\(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\) [197]

Optimal result

Integrand size = 36, antiderivative size = 270 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {a^{2/3} (A-i B) x}{2 \sqrt [3]{2}}-\frac {\sqrt {3} a^{2/3} (i A+B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac {a^{2/3} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {3 a^{2/3} (i A+B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d} \]

[Out]

1/4*a^(2/3)*(A-I*B)*x*2^(2/3)-1/4*a^(2/3)*(I*A+B)*ln(cos(d*x+c))*2^(2/3)/d-3/4*a^(2/3)*(I*A+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)*(I*A+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-9/8*B*(a+I*a*tan(d*x+c))^(2/3)/d+3/8*B*tan(d*x+c)^2*(a+I*a*tan(d*x+c))^
(2/3)/d-3/20*(4*I*A+B)*(a+I*a*tan(d*x+c))^(5/3)/a/d

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3678, 3673, 3608, 3562, 57, 631, 210, 31} \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {3} a^{2/3} (B+i A) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac {3 a^{2/3} (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac {a^{2/3} (B+i A) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {a^{2/3} x (A-i B)}{2 \sqrt [3]{2}}-\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d} \]

[In]

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

[Out]

(a^(2/3)*(A - I*B)*x)/(2*2^(1/3)) - (Sqrt[3]*a^(2/3)*(I*A + 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)*(I*A + B)*Log[Cos[c + d*x]])/(2*2^(1/3)*d) - (3*a^(2/3)*(I
*A + B)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(2*2^(1/3)*d) - (9*B*(a + I*a*Tan[c + d*x])^(2/3)
)/(8*d) + (3*B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(2/3))/(8*d) - (3*((4*I)*A + B)*(a + I*a*Tan[c + d*x])^(5
/3))/(20*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]

Rule 3678

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 \int \tan (c+d x) (a+i a \tan (c+d x))^{2/3} \left (-2 a B+\frac {2}{3} a (4 A-i B) \tan (c+d x)\right ) \, dx}{8 a} \\ & = \frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac {3 \int (a+i a \tan (c+d x))^{2/3} \left (-\frac {2}{3} a (4 A-i B)-2 a B \tan (c+d x)\right ) \, dx}{8 a} \\ & = -\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+(-A+i B) \int (a+i a \tan (c+d x))^{2/3} \, dx \\ & = -\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac {(a (i A+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} (A-i B) x}{2 \sqrt [3]{2}}-\frac {a^{2/3} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac {\left (3 a^{2/3} (i A+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 (i A+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} (A-i B) x}{2 \sqrt [3]{2}}-\frac {a^{2/3} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {3 a^{2/3} (i A+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 {9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac {\left (3 a^{2/3} (i A+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} (A-i B) x}{2 \sqrt [3]{2}}-\frac {\sqrt {3} a^{2/3} (i A+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} (i A+B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {3 a^{2/3} (i A+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 {9 B (a+i a \tan (c+d x))^{2/3}}{8 d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {3 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{20 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.25 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.73 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {10\ 2^{2/3} a^{2/3} (i A+B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )\right )+45 B (a+i a \tan (c+d x))^{2/3}-15 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}+\frac {6 (4 i A+B) (a+i a \tan (c+d x))^{5/3}}{a}}{40 d} \]

[In]

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

[Out]

-1/40*(10*2^(2/3)*a^(2/3)*(I*A + B)*(2*Sqrt[3]*ArcTan[(1 + (2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/a^(1/3))/Sqr
t[3]] - Log[I + Tan[c + d*x]] + 3*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)]) + 45*B*(a + I*a*Tan[c +
d*x])^(2/3) - 15*B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(2/3) + (6*((4*I)*A + B)*(a + I*a*Tan[c + d*x])^(5/3
))/a)/d

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {3 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+\frac {i a^{2} B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{3} \left (-i B +A \right )\right )}{d \,a^{2}}\)	\(223\)
default	\(\frac {3 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+\frac {i a^{2} B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}-\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{3} \left (-i B +A \right )\right )}{d \,a^{2}}\)	\(223\)
parts	\(\frac {3 i A \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}-\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) a^{2}\right )}{d a}+B \left (-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8 d \,a^{2}}+\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5 d a}-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 d}-\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{2 d}+\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{4 d}-\frac {a^{\frac {2}{3}} \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d}\right )\)	\(358\)

[In]

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

[Out]

3*I/d/a^2*(1/8*I*B*(a+I*a*tan(d*x+c))^(8/3)-1/5*I*B*a*(a+I*a*tan(d*x+c))^(5/3)-1/5*A*a*(a+I*a*tan(d*x+c))^(5/3
)+1/2*I*a^2*B*(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^3*(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. 660 vs. \(2 (201) = 402\).

