Integrand size = 17, antiderivative size = 177 \[ \int (a+i a \tan (c+d x))^{5/3} \, dx=-\frac {a^{5/3} x}{\sqrt [3]{2}}+\frac {i 2^{2/3} \sqrt {3} a^{5/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}+\frac {i a^{5/3} \log (\cos (c+d x))}{\sqrt [3]{2} d}+\frac {3 i a^{5/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{\sqrt [3]{2} d}+\frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d} \] Output:
-1/2*a^(5/3)*x*2^(2/3)+I*2^(2/3)*3^(1/2)*a^(5/3)*arctan(1/3*(a^(1/3)+2^(2/ 3)*(a+I*a*tan(d*x+c))^(1/3))*3^(1/2)/a^(1/3))/d+1/2*I*a^(5/3)*ln(cos(d*x+c ))*2^(2/3)/d+3/2*I*a^(5/3)*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^ (2/3)/d+3/2*I*a*(a+I*a*tan(d*x+c))^(2/3)/d
Time = 0.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.85 \[ \int (a+i a \tan (c+d x))^{5/3} \, dx=\frac {i a \left (2\ 2^{2/3} \sqrt {3} a^{2/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2^{2/3} a^{2/3} \log (i+\tan (c+d x))+3 \left (2^{2/3} a^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )+(a+i a \tan (c+d x))^{2/3}\right )\right )}{2 d} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(5/3),x]
Output:
((I/2)*a*(2*2^(2/3)*Sqrt[3]*a^(2/3)*ArcTan[(1 + (2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/a^(1/3))/Sqrt[3]] - 2^(2/3)*a^(2/3)*Log[I + Tan[c + d*x]] + 3 *(2^(2/3)*a^(2/3)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)] + (a + I*a*Tan[c + d*x])^(2/3))))/d
Time = 0.34 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.78, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 3959, 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.
| \(\displaystyle \int (a+i a \tan (c+d x))^{5/3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (c+d x))^{5/3}dx\) |
| \(\Big \downarrow \) 3959 |
| \(\displaystyle 2 a \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 a \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {2 i a^2 \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {2 i a^2 \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {2 i a^2 \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {2 i a^2 \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 i a (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {2 i a^2 \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(5/3),x]
Output:
((-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 + (((3*I)/2)*a*(a + I*a* Tan[c + d*x])^(2/3))/d
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[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]
Time = 0.84 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {3 i a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+2 a \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )\right )}{d}\) | \(153\) |
| default | \(\frac {3 i a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+2 a \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )\right )}{d}\) | \(153\) |
Input:
int((a+I*a*tan(d*x+c))^(5/3),x,method=_RETURNVERBOSE)
Output:
3*I/d*a*(1/2*(a+I*a*tan(d*x+c))^(2/3)+2*a*(1/6*2^(2/3)/a^(1/3)*ln((a+I*a*t an(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. 280 vs. \(2 (126) = 252\).
Time = 0.08 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.58 \[ \int (a+i a \tan (c+d x))^{5/3} \, dx=\frac {3 i \cdot 2^{\frac {2}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} + \left (-\frac {4 i \, a^{5}}{d^{3}}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} d - d\right )} \log \left (\frac {4 \cdot 2^{\frac {1}{3}} a^{3} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (i \, \sqrt {3} d^{2} - d^{2}\right )} \left (-\frac {4 i \, a^{5}}{d^{3}}\right )^{\frac {2}{3}}}{4 \, a^{3}}\right ) + \left (-\frac {4 i \, a^{5}}{d^{3}}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} d - d\right )} \log \left (\frac {4 \cdot 2^{\frac {1}{3}} a^{3} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (-i \, \sqrt {3} d^{2} - d^{2}\right )} \left (-\frac {4 i \, a^{5}}{d^{3}}\right )^{\frac {2}{3}}}{4 \, a^{3}}\right ) + 2 \, \left (-\frac {4 i \, a^{5}}{d^{3}}\right )^{\frac {1}{3}} d \log \left (\frac {2 \cdot 2^{\frac {1}{3}} a^{3} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + \left (-\frac {4 i \, a^{5}}{d^{3}}\right )^{\frac {2}{3}} d^{2}}{2 \, a^{3}}\right )}{2 \, d} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/3),x, algorithm="fricas")
