]
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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[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]
Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {3 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+\frac {i a^{2} B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{3} \left (-i B +A \right )\right )}{d \,a^{2}}\) | \(223\) |
| default | \(\frac {3 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+\frac {i a^{2} B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{3} \left (-i B +A \right )\right )}{d \,a^{2}}\) | \(223\) |
| parts | \(\frac {3 i A \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{2}\right )}{d a}+B \left (-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8 d \,a^{2}}+\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5 d a}-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 d}-\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}+\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}-\frac {a^{\frac {2}{3}} \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d}\right )\) | \(358\) |
3*I/d/a^2*(1/8*I*B*(a+I*a*tan(d*x+c))^(8/3)-1/5*I*B*a*(a+I*a*tan(d*x+c))^( 5/3)-1/5*A*a*(a+I*a*tan(d*x+c))^(5/3)+1/2*I*a^2*B*(a+I*a*tan(d*x+c))^(2/3) -(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^3*(A-I*B))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 660 vs. \(2 (201) = 402\).
Time = 0.26 (sec) , antiderivative size = 660, normalized size of antiderivative = 2.44 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {3 \cdot 2^{\frac {2}{3}} {\left (2 \, {\left (2 i \, A + 3 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (2 i \, A + 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 \, B\right )} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 10 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac {{\left (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} a \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 )} + 2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} d^{2} \left (\frac {{\left (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {2}{3}}}{{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}\right ) + 5 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (-i \, \sqrt {3} d + d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (-i \, \sqrt {3} d + d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {3} d + d\right )} \left (\frac {{\left (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} a \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 )} - \left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} d^{2} + d^{2}\right )} \left (\frac {{\left (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {2}{3}}}{{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}\right ) + 5 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (i \, \sqrt {3} d + d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (i \, \sqrt {3} d + d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {3} d + d\right )} \left (\frac {{\left (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} a \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 )} - \left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} d^{2} + d^{2}\right )} \left (\frac {{\left (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {2}{3}}}{{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}\right )}{10 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-1/10*(3*2^(2/3)*(2*(2*I*A + 3*B)*e^(4*I*d*x + 4*I*c) + 2*(2*I*A + 3*B)*e^ (2*I*d*x + 2*I*c) + 5*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)) + 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)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {2}{3}} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.39 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.78 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {i \, {\left (20 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 10 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 20 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 15 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {8}{3}} B a + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} {\left (A + i \, B\right )} a^{2} - 60 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} B a^{3}\right )}}{40 \, a^{3} d} \]
-1/40*I*(20*sqrt(3)*2^(2/3)*(A - I*B)*a^(11/3)*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)) - 10*2^(2/3)*( A - I*B)*a^(11/3)*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)) + 20*2^(2/3)*(A - I*B)*a^(11/3) *log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 15*I*(I*a*tan(d*x + c) + a)^(8/3)*B*a + 24*(I*a*tan(d*x + c) + a)^(5/3)*(A + I*B)*a^2 - 60*I *(I*a*tan(d*x + c) + a)^(2/3)*B*a^3)/(a^3*d)
\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \tan \left (d x + c\right )^{2} \,d x } \]
Time = 10.29 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.61 \[ \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\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}{2\,d}-\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}\,3{}\mathrm {i}}{5\,a\,d}+\frac {3\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}}{5\,a\,d}-\frac {3\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{8/3}}{8\,a^2\,d}-\frac {2^{2/3}\,B\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{2\,d}+\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,A\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{d}+\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,A\,a^{2/3}\,\ln \left (\frac {{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}}{2}-{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+\frac {{\left (-1\right )}^{5/6}\,2^{1/3}\,\sqrt {3}\,a^{1/3}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}-\frac {2^{2/3}\,B\,a^{2/3}\,\ln \left (\frac {9\,B^2\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}-\frac {9\,2^{1/3}\,B^2\,a^{7/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}+\frac {2^{2/3}\,B\,a^{2/3}\,\ln \left (\frac {9\,B^2\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}-\frac {9\,2^{1/3}\,B^2\,a^{7/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}-\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,A\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-\frac {{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}}{2}+\frac {{\left (-1\right )}^{5/6}\,2^{1/3}\,\sqrt {3}\,a^{1/3}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \]
(3*B*(a + a*tan(c + d*x)*1i)^(5/3))/(5*a*d) - (A*(a + a*tan(c + d*x)*1i)^( 5/3)*3i)/(5*a*d) - (3*B*(a + a*tan(c + d*x)*1i)^(2/3))/(2*d) - (3*B*(a + a *tan(c + d*x)*1i)^(8/3))/(8*a^2*d) - (2^(2/3)*B*a^(2/3)*log((a*(tan(c + d* x)*1i + 1))^(1/3) - 2^(1/3)*a^(1/3)))/(2*d) + ((1i/2)^(1/3)*A*a^(2/3)*log( (a*(tan(c + d*x)*1i + 1))^(1/3) + (-1)^(1/3)*2^(1/3)*a^(1/3)))/d + ((1i/2) ^(1/3)*A*a^(2/3)*log(((-1)^(1/3)*2^(1/3)*a^(1/3))/2 - (a*(tan(c + d*x)*1i + 1))^(1/3) + ((-1)^(5/6)*2^(1/3)*3^(1/2)*a^(1/3))/2)*((3^(1/2)*1i)/2 - 1/ 2))/d - (2^(2/3)*B*a^(2/3)*log((9*B^2*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d ^2 - (9*2^(1/3)*B^2*a^(7/3)*((3^(1/2)*1i)/2 - 1/2)^2)/d^2)*((3^(1/2)*1i)/2 - 1/2))/(2*d) + (2^(2/3)*B*a^(2/3)*log((9*B^2*a^2*(a + a*tan(c + d*x)*1i) ^(1/3))/d^2 - (9*2^(1/3)*B^2*a^(7/3)*((3^(1/2)*1i)/2 + 1/2)^2)/d^2)*((3^(1 /2)*1i)/2 + 1/2))/(2*d) - ((1i/2)^(1/3)*A*a^(2/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) - ((-1)^(1/3)*2^(1/3)*a^(1/3))/2 + ((-1)^(5/6)*2^(1/3)*3^(1/2) *a^(1/3))/2)*((3^(1/2)*1i)/2 + 1/2))/d