3.18 \(\int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=138 \[ \frac{2 a^3 (B+i A) \tan (c+d x)}{d}-\frac{4 a^3 (A-i B) \log (\cos (c+d x))}{d}-4 a^3 x (B+i A)+\frac{a (A-i B) (a+i a \tan (c+d x))^2}{2 d}+\frac{A (a+i a \tan (c+d x))^3}{3 d}-\frac{i B (a+i a \tan (c+d x))^4}{4 a d} \]

[Out]

-4*a^3*(I*A + B)*x - (4*a^3*(A - I*B)*Log[Cos[c + d*x]])/d + (2*a^3*(I*A + B)*Tan[c + d*x])/d + (a*(A - I*B)*(
a + I*a*Tan[c + d*x])^2)/(2*d) + (A*(a + I*a*Tan[c + d*x])^3)/(3*d) - ((I/4)*B*(a + I*a*Tan[c + d*x])^4)/(a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.134645, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3592, 3527, 3478, 3477, 3475} \[ \frac{2 a^3 (B+i A) \tan (c+d x)}{d}-\frac{4 a^3 (A-i B) \log (\cos (c+d x))}{d}-4 a^3 x (B+i A)+\frac{a (A-i B) (a+i a \tan (c+d x))^2}{2 d}+\frac{A (a+i a \tan (c+d x))^3}{3 d}-\frac{i B (a+i a \tan (c+d x))^4}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-4*a^3*(I*A + B)*x - (4*a^3*(A - I*B)*Log[Cos[c + d*x]])/d + (2*a^3*(I*A + B)*Tan[c + d*x])/d + (a*(A - I*B)*(
a + I*a*Tan[c + d*x])^2)/(2*d) + (A*(a + I*a*Tan[c + d*x])^3)/(3*d) - ((I/4)*B*(a + I*a*Tan[c + d*x])^4)/(a*d)

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 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 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[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] && G
tQ[n, 1]

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac{i B (a+i a \tan (c+d x))^4}{4 a d}+\int (a+i a \tan (c+d x))^3 (-B+A \tan (c+d x)) \, dx\\ &=\frac{A (a+i a \tan (c+d x))^3}{3 d}-\frac{i B (a+i a \tan (c+d x))^4}{4 a d}-(i A+B) \int (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{a (A-i B) (a+i a \tan (c+d x))^2}{2 d}+\frac{A (a+i a \tan (c+d x))^3}{3 d}-\frac{i B (a+i a \tan (c+d x))^4}{4 a d}-(2 a (i A+B)) \int (a+i a \tan (c+d x))^2 \, dx\\ &=-4 a^3 (i A+B) x+\frac{2 a^3 (i A+B) \tan (c+d x)}{d}+\frac{a (A-i B) (a+i a \tan (c+d x))^2}{2 d}+\frac{A (a+i a \tan (c+d x))^3}{3 d}-\frac{i B (a+i a \tan (c+d x))^4}{4 a d}+\left (4 a^3 (A-i B)\right ) \int \tan (c+d x) \, dx\\ &=-4 a^3 (i A+B) x-\frac{4 a^3 (A-i B) \log (\cos (c+d x))}{d}+\frac{2 a^3 (i A+B) \tan (c+d x)}{d}+\frac{a (A-i B) (a+i a \tan (c+d x))^2}{2 d}+\frac{A (a+i a \tan (c+d x))^3}{3 d}-\frac{i B (a+i a \tan (c+d x))^4}{4 a d}\\ \end{align*}

