Integrand size = 17, antiderivative size = 213 \[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=-\frac {x}{8 \sqrt [3]{2} a^{4/3}}+\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \] Output:
-1/16*x*2^(2/3)/a^(4/3)+1/8*I*3^(1/2)*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*t an(d*x+c))^(1/3))*3^(1/2)/a^(1/3))*2^(2/3)/a^(4/3)/d+1/16*I*ln(cos(d*x+c)) *2^(2/3)/a^(4/3)/d+3/16*I*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*I/d/(a+I*a*tan(d*x+c))^(4/3)+3/4*I/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 = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.23 \[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},1,-\frac {1}{3},\frac {1}{2} (1+i \tan (c+d x))\right )}{8 d (a+i a \tan (c+d x))^{4/3}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(-4/3),x]
Output:
(((3*I)/8)*Hypergeometric2F1[-4/3, 1, -1/3, (1 + I*Tan[c + d*x])/2])/(d*(a + I*a*Tan[c + d*x])^(4/3))
Time = 0.46 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.80, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {3042, 3960, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{4/3}}dx\) |
| \(\Big \downarrow \) 3960 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3960 |
| \(\displaystyle \frac {\frac {\int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}+\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int (i \tan (c+d x) a+a)^{2/3}dx}{2 a}+\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 d}}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}}{2 a}+\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(-4/3),x]
Output:
((3*I)/8)/(d*(a + I*a*Tan[c + d*x])^(4/3)) + (((-1/2*I)*(((-I)*Sqrt[3]*Arc Tanh[(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)))/(2*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]
Time = 0.92 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {3 i a \left (\frac {1}{4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {1}{8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}+\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{4 a^{2}}\right )}{d}\) | \(177\) |
| default | \(\frac {3 i a \left (\frac {1}{4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {1}{8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}+\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{4 a^{2}}\right )}{d}\) | \(177\) |
Input:
int(1/(a+I*a*tan(d*x+c))^(4/3),x,method=_RETURNVERBOSE)
Output:
3*I/d*a*(1/4/a^2/(a+I*a*tan(d*x+c))^(1/3)+1/8/a/(a+I*a*tan(d*x+c))^(4/3)+1 /4/a^2*(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))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (146) = 292\).
Time = 0.09 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.62 \[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {{\left (32 \, a^{2} d \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (32 \, a^{3} d^{2} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (-5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 16 \, {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-16 \, {\left (i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 16 \, {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-16 \, {\left (-i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (-\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")
Output:
1/32*(32*a^2*d*(-1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(32*a^3*d ^2*(-1/128*I/(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) *(-5*I*e^(4*I*d*x + 4*I*c) - 6*I*e^(2*I*d*x + 2*I*c) - I)*e^(4/3*I*d*x + 4 /3*I*c) - 16*(-I*sqrt(3)*a^2*d + a^2*d)*(-1/128*I/(a^4*d^3))^(1/3)*e^(4*I* d*x + 4*I*c)*log(-16*(I*sqrt(3)*a^3*d^2 + a^3*d^2)*(-1/128*I/(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)) - 16*(I*sqrt(3)*a^2*d + a^2*d)*(-1/128*I/(a^4*d^3))^(1/3)*e^(4*I*d*x + 4* I*c)*log(-16*(-I*sqrt(3)*a^3*d^2 + a^3*d^2)*(-1/128*I/(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)))*e^(-4 *I*d*x - 4*I*c)/(a^2*d)
\[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {4}{3}}}\, dx \] Input:
integrate(1/(a+I*a*tan(d*x+c))**(4/3),x)
Output:
Integral((I*a*tan(c + d*x) + a)**(-4/3), x)
Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {i \, {\left (\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - \frac {2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {1}{3}}} + \frac {2 \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {1}{3}}} + \frac {6 \, {\left (2 i \, a \tan \left (d x + c\right ) + 3 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\right )}}{16 \, a d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")
Output:
1/16*I*(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) + 3*a)/(I*a*tan(d*x + c) + a)^(4/3) )/(a*d)
Time = 0.34 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {i \, {\left (\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {4}{3}}} - \frac {2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {4}{3}}} + \frac {2 \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {4}{3}}} - \frac {6 \, {\left (-2 i \, a \tan \left (d x + c\right ) - 3 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a}\right )}}{16 \, d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")
Output:
1/16*I*(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^(4/3) - 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^(4/3) + 2*2^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^ (1/3))/a^(4/3) - 6*(-2*I*a*tan(d*x + c) - 3*a)/((I*a*tan(d*x + c) + a)^(4/ 3)*a))/d
Time = 1.65 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {\frac {3{}\mathrm {i}}{8\,d}+\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{4\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}-\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (9\,{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{a^{4/3}\,d}+\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d}-\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d} \] Input:
int(1/(a + a*tan(c + d*x)*1i)^(4/3),x)
Output:
(3i/(8*d) + ((a + a*tan(c + d*x)*1i)*3i)/(4*a*d))/(a + a*tan(c + d*x)*1i)^ (4/3) - ((1i/128)^(1/3)*log(9*(a*(tan(c + d*x)*1i + 1))^(1/3) + 9*(-1)^(1/ 3)*2^(1/3)*a^(1/3)))/(a^(4/3)*d) + ((1i/128)^(1/3)*log(- (9*(a + a*tan(c + d*x)*1i)^(1/3))/(16*a^2*d^2) - (9*(-1)^(1/3)*2^(1/3)*(3^(1/2)*1i - 1))/(3 2*a^(5/3)*d^2))*((3^(1/2)*1i)/2 + 1/2))/(a^(4/3)*d) - ((1i/128)^(1/3)*log( (9*(-1)^(1/3)*2^(1/3)*(3^(1/2)*1i + 1))/(32*a^(5/3)*d^2) - (9*(a + a*tan(c + d*x)*1i)^(1/3))/(16*a^2*d^2))*((3^(1/2)*1i)/2 - 1/2))/(a^(4/3)*d)
\[ \int \frac {1}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {\int \frac {1}{\left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}} \tan \left (d x +c \right ) i +\left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}}}d x}{a^{\frac {4}{3}}} \] Input:
int(1/(a+I*a*tan(d*x+c))^(4/3),x)
Output:
int(1/((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)