Integrand size = 28, antiderivative size = 201 \[ \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt [3]{a} (A-i B) x}{2\ 2^{2/3}}-\frac {\sqrt {3} \sqrt [3]{a} (i A+B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {\sqrt [3]{a} (i A+B) \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac {3 \sqrt [3]{a} (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d} \]
-1/4*a^(1/3)*(A-I*B)*x*2^(1/3)+1/4*a^(1/3)*(I*A+B)*ln(cos(d*x+c))*2^(1/3)/ d+3/4*a^(1/3)*(I*A+B)*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(1/3) /d-1/2*a^(1/3)*(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^(1/3)/d+3*B*(a+I*a*tan(d*x+c))^(1/3)/d
Time = 0.81 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.93 \[ \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {-i \sqrt [3]{2} \sqrt [3]{a} (A-i B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )+\log \left (2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} \sqrt [3]{a+i a \tan (c+d x)}+(a+i a \tan (c+d x))^{2/3}\right )\right )+12 B \sqrt [3]{a+i a \tan (c+d x)}}{4 d} \]
((-I)*2^(1/3)*a^(1/3)*(A - I*B)*(2*Sqrt[3]*ArcTan[(1 + (2^(2/3)*(a + I*a*T an[c + d*x])^(1/3))/a^(1/3))/Sqrt[3]] - 2*Log[2^(1/3)*a^(1/3) - (a + I*a*T an[c + d*x])^(1/3)] + Log[2^(2/3)*a^(2/3) + 2^(1/3)*a^(1/3)*(a + I*a*Tan[c + d*x])^(1/3) + (a + I*a*Tan[c + d*x])^(2/3)]) + 12*B*(a + I*a*Tan[c + d* x])^(1/3))/(4*d)
Time = 0.40 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.69, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4010, 3042, 3962, 69, 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.
| \(\displaystyle \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4010 |
| \(\displaystyle (A-i B) \int \sqrt [3]{i \tan (c+d x) a+a}dx+\frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (A-i B) \int \sqrt [3]{i \tan (c+d x) a+a}dx+\frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {i a (A-i B) \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}d(i a \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {i a (A-i B) \left (\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2\ 2^{2/3} a^{2/3}}+\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {i a (A-i B) \left (\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {i a (A-i B) \left (-\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 B \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {i a (A-i B) \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
((-I)*a*(A - I*B)*((I*Sqrt[3]*ArcTanh[(a*Tan[c + d*x])/Sqrt[3]])/(2^(2/3)* a^(2/3)) - (3*Log[2^(1/3)*a^(1/3) - I*a*Tan[c + d*x]])/(2*2^(2/3)*a^(2/3)) + Log[a - I*a*Tan[c + d*x]]/(2*2^(2/3)*a^(2/3))))/d + (3*B*(a + I*a*Tan[c + d*x])^(1/3))/d
3.2.96.3.1 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_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) 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]
Time = 0.07 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.79
| method | result | size |
| derivativedivides | \(\frac {3 i \left (-i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a \left (-i B +A \right )\right )}{d}\) | \(159\) |
| default | \(\frac {3 i \left (-i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a \left (-i B +A \right )\right )}{d}\) | \(159\) |
| parts | \(\frac {3 i A a \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right )}{d}+B \left (\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d}\right )\) | \(290\) |
3*I/d*(-I*B*(a+I*a*tan(d*x+c))^(1/3)+(1/6*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d* x+c))^(1/3)-2^(1/3)*a^(1/3))-1/12*2^(1/3)/a^(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))-1/6*2^(1/3)/ a^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/ 3)+1)))*a*(A-I*B))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (146) = 292\).
