Integrand size = 28, antiderivative size = 213 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {(A-i B) x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\sqrt {3} (i A+B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 (i A-B)}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \]
-1/8*(A-I*B)*x*2^(2/3)/a^(1/3)+1/8*(I*A+B)*ln(cos(d*x+c))*2^(2/3)/a^(1/3)/ d+3/8*(I*A+B)*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*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)/a^(1/3)/d+3/2*(I*A-B)/d/(a+I*a*tan(d*x+c))^(1/ 3)
Time = 0.89 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.70 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {\frac {2^{2/3} (i A+B) \left (2 \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )\right )}{\sqrt [3]{a}}+\frac {12 i (A+i B)}{\sqrt [3]{a+i a \tan (c+d x)}}}{8 d} \]
((2^(2/3)*(I*A + B)*(2*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3))] - Log[I + Tan[c + d*x]] + 3*Log[2^(1/3)*a^ (1/3) - (a + I*a*Tan[c + d*x])^(1/3)]))/a^(1/3) + ((12*I)*(A + I*B))/(a + I*a*Tan[c + d*x])^(1/3))/(8*d)
Time = 0.40 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.70, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4009, 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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4009 |
| \(\displaystyle \frac {(A-i B) \int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}+\frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A-i B) \int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}+\frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i (A-i B) \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 d}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i (A-i B) \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i (A-i B) \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i (A-i B) \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 (-B+i A)}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i (A-i B) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\) |
((-1/2*I)*(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 + (3*(I*A - B))/ (2*d*(a + I*a*Tan[c + d*x])^(1/3))
3.3.2.3.1 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[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[(-(b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a *f*m)), x] + Simp[(b*c + a*d)/(2*a*b) Int[(a + b*Tan[e + f*x])^(m + 1), x ], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && LtQ[m, 0]
Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {3 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (\frac {A}{2}-\frac {i B}{2}\right )\right )}{d}\) | \(166\) |
| default | \(\frac {3 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (\frac {A}{2}-\frac {i B}{2}\right )\right )}{d}\) | \(166\) |
| parts | \(\frac {3 i A a \left (\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{2 a}+\frac {1}{2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d}+\frac {B \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 a^{\frac {1}{3}}}-\frac {3}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d}\) | \(306\) |
3*I/d*(-(-1/2*A-1/2*I*B)/(a+I*a*tan(d*x+c))^(1/3)+(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*t an(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*ta n(d*x+c))^(1/3)+1)))*(1/2*A-1/2*I*B))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (156) = 312\).
Time = 0.26 (sec) , antiderivative size = 547, normalized size of antiderivative = 2.57 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {{\left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a d \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a d^{2} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} \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 )}}{A^{2} - 2 i \, A B - B^{2}}\right ) - \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a d + a d\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} \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} a d^{2} - a d^{2}\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {2}{3}}}{A^{2} - 2 i \, A B - B^{2}}\right ) - \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a d + a d\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} \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} a d^{2} - a d^{2}\right )} \left (\frac {-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}}{a d^{3}}\right )^{\frac {2}{3}}}{A^{2} - 2 i \, A B - B^{2}}\right ) - 3 \cdot 2^{\frac {2}{3}} {\left ({\left (-i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + 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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]
1/4*(2*(1/2)^(1/3)*a*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a*d^3))^(1/3 )*e^(2*I*d*x + 2*I*c)*log((2*(1/2)^(2/3)*a*d^2*((-I*A^3 - 3*A^2*B + 3*I*A* B^2 + B^3)/(a*d^3))^(2/3) + 2^(1/3)*(A^2 - 2*I*A*B - B^2)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c))/(A^2 - 2*I*A*B - B^2)) - (1/2 )^(1/3)*(I*sqrt(3)*a*d + a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a*d^3 ))^(1/3)*e^(2*I*d*x + 2*I*c)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*(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) *a*d^2 - a*d^2)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a*d^3))^(2/3))/(A^2 - 2*I*A*B - B^2)) - (1/2)^(1/3)*(-I*sqrt(3)*a*d + a*d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*(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)*a*d^2 - a*d^2)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)/(a*d^3))^(2/3))/(A^2 - 2*I*A*B - B^2)) - 3*2^(2/3)*((-I*A + B)*e^(2 *I*d*x + 2*I*c) - I*A + B)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*e^(4/3*I*d* x + 4/3*I*c))*e^(-2*I*d*x - 2*I*c)/(a*d)
\[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Time = 0.35 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.81 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {12 \, {\left (A + i \, B\right )} a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a d} \]
1/8*I*(2*sqrt(3)*2^(2/3)*(A - I*B)*a^(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)) - 2^(2/3)*(A - I*B )*a^(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)) + 2*2^(2/3)*(A - I*B)*a^(2/3)*log(-2^(1 /3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) + 12*(A + I*B)*a/(I*a*tan(d*x + c) + a)^(1/3))/(a*d)
\[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Time = 8.52 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.80 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {A\,3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}-\frac {3\,B}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,A\,\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 )}{a^{1/3}\,d}+\frac {4^{1/3}\,B\,\ln \left (18\,B^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,B^2\,a^{1/3}\,d\right )}{4\,a^{1/3}\,d}+\frac {4^{1/3}\,B\,\ln \left (18\,B^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,B^2\,a^{1/3}\,d\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{1/3}\,d}-\frac {4^{1/3}\,B\,\ln \left (18\,B^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,B^2\,a^{1/3}\,d\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,a^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,A\,\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 )}{a^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,A\,\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 )}{a^{1/3}\,d} \]
(A*3i)/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) - (3*B)/(2*d*(a + a*tan(c + d*x )*1i)^(1/3)) - ((1i/16)^(1/3)*A*log((a*(tan(c + d*x)*1i + 1))^(1/3) + (-1) ^(1/3)*2^(1/3)*a^(1/3)))/(a^(1/3)*d) + (4^(1/3)*B*log(18*B^2*d*(a + a*tan( c + d*x)*1i)^(1/3) - 9*4^(2/3)*B^2*a^(1/3)*d))/(4*a^(1/3)*d) + (4^(1/3)*B* log(18*B^2*d*(a + a*tan(c + d*x)*1i)^(1/3) - 9*4^(2/3)*B^2*a^(1/3)*d*((3^( 1/2)*1i)/2 - 1/2)^2)*((3^(1/2)*1i)/2 - 1/2))/(4*a^(1/3)*d) - (4^(1/3)*B*lo g(18*B^2*d*(a + a*tan(c + d*x)*1i)^(1/3) - 9*4^(2/3)*B^2*a^(1/3)*d*((3^(1/ 2)*1i)/2 + 1/2)^2)*((3^(1/2)*1i)/2 + 1/2))/(4*a^(1/3)*d) - ((1i/16)^(1/3)* A*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))/(a^(1/3)*d) + ((1i/16)^(1/3)*A*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))/(a^(1/3)*d)