Integrand size = 17, antiderivative size = 213 \[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=-\frac {x}{8\ 2^{2/3} a^{5/3}}-\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4\ 2^{2/3} a^{5/3} d}+\frac {i \log (\cos (c+d x))}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}} \] Output:
-1/16*x*2^(1/3)/a^(5/3)-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^(1/3)/a^(5/3)/d+1/16*I*ln(cos(d*x+c)) *2^(1/3)/a^(5/3)/d+3/16*I*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^( 1/3)/a^(5/3)/d+3/10*I/d/(a+I*a*tan(d*x+c))^(5/3)+3/8*I/a/d/(a+I*a*tan(d*x+ c))^(2/3)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.14 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.23 \[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},1,-\frac {2}{3},\frac {1}{2} (1+i \tan (c+d x))\right )}{10 d (a+i a \tan (c+d x))^{5/3}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(-5/3),x]
Output:
(((3*I)/10)*Hypergeometric2F1[-5/3, 1, -2/3, (1 + I*Tan[c + d*x])/2])/(d*( a + I*a*Tan[c + d*x])^(5/3))
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, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{5/3}}dx\) |
| \(\Big \downarrow \) 3960 |
| \(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^{2/3}}dx}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^{2/3}}dx}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 3960 |
| \(\displaystyle \frac {\frac {\int \sqrt [3]{i \tan (c+d x) a+a}dx}{2 a}+\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}}}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \sqrt [3]{i \tan (c+d x) a+a}dx}{2 a}+\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}}}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}}-\frac {i \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}d(i a \tan (c+d x))}{2 d}}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}}-\frac {i \left (\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2\ 2^{2/3} a^{2/3}}+\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{2 d}}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}}-\frac {i \left (\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{2 d}}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}}-\frac {i \left (-\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{2 d}}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {3 i}{4 d (a+i a \tan (c+d x))^{2/3}}-\frac {i \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{2 d}}{2 a}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(-5/3),x]
Output:
((3*I)/10)/(d*(a + I*a*Tan[c + d*x])^(5/3)) + (((-1/2*I)*((I*Sqrt[3]*ArcTa nh[(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)/4)/(d*(a + I*a*Tan[c + d*x])^(2/3)))/(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_))^(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[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 | \(\frac {3 i a \left (\frac {1}{8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {1}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}+\frac {\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}}{4 a^{2}}\right )}{d}\) | \(177\) |
| default | \(\frac {3 i a \left (\frac {1}{8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {1}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}+\frac {\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}}{4 a^{2}}\right )}{d}\) | \(177\) |
Input:
int(1/(a+I*a*tan(d*x+c))^(5/3),x,method=_RETURNVERBOSE)
Output:
3*I/d*a*(1/8/a^2/(a+I*a*tan(d*x+c))^(2/3)+1/10/a/(a+I*a*tan(d*x+c))^(5/3)+ 1/4/a^2*(1/6*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))- 1/12*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*ta n(d*x+c))^(1/3)+2^(2/3)*a^(2/3))-1/6*2^(1/3)/a^(2/3)*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. 336 vs. \(2 (146) = 292\).