Mathematica [B]  time = 7.61, size = 980, normalized size = 7.1 \[ \frac{x \left (-2 i A \cos ^3(c)-2 B \cos ^3(c)-8 A \sin (c) \cos ^2(c)+8 i B \sin (c) \cos ^2(c)+12 i A \sin ^2(c) \cos (c)+12 B \sin ^2(c) \cos (c)+2 i A \cos (c)+2 B \cos (c)+8 A \sin ^3(c)-8 i B \sin ^3(c)+4 A \sin (c)-4 i B \sin (c)-2 i A \sin ^3(c) \tan (c)-2 B \sin ^3(c) \tan (c)-2 i A \sin (c) \tan (c)-2 B \sin (c) \tan (c)+(A-i B) (4 \cos (3 c)-4 i \sin (3 c)) \tan (c)\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos ^4(c+d x)}{(\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{\left (A \cos \left (\frac{3 c}{2}\right )-i B \cos \left (\frac{3 c}{2}\right )-i A \sin \left (\frac{3 c}{2}\right )-B \sin \left (\frac{3 c}{2}\right )\right ) \left (2 i \log \left (\cos ^2(c+d x)\right ) \sin \left (\frac{3 c}{2}\right )-2 \cos \left (\frac{3 c}{2}\right ) \log \left (\cos ^2(c+d x)\right )\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(A-i B) (-4 i d x \cos (3 c)-4 d x \sin (3 c)) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{\left (\frac{1}{3} \cos (3 c)-\frac{1}{3} i \sin (3 c)\right ) (13 i A \sin (d x)+15 B \sin (d x)) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos ^3(c+d x)}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(-9 A \cos (c)+15 i B \cos (c)-2 i A \sin (c)-6 B \sin (c)) \left (\frac{1}{6} \cos (3 c)-\frac{1}{6} i \sin (3 c)\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos ^2(c+d x)}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{\left (\frac{1}{3} \cos (3 c)-\frac{1}{3} i \sin (3 c)\right ) (-i A \sin (d x)-3 B \sin (d x)) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos (c+d x)}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{\left (-\frac{1}{4} i B \cos (3 c)-\frac{1}{4} B \sin (3 c)\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x))}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^4*(A*Cos[(3*c)/2] - I*B*Cos[(3*c)/2] - I*A*Sin[(3*c)/2] - B*Sin[(3*c)/2])*(-2*Cos[(3*c)/2]*Log[C
os[c + d*x]^2] + (2*I)*Log[Cos[c + d*x]^2]*Sin[(3*c)/2])*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Co
s[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^2*(-9*A*Cos[c] + (15*I)*B*Cos[c] - (
2*I)*A*Sin[c] - 6*B*Sin[c])*(Cos[3*c]/6 - (I/6)*Sin[3*c])*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(C
os[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (((-I
/4)*B*Cos[3*c] - (B*Sin[3*c])/4)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(
A*Cos[c + d*x] + B*Sin[c + d*x])) + ((A - I*B)*Cos[c + d*x]^4*((-4*I)*d*x*Cos[3*c] - 4*d*x*Sin[3*c])*(a + I*a*
Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c
+ d*x]*(Cos[3*c]/3 - (I/3)*Sin[3*c])*((-I)*A*Sin[d*x] - 3*B*Sin[d*x])*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c +
d*x]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d
*x])) + (Cos[c + d*x]^3*(Cos[3*c]/3 - (I/3)*Sin[3*c])*((13*I)*A*Sin[d*x] + 15*B*Sin[d*x])*(a + I*a*Tan[c + d*x
])^3*(A + B*Tan[c + d*x]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c +
d*x] + B*Sin[c + d*x])) + (x*Cos[c + d*x]^4*((2*I)*A*Cos[c] + 2*B*Cos[c] - (2*I)*A*Cos[c]^3 - 2*B*Cos[c]^3 +
4*A*Sin[c] - (4*I)*B*Sin[c] - 8*A*Cos[c]^2*Sin[c] + (8*I)*B*Cos[c]^2*Sin[c] + (12*I)*A*Cos[c]*Sin[c]^2 + 12*B*
Cos[c]*Sin[c]^2 + 8*A*Sin[c]^3 - (8*I)*B*Sin[c]^3 - (2*I)*A*Sin[c]*Tan[c] - 2*B*Sin[c]*Tan[c] - (2*I)*A*Sin[c]
^3*Tan[c] - 2*B*Sin[c]^3*Tan[c] + (A - I*B)*(4*Cos[3*c] - (4*I)*Sin[3*c])*Tan[c])*(a + I*a*Tan[c + d*x])^3*(A
+ B*Tan[c + d*x]))/((Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 195, normalized size = 1.4 \begin{align*}{\frac{-{\frac{i}{4}}{a}^{3}B \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{{\frac{i}{3}}{a}^{3}A \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{2\,i{a}^{3}B \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{a}^{3}B \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{4\,i{a}^{3}A\tan \left ( dx+c \right ) }{d}}-{\frac{3\,{a}^{3}A \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+4\,{\frac{{a}^{3}B\tan \left ( dx+c \right ) }{d}}-{\frac{2\,i{a}^{3}B\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+2\,{\frac{{a}^{3}A\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-{\frac{4\,i{a}^{3}A\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-4\,{\frac{{a}^{3}B\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.69836, size = 155, normalized size = 1.12 \begin{align*} -\frac{3 i \, B a^{3} \tan \left (d x + c\right )^{4} + 4 \,{\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{3} +{\left (18 \, A - 24 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 48 \,{\left (d x + c\right )}{\left (i \, A + B\right )} a^{3} - 12 \,{\left (2 \, A - 2 i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 48 \,{\left (-i \, A - B\right )} a^{3} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*I*B*a^3*tan(d*x + c)^4 + 4*(I*A + 3*B)*a^3*tan(d*x + c)^3 + (18*A - 24*I*B)*a^3*tan(d*x + c)^2 + 48*(
d*x + c)*(I*A + B)*a^3 - 12*(2*A - 2*I*B)*a^3*log(tan(d*x + c)^2 + 1) + 48*(-I*A - B)*a^3*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.4785, size = 626, normalized size = 4.54 \begin{align*} -\frac{2 \,{\left (12 \,{\left (2 \, A - 3 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \,{\left (19 \, A - 23 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \,{\left (23 \, A - 27 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (13 \, A - 15 i \, B\right )} a^{3} + 6 \,{\left ({\left (A - i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \,{\left (A - i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \,{\left (A - i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (A - i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(12*(2*A - 3*I*B)*a^3*e^(6*I*d*x + 6*I*c) + 3*(19*A - 23*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 2*(23*A - 27*I*B)
*a^3*e^(2*I*d*x + 2*I*c) + (13*A - 15*I*B)*a^3 + 6*((A - I*B)*a^3*e^(8*I*d*x + 8*I*c) + 4*(A - I*B)*a^3*e^(6*I
*d*x + 6*I*c) + 6*(A - I*B)*a^3*e^(4*I*d*x + 4*I*c) + 4*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^3)*log
(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^
(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [A]  time = 25.5057, size = 223, normalized size = 1.62 \begin{align*} \frac{4 a^{3} \left (- A + i B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{\left (16 A a^{3} - 24 i B a^{3}\right ) e^{- 2 i c} e^{6 i d x}}{d} - \frac{\left (26 A a^{3} - 30 i B a^{3}\right ) e^{- 8 i c}}{3 d} - \frac{\left (38 A a^{3} - 46 i B a^{3}\right ) e^{- 4 i c} e^{4 i d x}}{d} - \frac{\left (92 A a^{3} - 108 i B a^{3}\right ) e^{- 6 i c} e^{2 i d x}}{3 d}}{e^{8 i d x} + 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} + 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a**3*(-A + I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-(16*A*a**3 - 24*I*B*a**3)*exp(-2*I*c)*exp(6*I*d*x)/d -
(26*A*a**3 - 30*I*B*a**3)*exp(-8*I*c)/(3*d) - (38*A*a**3 - 46*I*B*a**3)*exp(-4*I*c)*exp(4*I*d*x)/d - (92*A*a*
*3 - 108*I*B*a**3)*exp(-6*I*c)*exp(2*I*d*x)/(3*d))/(exp(8*I*d*x) + 4*exp(-2*I*c)*exp(6*I*d*x) + 6*exp(-4*I*c)*
exp(4*I*d*x) + 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