Time = 0.26 (sec) , antiderivative size = 660, normalized size of antiderivative = 2.44 \[ \int \tan ^2(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 (2 \, {\left (2 i \, A + 3 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (2 i \, A + 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 \, B\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 (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac {{\left (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - 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 (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - 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 (4 i \, d x + 4 i \, c\right )} + 2 \, {\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 (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - 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 (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - 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 (4 i \, d x + 4 i \, c\right )} + 2 \, {\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 (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - 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 (i \, A^{3} + 3 \, A^{2} B - 3 i \, A B^{2} - 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 (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(tan(d*x+c)^2*(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)*(2*(2*I*A + 3*B)*e^(4*I*d*x + 4*I*c) + 2*(2*I*A + 3*B)*e^(2*I*d*x + 2*I*c) + 5*B)*(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^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*
I*c) + d)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - 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*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - 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^(4*I*d*x + 4*I*c) + 2*(-I*sqrt(3)
*d + d)*e^(2*I*d*x + 2*I*c) - I*sqrt(3)*d + d)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - 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)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - 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^(4*I*d*x + 4*I*c) + 2*(I*sqrt(3)*d + d)*e^(2*I*d*x + 2*I*c) + I*sqrt(3)*d + d)*((I*A^3 + 3*
A^2*B - 3*I*A*B^2 - 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)*((I*A^3 + 3*A^2*B - 3*I*A*B^2 - B^3)*a^2/d^3)^
(2/3))/((A^2 - 2*I*A*B - B^2)*a)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

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

[In]

integrate(tan(d*x+c)**2*(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)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.78 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {i \, {\left (20 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 10 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 20 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 15 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {8}{3}} B a + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} {\left (A + i \, B\right )} a^{2} - 60 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} B a^{3}\right )}}{40 \, a^{3} d} \]

[In]

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

[Out]

-1/40*I*(20*sqrt(3)*2^(2/3)*(A - I*B)*a^(11/3)*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)) - 10*2^(2/3)*(A - I*B)*a^(11/3)*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)) + 20*2^(2/3)*(A - I*B)*a^(11/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d
*x + c) + a)^(1/3)) - 15*I*(I*a*tan(d*x + c) + a)^(8/3)*B*a + 24*(I*a*tan(d*x + c) + a)^(5/3)*(A + I*B)*a^2 -
60*I*(I*a*tan(d*x + c) + a)^(2/3)*B*a^3)/(a^3*d)

Giac [F]

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

[In]

integrate(tan(d*x+c)^2*(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)^2, x)

Mupad [B] (verification not implemented)

Time = 10.29 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.61 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {3\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}{2\,d}-\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}\,3{}\mathrm {i}}{5\,a\,d}+\frac {3\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}}{5\,a\,d}-\frac {3\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{8/3}}{8\,a^2\,d}-\frac {2^{2/3}\,B\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{2\,d}+\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,A\,a^{2/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{d}+\frac {{\left (\frac {1}{2}{}\mathrm {i}\right )}^{1/3}\,A\,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}\,B\,a^{2/3}\,\ln \left (\frac {9\,B^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}\,B^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}\,B\,a^{2/3}\,\ln \left (\frac {9\,B^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}\,B^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}\,A\,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)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(2/3),x)

[Out]

(3*B*(a + a*tan(c + d*x)*1i)^(5/3))/(5*a*d) - (A*(a + a*tan(c + d*x)*1i)^(5/3)*3i)/(5*a*d) - (3*B*(a + a*tan(c
+ d*x)*1i)^(2/3))/(2*d) - (3*B*(a + a*tan(c + d*x)*1i)^(8/3))/(8*a^2*d) - (2^(2/3)*B*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)*A*a^(2/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) +
(-1)^(1/3)*2^(1/3)*a^(1/3)))/d + ((1i/2)^(1/3)*A*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)*B*a^(2/3)*log((
9*B^2*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d^2 - (9*2^(1/3)*B^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)*B*a^(2/3)*log((9*B^2*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d^2 - (9*2^(1/3)*B^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)*A*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