Integrand size = 26, antiderivative size = 251 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=-\frac {i a^{4/3} x}{2^{2/3}}+\frac {\sqrt [3]{2} \sqrt {3} a^{4/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}-\frac {a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac {3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}-\frac {3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {9 (a+i a \tan (c+d x))^{4/3}}{20 d}+\frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {6 (a+i a \tan (c+d x))^{7/3}}{35 a d} \]
-1/2*I*a^(4/3)*x*2^(1/3)-1/2*a^(4/3)*ln(cos(d*x+c))*2^(1/3)/d-3/2*a^(4/3)* ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(1/3)/d+2^(1/3)*a^(4/3)*arc tan(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 )/d-3*a*(a+I*a*tan(d*x+c))^(1/3)/d-9/20*(a+I*a*tan(d*x+c))^(4/3)/d+3/10*ta n(d*x+c)^2*(a+I*a*tan(d*x+c))^(4/3)/d-6/35*(a+I*a*tan(d*x+c))^(7/3)/a/d
Time = 1.61 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.12 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\frac {140 \sqrt [3]{2} \sqrt {3} a^{4/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-140 \sqrt [3]{2} a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )+70 \sqrt [3]{2} a^{4/3} \log \left (2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} \sqrt [3]{a+i a \tan (c+d x)}+(a+i a \tan (c+d x))^{2/3}\right )-507 a \sqrt [3]{a+i a \tan (c+d x)}-111 i a \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)}+66 a \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)}+42 i a \tan ^3(c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{140 d} \]
(140*2^(1/3)*Sqrt[3]*a^(4/3)*ArcTan[(1 + (2^(2/3)*(a + I*a*Tan[c + d*x])^( 1/3))/a^(1/3))/Sqrt[3]] - 140*2^(1/3)*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a + I *a*Tan[c + d*x])^(1/3)] + 70*2^(1/3)*a^(4/3)*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)] - 50 7*a*(a + I*a*Tan[c + d*x])^(1/3) - (111*I)*a*Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(1/3) + 66*a*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(1/3) + (42*I)*a *Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^(1/3))/(140*d)
Time = 0.90 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.92, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {3042, 4043, 27, 3042, 4075, 3042, 4010, 3042, 3959, 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.
\(\displaystyle \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 (a+i a \tan (c+d x))^{4/3}dx\) |
\(\Big \downarrow \) 4043 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {3 \int \frac {2}{3} \tan (c+d x) (i \tan (c+d x) a+a)^{4/3} (2 i \tan (c+d x) a+3 a)dx}{10 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {\int \tan (c+d x) (i \tan (c+d x) a+a)^{4/3} (2 i \tan (c+d x) a+3 a)dx}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {\int \tan (c+d x) (i \tan (c+d x) a+a)^{4/3} (2 i \tan (c+d x) a+3 a)dx}{5 a}\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {\int (i \tan (c+d x) a+a)^{4/3} (3 a \tan (c+d x)-2 i a)dx+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {\int (i \tan (c+d x) a+a)^{4/3} (3 a \tan (c+d x)-2 i a)dx+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}}{5 a}\) |
\(\Big \downarrow \) 4010 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \int (i \tan (c+d x) a+a)^{4/3}dx+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \int (i \tan (c+d x) a+a)^{4/3}dx+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \left (2 a \int \sqrt [3]{i \tan (c+d x) a+a}dx+\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}\right )+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \left (2 a \int \sqrt [3]{i \tan (c+d x) a+a}dx+\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}\right )+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 3962 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \left (\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {2 i a^2 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}d(i a \tan (c+d x))}{d}\right )+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 69 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \left (\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {2 i a^2 \left (\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2\ 2^{2/3} a^{2/3}}+\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\right )+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \left (\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {2 i a^2 \left (\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\right )+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \left (\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {2 i a^2 \left (-\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\right )+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {3 \tan ^2(c+d x) (a+i a \tan (c+d x))^{4/3}}{10 d}-\frac {-5 i a \left (\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {2 i a^2 \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\right )+\frac {6 (a+i a \tan (c+d x))^{7/3}}{7 d}+\frac {9 a (a+i a \tan (c+d x))^{4/3}}{4 d}}{5 a}\) |
(3*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(4/3))/(10*d) - ((9*a*(a + I*a*Ta n[c + d*x])^(4/3))/(4*d) + (6*(a + I*a*Tan[c + d*x])^(7/3))/(7*d) - (5*I)* a*(((-2*I)*a^2*((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*I)*a*(a + I*a*Tan[ c + d*x])^(1/3))/d))/(5*a)
3.3.80.3.1 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_))^(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*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[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] && GtQ[n , 1]
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_)*((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[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]
Time = 0.55 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {10}{3}}}{10 d \,a^{2}}+\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{3}}}{7 a d}-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4 d}-\frac {3 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}-\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{d}+\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{2 d}+\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{d}\) | \(217\) |
default | \(-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {10}{3}}}{10 d \,a^{2}}+\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{3}}}{7 a d}-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4 d}-\frac {3 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}-\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{d}+\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{2 d}+\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{d}\) | \(217\) |
-3/10/d/a^2*(a+I*a*tan(d*x+c))^(10/3)+3/7*(a+I*a*tan(d*x+c))^(7/3)/a/d-3/4 *(a+I*a*tan(d*x+c))^(4/3)/d-3*a*(a+I*a*tan(d*x+c))^(1/3)/d-1/d*a^(4/3)*2^( 1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))+1/2/d*a^(4/3)*2^(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/d*a^(4/3)*2^(1/3)*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))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 501 vs. \(2 (190) = 380\).
