Integrand size = 24, antiderivative size = 205 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \] Output:
1/16*I*x*2^(2/3)/a^(4/3)+1/8*3^(1/2)*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*ta n(d*x+c))^(1/3))*3^(1/2)/a^(1/3))*2^(2/3)/a^(4/3)/d+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-3/8/d/(a+I*a*tan(d*x+c))^(4/3)+3/4/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 = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.31 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {3 \left (-1+\operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},\frac {1}{2} (1+i \tan (c+d x))\right ) (2+2 i \tan (c+d x))\right )}{8 d (a+i a \tan (c+d x))^{4/3}} \] Input:
Integrate[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^(4/3),x]
Output:
(3*(-1 + Hypergeometric2F1[-1/3, 1, 2/3, (1 + I*Tan[c + d*x])/2]*(2 + (2*I )*Tan[c + d*x])))/(8*d*(a + I*a*Tan[c + d*x])^(4/3))
Time = 0.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.83, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4009, 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.
| \(\displaystyle \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}}dx\) |
| \(\Big \downarrow \) 4009 |
| \(\displaystyle -\frac {i \int \frac {1}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {i \int \frac {1}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3960 |
| \(\displaystyle -\frac {i \left (\frac {\int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}+\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\right )}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {i \left (\frac {\int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}+\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}\right )}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle -\frac {i \left (\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 d}\right )}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle -\frac {i \left (\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {i \left (\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {i \left (\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {i \left (\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\right )}{2 a}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
Input:
Int[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^(4/3),x]
Output:
-3/(8*d*(a + I*a*Tan[c + d*x])^(4/3)) - ((I/2)*(((-1/2*I)*(((-I)*Sqrt[3]*A rcTanh[(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))))/a
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_)]), 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 = 0.90 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(-\frac {3}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}+\frac {3}{4 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}\) | \(176\) |
| default | \(-\frac {3}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}+\frac {3}{4 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}\) | \(176\) |
Input:
int(tan(d*x+c)/(a+I*a*tan(d*x+c))^(4/3),x,method=_RETURNVERBOSE)
Output:
-3/8/d/(a+I*a*tan(d*x+c))^(4/3)+3/4/a/d/(a+I*a*tan(d*x+c))^(1/3)+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))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (146) = 292\).
Time = 0.09 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.76 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {{\left (8 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a^{2} d \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a^{3} d^{2} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \] Input:
integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")
Output:
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)*(3*e^(4*I*d*x + 4*I*c) + 2*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)
\[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \] Input:
integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))**(4/3),x)
Output:
Integral(tan(c + d*x)/(I*a*(tan(c + d*x) - I))**(4/3), x)
Time = 0.13 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.83 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {6 \, {\left (2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}}{16 \, a^{2} d} \] Input:
integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")
Output:
1/16*(2*sqrt(3)*2^(2/3)*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^(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^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) + 6*(2*(I*a*tan(d*x + c) + a)*a - a^2)/(I*a*tan(d*x + c) + a) ^(4/3))/(a^2*d)
Time = 0.41 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.79 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - \frac {2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {1}{3}}} + \frac {2 \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {1}{3}}} + \frac {6 \, {\left (2 i \, a \tan \left (d x + c\right ) + a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}}{16 \, a d} \] Input:
integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")
Output:
1/16*(2*sqrt(3)*2^(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))/a^(1/3) - 2^(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))/a^(1/3) + 2*2^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1 /3))/a^(1/3) + 6*(2*I*a*tan(d*x + c) + a)/(I*a*tan(d*x + c) + a)^(4/3))/(a *d)
Time = 0.17 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.94 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {\frac {3\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{4\,a}-\frac {3}{8}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}+\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-18\,4^{2/3}\,a^{4/3}\,d\right )}{8\,a^{4/3}\,d}+\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d}-\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d} \] Input:
int(tan(c + d*x)/(a + a*tan(c + d*x)*1i)^(4/3),x)
Output:
((3*(a + a*tan(c + d*x)*1i))/(4*a) - 3/8)/(d*(a + a*tan(c + d*x)*1i)^(4/3) ) + (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) + (4^(1/3)*log(36*a*d*(a + a*tan(c + d*x)*1i)^(1/3) - 1 152*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) - 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)
\[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {\int \frac {\tan \left (d x +c \right )}{\left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}} \tan \left (d x +c \right ) i +\left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}}}d x}{a^{\frac {4}{3}}} \] Input:
int(tan(d*x+c)/(a+I*a*tan(d*x+c))^(4/3),x)
Output:
int(tan(c + d*x)/((tan(c + d*x)*i + 1)**(1/3)*tan(c + d*x)*i + (tan(c + d* x)*i + 1)**(1/3)),x)/(a**(1/3)*a)