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

Optimal. Leaf size=270 \[ -\frac {\sqrt {3} a^{2/3} (B+i A) \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac {3 a^{2/3} (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac {a^{2/3} (B+i A) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {a^{2/3} x (A-i B)}{2 \sqrt [3]{2}}-\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d} \]

[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

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Rubi [A]  time = 0.44, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3597, 3592, 3527, 3481, 55, 617, 204, 31} \[ -\frac {\sqrt {3} a^{2/3} (B+i A) \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac {3 a^{2/3} (B+i A) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d}-\frac {a^{2/3} (B+i A) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}+\frac {a^{2/3} x (A-i B)}{2 \sqrt [3]{2}}-\frac {3 (B+4 i A) (a+i a \tan (c+d x))^{5/3}}{20 a d}+\frac {3 B \tan ^2(c+d x) (a+i a \tan (c+d x))^{2/3}}{8 d}-\frac {9 B (a+i a \tan (c+d x))^{2/3}}{8 d} \]

Antiderivative was successfully verified.

[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 55

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

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 3481

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 3527

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 3592

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 3597

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*} \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 \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)) \operatorname {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 ) \operatorname {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)) \operatorname {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 ) \operatorname {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) \tan ^{-1}\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*}

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Mathematica [C]  time = 2.96, size = 104, normalized size = 0.39 \[ \frac {3 (a+i a \tan (c+d x))^{2/3} \left (10 (B+i A) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )+(8 A-2 i B) \tan (c+d x)-8 i A+5 B \sec ^2(c+d x)-22 B\right )}{40 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + I*a*Tan[c + d*x])^(2/3)*((-8*I)*A - 22*B + 10*(I*A + B)*Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d
*x))/(1 + E^((2*I)*(c + d*x)))] + 5*B*Sec[c + d*x]^2 + (8*A - (2*I)*B)*Tan[c + d*x]))/(40*d)

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fricas [B]  time = 1.52, size = 665, normalized size = 2.46 \[ \frac {2^{\frac {2}{3}} {\left ({\left (-12 i \, A - 18 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-12 i \, A - 18 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 15 \, 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 ) + \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (5 i \, \sqrt {3} d - 5 \, d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (10 i \, \sqrt {3} d - 10 \, d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, \sqrt {3} d - 5 \, 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 ) + \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left ({\left (-5 i \, \sqrt {3} d - 5 \, d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-10 i \, \sqrt {3} d - 10 \, d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, \sqrt {3} d - 5 \, 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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*(2^(2/3)*((-12*I*A - 18*B)*e^(4*I*d*x + 4*I*c) + (-12*I*A - 18*B)*e^(2*I*d*x + 2*I*c) - 15*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)) + (1/2)^(1/3)*((5*I*sqrt(3)*d - 5*d)*e^(4*I*d*x + 4*I*c) + (10*I*sqrt(3
)*d - 10*d)*e^(2*I*d*x + 2*I*c) + 5*I*sqrt(3)*d - 5*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)) + (1/2)^(1
/3)*((-5*I*sqrt(3)*d - 5*d)*e^(4*I*d*x + 4*I*c) + (-10*I*sqrt(3)*d - 10*d)*e^(2*I*d*x + 2*I*c) - 5*I*sqrt(3)*d
- 5*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \tan \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.25, size = 367, normalized size = 1.36 \[ -\frac {3 B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8 d \,a^{2}}+\frac {3 B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5 d a}-\frac {3 i A \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5 d a}-\frac {3 B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 d}-\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) B}{2 d}-\frac {i a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right ) A}{2 d}+\frac {a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) B}{4 d}+\frac {i a^{\frac {2}{3}} 2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right ) A}{4 d}-\frac {a^{\frac {2}{3}} \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) B}{2 d}-\frac {i a^{\frac {2}{3}} \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right ) A}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.69, size = 210, normalized size = 0.78 \[ -\frac {i \, {\left (20 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 10 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 20 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {11}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 15 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {8}{3}} B a + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} {\left (A + i \, B\right )} a^{2} - 60 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} B a^{3}\right )}}{40 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 7.75, size = 436, normalized size = 1.61 \[ -\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} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {2}{3}} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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