Time = 0.27 (sec) , antiderivative size = 501, normalized size of antiderivative = 2.00 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=-\frac {3 \cdot 2^{\frac {1}{3}} {\left (121 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 240 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 245 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 70 \, a\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 )} + 35 \cdot 2^{\frac {1}{3}} {\left ({\left (-i \, \sqrt {3} d + d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (-i \, \sqrt {3} d + d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\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 {a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 2^{\frac {1}{3}} 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^{\frac {1}{3}} {\left (i \, \sqrt {3} d - d\right )} \left (-\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + 35 \cdot 2^{\frac {1}{3}} {\left ({\left (i \, \sqrt {3} d + d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (i \, \sqrt {3} d + d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\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 {a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 2^{\frac {1}{3}} 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^{\frac {1}{3}} {\left (-i \, \sqrt {3} d - d\right )} \left (-\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) - 70 \cdot 2^{\frac {1}{3}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (-\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} 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^{\frac {1}{3}} \left (-\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}} d}{a}\right )}{70 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-1/70*(3*2^(1/3)*(121*a*e^(6*I*d*x + 6*I*c) + 240*a*e^(4*I*d*x + 4*I*c) + 245*a*e^(2*I*d*x + 2*I*c) + 70*a)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2 /3*I*d*x + 2/3*I*c) + 35*2^(1/3)*((-I*sqrt(3)*d + d)*e^(6*I*d*x + 6*I*c) + 3*(-I*sqrt(3)*d + d)*e^(4*I*d*x + 4*I*c) + 3*(-I*sqrt(3)*d + d)*e^(2*I*d* x + 2*I*c) - I*sqrt(3)*d + d)*(-a^4/d^3)^(1/3)*log(1/2*(2*2^(1/3)*a*(a/(e^ (2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + 2^(1/3)*(I*sqrt(3) *d - d)*(-a^4/d^3)^(1/3))/a) + 35*2^(1/3)*((I*sqrt(3)*d + d)*e^(6*I*d*x + 6*I*c) + 3*(I*sqrt(3)*d + d)*e^(4*I*d*x + 4*I*c) + 3*(I*sqrt(3)*d + d)*e^( 2*I*d*x + 2*I*c) + I*sqrt(3)*d + d)*(-a^4/d^3)^(1/3)*log(1/2*(2*2^(1/3)*a* (a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + 2^(1/3)*(-I* sqrt(3)*d - d)*(-a^4/d^3)^(1/3))/a) - 70*2^(1/3)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*(-a^4/d^3)^(1/3)*lo g((2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + 2^(1/3)*(-a^4/d^3)^(1/3)*d)/a))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}} \tan ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.31 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.83 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\frac {140 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {16}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 70 \cdot 2^{\frac {1}{3}} a^{\frac {16}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 140 \cdot 2^{\frac {1}{3}} a^{\frac {16}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 42 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {10}{3}} a^{2} + 60 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{3}} a^{3} - 105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a^{4} - 420 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{5}}{140 \, a^{4} d} \]
1/140*(140*sqrt(3)*2^(1/3)*a^(16/3)*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)) + 70*2^(1/3)*a^(16/3)*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)) - 140*2^(1/3)*a^(16/3)*log(-2^(1/3)*a^(1/3) + (I*a*t an(d*x + c) + a)^(1/3)) - 42*(I*a*tan(d*x + c) + a)^(10/3)*a^2 + 60*(I*a*t an(d*x + c) + a)^(7/3)*a^3 - 105*(I*a*tan(d*x + c) + a)^(4/3)*a^4 - 420*(I *a*tan(d*x + c) + a)^(1/3)*a^5)/(a^4*d)
\[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} \tan \left (d x + c\right )^{3} \,d x } \]
Time = 5.30 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.97 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=-\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}{4\,d}+\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/3}}{7\,a\,d}-\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{10/3}}{10\,a^2\,d}-\frac {3\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {18\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,a^{7/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {18\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,a^{7/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \]
(3*(a + a*tan(c + d*x)*1i)^(7/3))/(7*a*d) - (3*(a + a*tan(c + d*x)*1i)^(4/ 3))/(4*d) - (3*(a + a*tan(c + d*x)*1i)^(10/3))/(10*a^2*d) - (3*a*(a + a*ta n(c + d*x)*1i)^(1/3))/d - (2^(1/3)*a^(4/3)*log((a*(tan(c + d*x)*1i + 1))^( 1/3) - 2^(1/3)*a^(1/3)))/d - (2^(1/3)*a^(4/3)*log((18*a^2*(a + a*tan(c + d *x)*1i)^(1/3))/d - (9*2^(1/3)*a^(7/3)*(3^(1/2)*1i - 1))/d)*((3^(1/2)*1i)/2 - 1/2))/d + (2^(1/3)*a^(4/3)*log((18*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d + (9*2^(1/3)*a^(7/3)*(3^(1/2)*1i + 1))/d)*((3^(1/2)*1i)/2 + 1/2))/d