________________________________________________________________________________________

Giac [B]  time = 1.39743, size = 551, normalized size = 3.99 \begin{align*} -\frac{12 \, A a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, B a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 48 i \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 72 \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 72 i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 48 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 72 i \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 114 \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 138 i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 92 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 108 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 12 \, A a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, B a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 26 \, A a^{3} - 30 i \, B a^{3}}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(12*A*a^3*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*B*a^3*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*
x + 2*I*c) + 1) + 48*A*a^3*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 48*I*B*a^3*e^(6*I*d*x + 6*I*c)*l
og(e^(2*I*d*x + 2*I*c) + 1) + 72*A*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 72*I*B*a^3*e^(4*I*d*
x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 48*A*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 48*I*B*a
^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 48*A*a^3*e^(6*I*d*x + 6*I*c) - 72*I*B*a^3*e^(6*I*d*x + 6
*I*c) + 114*A*a^3*e^(4*I*d*x + 4*I*c) - 138*I*B*a^3*e^(4*I*d*x + 4*I*c) + 92*A*a^3*e^(2*I*d*x + 2*I*c) - 108*I
*B*a^3*e^(2*I*d*x + 2*I*c) + 12*A*a^3*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*B*a^3*log(e^(2*I*d*x + 2*I*c) + 1) +
26*A*a^3 - 30*I*B*a^3)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*
I*d*x + 2*I*c) + d)