Integrand size = 26, antiderivative size = 237 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=-\frac {i x}{8 \sqrt [3]{2} a^{4/3}}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {15}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {27}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \]
-1/16*I*x*2^(2/3)/a^(4/3)-1/16*ln(cos(d*x+c))*2^(2/3)/a^(4/3)/d-3/16*ln(2^ (1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(2/3)/a^(4/3)/d-1/8*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^(4/3)/d+15/8/d/(a+I*a*tan(d*x+c))^(4/3)+3/2*tan(d*x+c)^2/d/(a+I*a*tan(d *x+c))^(4/3)-27/4/a/d/(a+I*a*tan(d*x+c))^(1/3)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.60 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.50 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {3 \sec ^2(c+d x) \left (9 i+17 i \cos (2 (c+d x))-18 \sin (2 (c+d x))+\operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {1}{2} (1+i \tan (c+d x))\right ) (-i \cos (2 (c+d x))+\sin (2 (c+d x)))\right )}{16 a d (-i+\tan (c+d x)) \sqrt [3]{a+i a \tan (c+d x)}} \]
(3*Sec[c + d*x]^2*(9*I + (17*I)*Cos[2*(c + d*x)] - 18*Sin[2*(c + d*x)] + H ypergeometric2F1[2/3, 1, 5/3, (1 + I*Tan[c + d*x])/2]*((-I)*Cos[2*(c + d*x )] + Sin[2*(c + d*x)])))/(16*a*d*(-I + Tan[c + d*x])*(a + I*a*Tan[c + d*x] )^(1/3))
Time = 0.78 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4043, 27, 3042, 4073, 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.
| \(\displaystyle \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3}{(a+i a \tan (c+d x))^{4/3}}dx\) |
| \(\Big \downarrow \) 4043 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {3 \int \frac {2 \tan (c+d x) (3 a-2 i a \tan (c+d x))}{3 (i \tan (c+d x) a+a)^{4/3}}dx}{2 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {\int \frac {\tan (c+d x) (3 a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{4/3}}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {\int \frac {\tan (c+d x) (3 a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{4/3}}dx}{a}\) |
| \(\Big \downarrow \) 4073 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \int \frac {5 a^2-4 i a^2 \tan (c+d x)}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \int \frac {5 a^2-4 i a^2 \tan (c+d x)}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 4009 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \left (\frac {1}{2} a \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {27 i a^2}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \left (\frac {1}{2} a \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {27 i a^2}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \left (\frac {27 i a^2}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i a^2 \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 d}\right )}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \left (\frac {27 i a^2}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i a^2 \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \left (\frac {27 i a^2}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i a^2 \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \left (\frac {27 i a^2}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i a^2 \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 \tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))^{4/3}}-\frac {-\frac {i \left (\frac {27 i a^2}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i a^2 \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a^2}-\frac {15 a}{8 d (a+i a \tan (c+d x))^{4/3}}}{a}\) |
(3*Tan[c + d*x]^2)/(2*d*(a + I*a*Tan[c + d*x])^(4/3)) - ((-15*a)/(8*d*(a + I*a*Tan[c + d*x])^(4/3)) - ((I/2)*(((-1/2*I)*a^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 + (((27*I)/2)*a^2)/(d*(a + I*a*Tan[c + d*x])^(1/3))))/a^2)/a
3.4.1.3.1 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[(-(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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[1/(a*(m + n - 1)) Int[(a + b *Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m - a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-( A*b - a*B))*(a*c + b*d)*((a + b*Tan[e + f*x])^m/(2*a^2*f*m)), x] + Simp[1/( 2*a*b) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[A*b*c + a*B*c + a*A*d + b*B* d + 2*a*B*d*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 + b^2, 0]
Time = 0.55 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 d \,a^{2}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}-\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {15}{4 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {3}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\) | \(198\) |
| default | \(-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 d \,a^{2}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}-\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {15}{4 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {3}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\) | \(198\) |
-3/2/d/a^2*(a+I*a*tan(d*x+c))^(2/3)-1/8/d/a^(4/3)*2^(2/3)*ln((a+I*a*tan(d* x+c))^(1/3)-2^(1/3)*a^(1/3))+1/16/d/a^(4/3)*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))-1/8/d/a^(4 /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))-15/4/a/d/(a+I*a*tan(d*x+c))^(1/3)+3/8/d/(a+I*a*tan(d*x+c))^(4/3 )
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (172) = 344\).
Time = 0.25 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.54 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {{\left (8 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a^{2} d \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a^{3} d^{2} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (-\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (35 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]
1/32*(8*(1/2)^(1/3)*a^2*d*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-2* (1/2)^(2/3)*a^3*d^2*(-1/(a^4*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)) - 4*(1/2)^(1/3)*(I*sqrt(3)*a^2*d + a ^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-(1/2)^(2/3)*(I*sqrt(3) *a^3*d^2 - a^3*d^2)*(-1/(a^4*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)) - 4*(1/2)^(1/3)*(-I*sqrt(3)*a^2*d + a^2*d)*(-1/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(-(1/2)^(2/3)*(-I*sqrt( 3)*a^3*d^2 - a^3*d^2)*(-1/(a^4*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)*(35*e^(4*I*d*x + 4*I*c) + 18*e^(2*I*d*x + 2*I*c) - 1)*e^(4/3 *I*d*x + 4/3*I*c))*e^(-4*I*d*x - 4*I*c)/(a^2*d)
\[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=-\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {8}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{2} + \frac {6 \, {\left (10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} - a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}}{16 \, a^{4} d} \]
-1/16*(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)) + 24*(I*a*tan(d*x + c) + a)^(2/3)*a^2 + 6*(10*(I*a*tan(d*x + c) + a)*a^3 - a^4)/(I*a*tan(d*x + c) + a)^(4/3))/(a^4*d)
\[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int { \frac {\tan \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
Time = 0.20 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=-\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}{2\,a^2\,d}-\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-18\,4^{2/3}\,a^{4/3}\,d\right )}{8\,a^{4/3}\,d}-\frac {\frac {27\,a}{8}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,15{}\mathrm {i}}{4}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}-\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d}+\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d} \]
(4^(1/3)*log(36*a*d*(a + a*tan(c + d*x)*1i)^(1/3) - 1152*4^(2/3)*a^(4/3)*d *((3^(1/2)*1i)/16 + 1/16)^2)*((3^(1/2)*1i)/16 + 1/16))/(a^(4/3)*d) - (4^(1 /3)*log(36*a*d*(a + a*tan(c + d*x)*1i)^(1/3) - 18*4^(2/3)*a^(4/3)*d))/(8*a ^(4/3)*d) - ((27*a)/8 + (a*tan(c + d*x)*15i)/4)/(a*d*(a + a*tan(c + d*x)*1 i)^(4/3)) - (4^(1/3)*log(36*a*d*(a + a*tan(c + d*x)*1i)^(1/3) - 1152*4^(2/ 3)*a^(4/3)*d*((3^(1/2)*1i)/16 - 1/16)^2)*((3^(1/2)*1i)/16 - 1/16))/(a^(4/3 )*d) - (3*(a + a*tan(c + d*x)*1i)^(2/3))/(2*a^2*d)