\(\int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\) [198]

Optimal result

Integrand size = 34, antiderivative size = 232 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac {\sqrt {3} a^{2/3} (A-i B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac {3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d} \]

[Out]

1/4*a^(2/3)*(I*A+B)*x*2^(2/3)+1/4*a^(2/3)*(A-I*B)*ln(cos(d*x+c))*2^(2/3)/d+3/4*a^(2/3)*(A-I*B)*ln(2^(1/3)*a^(1
/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(2/3)/d+1/2*a^(2/3)*(A-I*B)*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*tan(d*x+c))^(1/
3))/a^(1/3)*3^(1/2))*3^(1/2)*2^(2/3)/d+3/2*A*(a+I*a*tan(d*x+c))^(2/3)/d-3/5*I*B*(a+I*a*tan(d*x+c))^(5/3)/a/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3673, 3608, 3562, 57, 631, 210, 31} \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {3} a^{2/3} (A-i B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}+\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {a^{2/3} x (B+i A)}{2 \sqrt [3]{2}}+\frac {3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d} \]

[In]

Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]

[Out]

(a^(2/3)*(I*A + B)*x)/(2*2^(1/3)) + (Sqrt[3]*a^(2/3)*(A - I*B)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x]
)^(1/3))/(Sqrt[3]*a^(1/3))])/(2^(1/3)*d) + (a^(2/3)*(A - I*B)*Log[Cos[c + d*x]])/(2*2^(1/3)*d) + (3*a^(2/3)*(A
- I*B)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(2*2^(1/3)*d) + (3*A*(a + I*a*Tan[c + d*x])^(2/3)
)/(2*d) - (((3*I)/5)*B*(a + I*a*Tan[c + d*x])^(5/3))/(a*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[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]

Rule 210

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])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3562

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[-b/d, Subst[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]

Rule 3608

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] + Dist[(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]

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}+\int (a+i a \tan (c+d x))^{2/3} (-B+A \tan (c+d x)) \, dx \\ & = \frac {3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-(i A+B) \int (a+i a \tan (c+d x))^{2/3} \, dx \\ & = \frac {3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-\frac {(a (A-i B)) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = \frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-\frac {\left (3 a^{2/3} (A-i B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac {(3 a (A-i B)) \text {Subst}\left (\int \frac {1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d} \\ & = \frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac {3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d}-\frac {\left (3 a^{2/3} (A-i B)\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{\sqrt [3]{2} d} \\ & = \frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac {\sqrt {3} a^{2/3} (A-i B) \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{2} d}+\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}+\frac {3 A (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {3 i B (a+i a \tan (c+d x))^{5/3}}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.71 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {5\ 2^{2/3} a^{5/3} (A-i B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )\right )+30 a A (a+i a \tan (c+d x))^{2/3}-12 i B (a+i a \tan (c+d x))^{5/3}}{20 a d} \]

[In]

Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]

[Out]

(5*2^(2/3)*a^(5/3)*(A - I*B)*(2*Sqrt[3]*ArcTan[(1 + (2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/a^(1/3))/Sqrt[3]] -
Log[I + Tan[c + d*x]] + 3*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)]) + 30*a*A*(a + I*a*Tan[c + d*x]
)^(2/3) - (12*I)*B*(a + I*a*Tan[c + d*x])^(5/3))/(20*a*d)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {-\frac {3 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+\frac {3 a A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+3 \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{2} \left (-i B +A \right )}{a d}\)	\(181\)
default	\(\frac {-\frac {3 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+\frac {3 a A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+3 \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{2} \left (-i B +A \right )}{a d}\)	\(181\)
parts	\(A \left (\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 d}+\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}-\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}+\frac {a^{\frac {2}{3}} \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d}\right )+\frac {3 i B \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{2}\right )}{d a}\)	\(314\)

[In]

int(tan(d*x+c)*(a+I*a*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

3/d/a*(-1/5*I*B*(a+I*a*tan(d*x+c))^(5/3)+1/2*a*A*(a+I*a*tan(d*x+c))^(2/3)+(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)))*a^2*(A-I*B))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (168) = 336\).