Time = 0.09 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=\frac {{\left (80 \, a^{2} d \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (8 i \, a^{2} d \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 40 \, {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - 4 \, {\left (\sqrt {3} a^{2} d + i \, a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}}\right ) - 40 \, {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + 4 \, {\left (\sqrt {3} a^{2} d - i \, a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}}\right ) - 3 \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} {\left (-7 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{80 \, a^{2} d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^(5/3),x, algorithm="fricas")
Output:
1/80*(80*a^2*d*(-1/256*I/(a^5*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(8*I*a^2* d*(-1/256*I/(a^5*d^3))^(1/3) + 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)) - 40*(-I*sqrt(3)*a^2*d + a^2*d)*(-1/256*I/(a^5*d ^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(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*(sqrt(3)*a^2*d + I*a^2*d)*(-1/256*I/(a^5* d^3))^(1/3)) - 40*(I*sqrt(3)*a^2*d + a^2*d)*(-1/256*I/(a^5*d^3))^(1/3)*e^( 4*I*d*x + 4*I*c)*log(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*(sqrt(3)*a^2*d - I*a^2*d)*(-1/256*I/(a^5*d^3))^(1/3)) - 3*2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*(-7*I*e^(4*I*d*x + 4*I*c) - 9*I*e^(2*I*d*x + 2*I*c) - 2*I)*e^(2/3*I*d*x + 2/3*I*c))*e^(-4*I*d*x - 4*I *c)/(a^2*d)
\[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=\int \frac {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {5}{3}}}\, dx \] Input:
integrate(1/(a+I*a*tan(d*x+c))**(5/3),x)
Output:
Integral((I*a*tan(c + d*x) + a)**(-5/3), x)
Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=-\frac {i \, {\left (\frac {10 \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {2}{3}}} + \frac {5 \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {2}{3}}} - \frac {10 \cdot 2^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {2}{3}}} - \frac {6 \, {\left (5 i \, a \tan \left (d x + c\right ) + 9 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\right )}}{80 \, a d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^(5/3),x, algorithm="maxima")
Output:
-1/80*I*(10*sqrt(3)*2^(1/3)*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^(2/3) + 5*2^(1/3)*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^(2/3) - 10*2^(1/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3))/a^(2/3) - 6*(5*I*a*tan(d*x + c) + 9*a)/(I*a*tan(d*x + c) + a)^ (5/3))/(a*d)
Time = 0.39 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=-\frac {i \, {\left (\frac {10 \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}}} + \frac {5 \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {5}{3}}} - \frac {10 \cdot 2^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {5}{3}}} - \frac {6 \, {\left (5 i \, a \tan \left (d x + c\right ) + 9 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a}\right )}}{80 \, d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^(5/3),x, algorithm="giac")
Output:
-1/80*I*(10*sqrt(3)*2^(1/3)*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^(5/3) + 5*2^(1/3)*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^(5/3) - 10*2^(1/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3))/a^(5/3) - 6*(5*I*a*tan(d*x + c) + 9*a)/((I*a*tan(d*x + c) + a) ^(5/3)*a))/d
Time = 0.20 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.09 \[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=\frac {\frac {3{}\mathrm {i}}{10\,d}+\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}}+\frac {{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,144{}\mathrm {i}-1152\,{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{7/3}\,d^2\right )}{{\left (-a\right )}^{5/3}\,d}+\frac {{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,144{}\mathrm {i}-1152\,{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{7/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{5/3}\,d}-\frac {{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,144{}\mathrm {i}+1152\,{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{7/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{5/3}\,d} \] Input:
int(1/(a + a*tan(c + d*x)*1i)^(5/3),x)
Output:
(3i/(10*d) + ((a + a*tan(c + d*x)*1i)*3i)/(8*a*d))/(a + a*tan(c + d*x)*1i) ^(5/3) + ((1i/256)^(1/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*144i - 1152*(1i/256)^(1/3)*(-a)^(7/3)*d^2))/((-a)^(5/3)*d) + ((1i/256)^(1/3)*log( a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*144i - 1152*(1i/256)^(1/3)*(-a)^(7/3 )*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/((-a)^(5/3)*d) - ((1 i/256)^(1/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*144i + 1152*(1i/256 )^(1/3)*(-a)^(7/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/((- a)^(5/3)*d)
\[ \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx=\frac {\int \frac {1}{\left (\tan \left (d x +c \right ) i +1\right )^{\frac {2}{3}} \tan \left (d x +c \right ) i +\left (\tan \left (d x +c \right ) i +1\right )^{\frac {2}{3}}}d x}{a^{\frac {5}{3}}} \] Input:
int(1/(a+I*a*tan(d*x+c))^(5/3),x)
Output:
int(1/((tan(c + d*x)*i + 1)**(2/3)*tan(c + d*x)*i + (tan(c + d*x)*i + 1)** (2/3)),x)/(a**(2/3)*a)