Output:
1/2*(3*I*2^(2/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*e^(4/3*I*d*x + 4/3* I*c) + (-4*I*a^5/d^3)^(1/3)*(-I*sqrt(3)*d - d)*log(1/4*(4*2^(1/3)*a^3*(a/( e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (I*sqrt(3)*d^2 - d^2)*(-4*I*a^5/d^3)^(2/3))/a^3) + (-4*I*a^5/d^3)^(1/3)*(I*sqrt(3)*d - d)* log(1/4*(4*2^(1/3)*a^3*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (-I*sqrt(3)*d^2 - d^2)*(-4*I*a^5/d^3)^(2/3))/a^3) + 2*(-4*I*a^5 /d^3)^(1/3)*d*log(1/2*(2*2^(1/3)*a^3*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e ^(2/3*I*d*x + 2/3*I*c) + (-4*I*a^5/d^3)^(2/3)*d^2)/a^3))/d
\[ \int (a+i a \tan (c+d x))^{5/3} \, dx=\int \left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {5}{3}}\, dx \] Input:
integrate((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 = 154, normalized size of antiderivative = 0.87 \[ \int (a+i a \tan (c+d x))^{5/3} \, dx=\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {8}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {8}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{2}\right )}}{2 \, a d} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/3),x, algorithm="maxima")
Output:
1/2*I*(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)) + 3*(I*a*tan(d*x + c) + a)^(2/3)*a^2)/(a*d)
Time = 0.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.84 \[ \int (a+i a \tan (c+d x))^{5/3} \, dx=\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )} a}{2 \, d} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/3),x, algorithm="giac")
Output:
1/2*I*(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)) + 3*(I*a*tan(d*x + c) + a)^(2/3))*a/d
Time = 1.78 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.18 \[ \int (a+i a \tan (c+d x))^{5/3} \, dx=\frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}\,3{}\mathrm {i}}{2\,d}+\frac {{\left (4{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{5/3}\,\ln \left (36\,a^4\,{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-36\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{13/3}\right )}{d}-\frac {{\left (4{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{5/3}\,\ln \left (-\frac {36\,a^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}+\frac {18\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{13/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {{\left (4{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{5/3}\,\ln \left (-\frac {36\,a^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}-\frac {18\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{13/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \] Input:
int((a + a*tan(c + d*x)*1i)^(5/3),x)
Output:
(a*(a + a*tan(c + d*x)*1i)^(2/3)*3i)/(2*d) + (4i^(1/3)*(-a)^(5/3)*log(36*a ^4*(a*(tan(c + d*x)*1i + 1))^(1/3) - 36*(-1)^(1/3)*2^(1/3)*(-a)^(13/3)))/d - (4i^(1/3)*(-a)^(5/3)*log((18*(-1)^(1/3)*2^(1/3)*(-a)^(13/3)*(3^(1/2)*1i - 1))/d^2 - (36*a^4*(a + a*tan(c + d*x)*1i)^(1/3))/d^2)*((3^(1/2)*1i)/2 + 1/2))/d + (4i^(1/3)*(-a)^(5/3)*log(- (36*a^4*(a + a*tan(c + d*x)*1i)^(1/3 ))/d^2 - (18*(-1)^(1/3)*2^(1/3)*(-a)^(13/3)*(3^(1/2)*1i + 1))/d^2)*((3^(1/ 2)*1i)/2 - 1/2))/d
\[ \int (a+i a \tan (c+d x))^{5/3} \, dx=a^{\frac {5}{3}} \left (\int \left (\tan \left (d x +c \right ) i +1\right )^{\frac {2}{3}}d x +\left (\int \left (\tan \left (d x +c \right ) i +1\right )^{\frac {2}{3}} \tan \left (d x +c \right )d x \right ) i \right ) \] Input:
int((a+I*a*tan(d*x+c))^(5/3),x)
Output:
a**(2/3)*a*(int((tan(c + d*x)*i + 1)**(2/3),x) + int((tan(c + d*x)*i + 1)* *(2/3)*tan(c + d*x),x)*i)