Integrand size = 24, antiderivative size = 174 \[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {i \sqrt [3]{a} x}{2\ 2^{2/3}}-\frac {\sqrt {3} \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac {\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac {3 \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d} \] Output:
1/4*I*a^(1/3)*x*2^(1/3)-1/2*3^(1/2)*a^(1/3)*arctan(1/3*(a^(1/3)+2^(2/3)*(a +I*a*tan(d*x+c))^(1/3))*3^(1/2)/a^(1/3))*2^(1/3)/d+1/4*a^(1/3)*ln(cos(d*x+ c))*2^(1/3)/d+3/4*a^(1/3)*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^( 1/3)/d+3*(a+I*a*tan(d*x+c))^(1/3)/d
Time = 0.14 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.12 \[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {-2 \sqrt [3]{2} \sqrt {3} \sqrt [3]{a} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 \sqrt [3]{2} \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )-\sqrt [3]{2} \sqrt [3]{a} \log \left (2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} \sqrt [3]{a+i a \tan (c+d x)}+(a+i a \tan (c+d x))^{2/3}\right )+12 \sqrt [3]{a+i a \tan (c+d x)}}{4 d} \] Input:
Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(1/3),x]
Output:
(-2*2^(1/3)*Sqrt[3]*a^(1/3)*ArcTan[(1 + (2^(2/3)*(a + I*a*Tan[c + d*x])^(1 /3))/a^(1/3))/Sqrt[3]] + 2*2^(1/3)*a^(1/3)*Log[2^(1/3)*a^(1/3) - (a + I*a* Tan[c + d*x])^(1/3)] - 2^(1/3)*a^(1/3)*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)] + 12*(a + I*a*Tan[c + d*x])^(1/3))/(4*d)
Time = 0.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4010, 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.
| \(\displaystyle \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)}dx\) |
| \(\Big \downarrow \) 4010 |
| \(\displaystyle \frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-i \int \sqrt [3]{i \tan (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-i \int \sqrt [3]{i \tan (c+d x) a+a}dx\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {a \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}d(i a \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {a \left (\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2\ 2^{2/3} a^{2/3}}+\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {a \left (\frac {3 \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {a \left (-\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac {a \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a-i a \tan (c+d x))}{2\ 2^{2/3} a^{2/3}}\right )}{d}\) |
Input:
Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(1/3),x]
Output:
-((a*((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*(a + I*a*Tan[c + d*x])^(1/3) )/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_))^(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/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]
Time = 0.90 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d}\) | \(154\) |
| default | \(\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}-\frac {a^{\frac {1}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d}\) | \(154\) |
Input:
int(tan(d*x+c)*(a+I*a*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
Output:
3*(a+I*a*tan(d*x+c))^(1/3)/d+1/2/d*a^(1/3)*2^(1/3)*ln((a+I*a*tan(d*x+c))^( 1/3)-2^(1/3)*a^(1/3))-1/4/d*a^(1/3)*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))-1/2/d*a^(1/3)*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 ))
Time = 0.09 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.37 \[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {\left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} d - d\right )} \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} \log \left (\left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} d + d\right )} \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + \left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} d - d\right )} \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} \log \left (\left (\frac {1}{4}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} d + d\right )} \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} \log \left (-2 \, \left (\frac {1}{4}\right )^{\frac {1}{3}} d \left (\frac {a}{d^{3}}\right )^{\frac {1}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 6 \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}}{2 \, d} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")
Output:
1/2*((1/4)^(1/3)*(-I*sqrt(3)*d - d)*(a/d^3)^(1/3)*log((1/4)^(1/3)*(I*sqrt( 3)*d + d)*(a/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)) + (1/4)^(1/3)*(I*sqrt(3)*d - d)*(a/d^3)^(1/3)*log((1/ 4)^(1/3)*(-I*sqrt(3)*d + d)*(a/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)) + 2*(1/4)^(1/3)*d*(a/d^3)^(1/3)*log (-2*(1/4)^(1/3)*d*(a/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)) + 6*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))/d
\[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\int \sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )} \tan {\left (c + d x \right )}\, dx \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))**(1/3),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(1/3)*tan(c + d*x), x)
Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.88 \[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=-\frac {2 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {7}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{2}}{4 \, a^{2} d} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")
Output:
-1/4*(2*sqrt(3)*2^(1/3)*a^(7/3)*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^(1/3)*a^(7/3)*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^(1/3)*a^(7/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 12*(I*a*tan(d*x + c) + a)^(1/3)*a^2)/(a^2*d)
Time = 0.21 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.87 \[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (2 i \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + i \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 i \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 12 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a\right )}}{4 \, a d} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")
Output:
1/4*I*(2*I*sqrt(3)*2^(1/3)*a^(4/3)*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)) + I*2^(1/3)*a^(4/3)*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*I*2^(1/3)*a^(4/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d *x + c) + a)^(1/3)) - 12*I*(I*a*tan(d*x + c) + a)^(1/3)*a)/(a*d)
Time = 1.50 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01 \[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {2^{1/3}\,a^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{2\,d}+\frac {4^{2/3}\,a^{1/3}\,\ln \left (\frac {9\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,a^{4/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d}-\frac {4^{2/3}\,a^{1/3}\,\ln \left (\frac {9\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,a^{4/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,d} \] Input:
int(tan(c + d*x)*(a + a*tan(c + d*x)*1i)^(1/3),x)
Output:
(3*(a + a*tan(c + d*x)*1i)^(1/3))/d + (2^(1/3)*a^(1/3)*log((a*(tan(c + d*x )*1i + 1))^(1/3) - 2^(1/3)*a^(1/3)))/(2*d) + (4^(2/3)*a^(1/3)*log((9*a*(a + a*tan(c + d*x)*1i)^(1/3))/d - (9*2^(1/3)*a^(4/3)*(3^(1/2)*1i - 1))/(2*d) )*((3^(1/2)*1i)/2 - 1/2))/(4*d) - (4^(2/3)*a^(1/3)*log((9*a*(a + a*tan(c + d*x)*1i)^(1/3))/d + (9*2^(1/3)*a^(4/3)*(3^(1/2)*1i + 1))/(2*d))*((3^(1/2) *1i)/2 + 1/2))/(4*d)
\[ \int \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=a^{\frac {1}{3}} \left (\int \left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}} \tan \left (d x +c \right )d x \right ) \] Input:
int(tan(d*x+c)*(a+I*a*tan(d*x+c))^(1/3),x)
Output:
a**(1/3)*int((tan(c + d*x)*i + 1)**(1/3)*tan(c + d*x),x)