Time = 0.25 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.46 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {3 \cdot 2^{\frac {2}{3}} {\left ({\left (5 \, A - 4 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 \, A\right )} \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 )} + 10 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac {{\left (A^{3} - 3 i \, A^{2} B - 3 \, A B^{2} + i \, B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} a \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 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} d^{2} \left (\frac {{\left (A^{3} - 3 i \, A^{2} B - 3 \, A B^{2} + i \, B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {2}{3}}}{{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}\right ) - 5 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (i \, \sqrt {3} d + d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {3} d + d\right )} \left (\frac {{\left (A^{3} - 3 i \, A^{2} B - 3 \, A B^{2} + i \, B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} a \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 {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} d^{2} - d^{2}\right )} \left (\frac {{\left (A^{3} - 3 i \, A^{2} B - 3 \, A B^{2} + i \, B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {2}{3}}}{{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}\right ) - 5 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (-i \, \sqrt {3} d + d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {3} d + d\right )} \left (\frac {{\left (A^{3} - 3 i \, A^{2} B - 3 \, A B^{2} + i \, B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} {\left (A^{2} - 2 i \, A B - B^{2}\right )} a \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 {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} d^{2} - d^{2}\right )} \left (\frac {{\left (A^{3} - 3 i \, A^{2} B - 3 \, A B^{2} + i \, B^{3}\right )} a^{2}}{d^{3}}\right )^{\frac {2}{3}}}{{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}\right )}{10 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/10*(3*2^(2/3)*((5*A - 4*I*B)*e^(2*I*d*x + 2*I*c) + 5*A)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*e^(4/3*I*d*x + 4
/3*I*c) + 10*(1/2)^(1/3)*(d*e^(2*I*d*x + 2*I*c) + d)*((A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(1/3)*log((
2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - 2*(1/2)^(2/3)*d^
2*((A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) - 5*(1/2)^(1/3)*((I*sqrt(3)*
d + d)*e^(2*I*d*x + 2*I*c) + I*sqrt(3)*d + d)*((A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(1/3)*log((2^(1/3)
*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(I*sqrt(3)*
d^2 - d^2)*((A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) - 5*(1/2)^(1/3)*((-
I*sqrt(3)*d + d)*e^(2*I*d*x + 2*I*c) - I*sqrt(3)*d + d)*((A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(1/3)*lo
g((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(
-I*sqrt(3)*d^2 - d^2)*((A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)))/(d*e^(2
*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {2}{3}} \left (A + B \tan {\left (c + d x \right )}\right ) \tan {\left (c + d x \right )}\, dx \]

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))**(2/3)*(A+B*tan(d*x+c)),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(2/3)*(A + B*tan(c + d*x))*tan(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.81 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {10 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {8}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 5 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {8}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 10 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {8}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 12 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} B a + 30 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} A a^{2}}{20 \, a^{2} d} \]

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/20*(10*sqrt(3)*2^(2/3)*(A - I*B)*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)) - 5*2^(2/3)*(A - I*B)*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)) + 10*2^(2/3)*(A - I*B)*a^(8/3)*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)^(5/3)*B*a + 30*(I*a*tan(d*x + c) + a)^(2/3)*A*a^2)/(a^2*d)

Giac [F]

\[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \tan \left (d x + c\right ) \,d x } \]

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(2/3)*tan(d*x + c), x)

Mupad [B] (verification not implemented)

Time = 9.57 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.68 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {3\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}{2\,d}-\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}\,3{}\mathrm {i}}{5\,a\,d}+\frac {2^{2/3}\,A\,a^{2/3}\,\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 {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,B\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{d}+\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,B\,a^{2/3}\,\ln \left (\frac {{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}}{2}-{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+\frac {{\left (-1\right )}^{5/6}\,2^{1/3}\,\sqrt {3}\,a^{1/3}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {2^{2/3}\,A\,a^{2/3}\,\ln \left (\frac {9\,A^2\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}-\frac {9\,2^{1/3}\,A^2\,a^{7/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}-\frac {2^{2/3}\,A\,a^{2/3}\,\ln \left (\frac {9\,A^2\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d^2}-\frac {9\,2^{1/3}\,A^2\,a^{7/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}-\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,B\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-\frac {{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}}{2}+\frac {{\left (-1\right )}^{5/6}\,2^{1/3}\,\sqrt {3}\,a^{1/3}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \]

[In]

int(tan(c + d*x)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(2/3),x)

[Out]

(3*A*(a + a*tan(c + d*x)*1i)^(2/3))/(2*d) - (B*(a + a*tan(c + d*x)*1i)^(5/3)*3i)/(5*a*d) + (2^(2/3)*A*a^(2/3)*
log((a*(tan(c + d*x)*1i + 1))^(1/3) - 2^(1/3)*a^(1/3)))/(2*d) + ((1i/2)^(1/3)*B*a^(2/3)*log((a*(tan(c + d*x)*1
i + 1))^(1/3) + (-1)^(1/3)*2^(1/3)*a^(1/3)))/d + ((1i/2)^(1/3)*B*a^(2/3)*log(((-1)^(1/3)*2^(1/3)*a^(1/3))/2 -
(a*(tan(c + d*x)*1i + 1))^(1/3) + ((-1)^(5/6)*2^(1/3)*3^(1/2)*a^(1/3))/2)*((3^(1/2)*1i)/2 - 1/2))/d + (2^(2/3)
*A*a^(2/3)*log((9*A^2*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d^2 - (9*2^(1/3)*A^2*a^(7/3)*((3^(1/2)*1i)/2 - 1/2)^2
)/d^2)*((3^(1/2)*1i)/2 - 1/2))/(2*d) - (2^(2/3)*A*a^(2/3)*log((9*A^2*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d^2 -
(9*2^(1/3)*A^2*a^(7/3)*((3^(1/2)*1i)/2 + 1/2)^2)/d^2)*((3^(1/2)*1i)/2 + 1/2))/(2*d) - ((1i/2)^(1/3)*B*a^(2/3)*
log((a*(tan(c + d*x)*1i + 1))^(1/3) - ((-1)^(1/3)*2^(1/3)*a^(1/3))/2 + ((-1)^(5/6)*2^(1/3)*3^(1/2)*a^(1/3))/2)
*((3^(1/2)*1i)/2 + 1/2))/d