Integrand size = 26, antiderivative size = 213 \[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d} \] Output:
1/8*x*2^(2/3)/a^(1/3)-1/4*I*3^(1/2)*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*tan (d*x+c))^(1/3))*3^(1/2)/a^(1/3))*2^(2/3)/a^(1/3)/d-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-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
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.29 \[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {3 i \left (1+\operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},\frac {1}{2} (1+i \tan (c+d x))\right )+i \tan (c+d x)\right )}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \] Input:
Integrate[Tan[c + d*x]^2/(a + I*a*Tan[c + d*x])^(1/3),x]
Output:
(((-3*I)/2)*(1 + Hypergeometric2F1[-1/3, 1, 2/3, (1 + I*Tan[c + d*x])/2] + I*Tan[c + d*x]))/(d*(a + I*a*Tan[c + d*x])^(1/3))
Time = 0.47 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.77, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 4026, 25, 3042, 3960, 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.
| \(\displaystyle \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^2}{\sqrt [3]{a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4026 |
| \(\displaystyle \int -\frac {1}{\sqrt [3]{i \tan (c+d x) a+a}}dx-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\sqrt [3]{i \tan (c+d x) a+a}}dx-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\int \frac {1}{\sqrt [3]{i \tan (c+d x) a+a}}dx-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}\) |
| \(\Big \downarrow \) 3960 |
| \(\displaystyle -\frac {\int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {i \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 d}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {i \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {i \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\) |
Input:
Int[Tan[c + d*x]^2/(a + I*a*Tan[c + d*x])^(1/3),x]
Output:
((I/2)*(((-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)/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)
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[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[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]
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_)])^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])
Time = 0.98 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {3 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\frac {a \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )}{2}-\frac {a}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d a}\) | \(172\) |
| default | \(\frac {3 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\frac {a \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )}{2}-\frac {a}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d a}\) | \(172\) |
Input:
int(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
Output:
3*I/d/a*(-1/2*(a+I*a*tan(d*x+c))^(2/3)-1/2*a*(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)))-1/2*a/(a+I*a*tan(d*x+c))^(1/3))
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 = 0.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.48 \[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {{\left (4 \, a d \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (8 \, a d^{2} \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 2 \, {\left (-i \, \sqrt {3} a d + a d\right )} \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 2 \, {\left (i \, \sqrt {3} a d + a d\right )} \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (-i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \] Input:
integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")
Output:
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)
\[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int \frac {\tan ^{2}{\left (c + d x \right )}}{\sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:
integrate(tan(d*x+c)**2/(a+I*a*tan(d*x+c))**(1/3),x)
Output:
Integral(tan(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(1/3), x)
Time = 0.12 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {8}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{2} + \frac {12 \, a^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a^{3} d} \] Input:
integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")
Output:
-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)
Time = 0.29 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {-2 i \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {5}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + i \cdot 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 i \cdot 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 12 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a - \frac {12 i \, a^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}}{8 \, a^{2} d} \] Input:
integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")
Output:
1/8*(-2*I*sqrt(3)*2^(2/3)*a^(5/3)*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)) + I*2^(2/3)*a^(5/3)*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*I*2^(2/3)*a^(5/3)*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)^(2/3)*a - 12*I*a^2/(I*a*t an(d*x + c) + a)^(1/3))/(a^2*d)
Time = 1.77 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}\,3{}\mathrm {i}}{2\,a\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (9\,{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{a^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{1/3}\,d} \] Input:
int(tan(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(1/3),x)
Output:
((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*tan(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)
\[ \int \frac {\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {\int \frac {\tan \left (d x +c \right )^{2}}{\left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}}}d x}{a^{\frac {1}{3}}} \] Input:
int(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x)
Output:
int(tan(c + d*x)**2/(tan(c + d*x)*i + 1)**(1/3),x)/a**(1/3)