Time = 0.24 (sec) , antiderivative size = 405, normalized size of antiderivative = 2.01 \[ \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {6 \cdot 2^{\frac {1}{3}} B \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}{4}\right )^{\frac {1}{3}} {\left (-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}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (i \, A + B\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}{4}\right )^{\frac {1}{3}} {\left (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}{d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right ) + \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (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}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (i \, A + B\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}{4}\right )^{\frac {1}{3}} {\left (-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}{d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right ) + 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (i \, A + B\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 )} - 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {{\left (-i \, A^{3} - 3 \, A^{2} B + 3 i \, A B^{2} + B^{3}\right )} a}{d^{3}}\right )^{\frac {1}{3}}}{i \, A + B}\right )}{2 \, d} \]
1/2*(6*2^(1/3)*B*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I* c) + (1/4)^(1/3)*(-I*sqrt(3)*d - d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)* a/d^3)^(1/3)*log((2^(1/3)*(I*A + B)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^ (2/3*I*d*x + 2/3*I*c) + (1/4)^(1/3)*(I*sqrt(3)*d + d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3))/(I*A + B)) + (1/4)^(1/3)*(I*sqrt(3)*d - d) *((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3)*log((2^(1/3)*(I*A + B) *(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (1/4)^(1/3) *(-I*sqrt(3)*d + d)*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1/3))/(I *A + B)) + 2*(1/4)^(1/3)*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3)^(1 /3)*log((2^(1/3)*(I*A + B)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d* x + 2/3*I*c) - 2*(1/4)^(1/3)*d*((-I*A^3 - 3*A^2*B + 3*I*A*B^2 + B^3)*a/d^3 )^(1/3))/(I*A + B)))/d
\[ \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right )\, dx \]
Time = 0.39 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.83 \[ \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} {\left (A - i \, B\right )} a^{\frac {4}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 12 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} B a\right )}}{4 \, a d} \]
-1/4*I*(2*sqrt(3)*2^(1/3)*(A - I*B)*a^(4/3)*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^(1/3)*(A - I* B)*a^(4/3)*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^(1/3)*(A - I*B)*a^(4/3)*log(-2^( 1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) + 12*I*(I*a*tan(d*x + c) + a) ^(1/3)*B*a)/(a*d)
\[ \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
Time = 1.25 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.82 \[ \int \sqrt [3]{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {3\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {2^{1/3}\,B\,a^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{2\,d}-\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,\ln \left (A\,a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}+18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{4/3}\,d^2\right )}{d}-\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,\ln \left (A\,a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}+18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{4/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{1/3}\,\ln \left (A\,a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}-18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,A\,a^{4/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {4^{2/3}\,B\,a^{1/3}\,\ln \left (\frac {9\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,B\,a^{4/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d}-\frac {4^{2/3}\,B\,a^{1/3}\,\ln \left (\frac {9\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,B\,a^{4/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d} \]
(3*B*(a + a*tan(c + d*x)*1i)^(1/3))/d + (2^(1/3)*B*a^(1/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) - 2^(1/3)*a^(1/3)))/(2*d) - ((1i/4)^(1/3)*A*a^(1/3)*l og(A*a*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*9i + 18*(1i/4)^(1/3)*A*a^(4/3)*d^ 2))/d - ((1i/4)^(1/3)*A*a^(1/3)*log(A*a*d^2*(a + a*tan(c + d*x)*1i)^(1/3)* 9i + 18*(1i/4)^(1/3)*A*a^(4/3)*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/d + ((1i/4)^(1/3)*A*a^(1/3)*log(A*a*d^2*(a + a*tan(c + d*x)*1i)^( 1/3)*9i - 18*(1i/4)^(1/3)*A*a^(4/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)* 1i)/2 + 1/2))/d + (4^(2/3)*B*a^(1/3)*log((9*B*a*(a + a*tan(c + d*x)*1i)^(1 /3))/d - (9*2^(1/3)*B*a^(4/3)*(3^(1/2)*1i - 1))/(2*d))*((3^(1/2)*1i)/2 - 1 /2))/(4*d) - (4^(2/3)*B*a^(1/3)*log((9*B*a*(a + a*tan(c + d*x)*1i)^(1/3))/ d + (9*2^(1/3)*B*a^(4/3)*(3^(1/2)*1i + 1))/(2*d))*((3^(1/2)*1i)/2 + 1/2))/ (4*d)