\(\int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [18]

Optimal result

Integrand size = 32, antiderivative size = 138 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-4 a^3 (i A+B) x-\frac {4 a^3 (A-i B) \log (\cos (c+d x))}{d}+\frac {2 a^3 (i A+B) \tan (c+d x)}{d}+\frac {a (A-i B) (a+i a \tan (c+d x))^2}{2 d}+\frac {A (a+i a \tan (c+d x))^3}{3 d}-\frac {i B (a+i a \tan (c+d x))^4}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3673, 3608, 3559, 3558, 3556} \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 a^3 (B+i A) \tan (c+d x)}{d}-\frac {4 a^3 (A-i B) \log (\cos (c+d x))}{d}-4 a^3 x (B+i A)+\frac {a (A-i B) (a+i a \tan (c+d x))^2}{2 d}+\frac {A (a+i a \tan (c+d x))^3}{3 d}-\frac {i B (a+i a \tan (c+d x))^4}{4 a d} \]

[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 3556

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

Rule 3558

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 3559

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 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 {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 [A] (verified)

Time = 1.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^3 \left (4 A-3 i B+48 (A-i B) \log (i+\tan (c+d x))+48 (i A+B) \tan (c+d x)-6 (3 A-4 i B) \tan ^2(c+d x)-4 i (A-3 i B) \tan ^3(c+d x)-3 i B \tan ^4(c+d x)\right )}{12 d} \]

[In]

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

[Out]

(a^3*(4*A - (3*I)*B + 48*(A - I*B)*Log[I + Tan[c + d*x]] + 48*(I*A + B)*Tan[c + d*x] - 6*(3*A - (4*I)*B)*Tan[c
+ d*x]^2 - (4*I)*(A - (3*I)*B)*Tan[c + d*x]^3 - (3*I)*B*Tan[c + d*x]^4))/(12*d)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {i B \left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {i A \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 i B \left (\tan ^{2}\left (d x +c \right )\right )-B \left (\tan ^{3}\left (d x +c \right )\right )+4 i A \tan \left (d x +c \right )-\frac {3 A \left (\tan ^{2}\left (d x +c \right )\right )}{2}+4 B \tan \left (d x +c \right )+\frac {\left (-4 i B +4 A \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-4 i A -4 B \right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(123\)
default	\(\frac {a^{3} \left (-\frac {i B \left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {i A \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 i B \left (\tan ^{2}\left (d x +c \right )\right )-B \left (\tan ^{3}\left (d x +c \right )\right )+4 i A \tan \left (d x +c \right )-\frac {3 A \left (\tan ^{2}\left (d x +c \right )\right )}{2}+4 B \tan \left (d x +c \right )+\frac {\left (-4 i B +4 A \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-4 i A -4 B \right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(123\)
norman	\(\left (-4 i A \,a^{3}-4 B \,a^{3}\right ) x -\frac {\left (i A \,a^{3}+3 B \,a^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {\left (-4 i B \,a^{3}+3 A \,a^{3}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {4 \left (i A \,a^{3}+B \,a^{3}\right ) \tan \left (d x +c \right )}{d}-\frac {i B \,a^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {2 \left (-i B \,a^{3}+A \,a^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(143\)
parallelrisch	\(-\frac {3 i B \,a^{3} \left (\tan ^{4}\left (d x +c \right )\right )+4 i A \left (\tan ^{3}\left (d x +c \right )\right ) a^{3}+48 i A x \,a^{3} d -24 i B \left (\tan ^{2}\left (d x +c \right )\right ) a^{3}+12 B \left (\tan ^{3}\left (d x +c \right )\right ) a^{3}-48 i A \tan \left (d x +c \right ) a^{3}+18 A \left (\tan ^{2}\left (d x +c \right )\right ) a^{3}+24 i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3}+48 B x \,a^{3} d -24 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3}-48 B \tan \left (d x +c \right ) a^{3}}{12 d}\)	\(157\)
risch	\(\frac {8 a^{3} B c}{d}+\frac {8 i a^{3} A c}{d}+\frac {2 i a^{3} \left (24 i A \,{\mathrm e}^{6 i \left (d x +c \right )}+36 B \,{\mathrm e}^{6 i \left (d x +c \right )}+57 i A \,{\mathrm e}^{4 i \left (d x +c \right )}+69 B \,{\mathrm e}^{4 i \left (d x +c \right )}+46 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+54 B \,{\mathrm e}^{2 i \left (d x +c \right )}+13 i A +15 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{d}\)	\(171\)
parts	\(\frac {\left (-i A \,a^{3}-3 B \,a^{3}\right ) \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {\left (3 i B \,a^{3}-3 A \,a^{3}\right ) \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3}}{2 d}-\frac {i B \,a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(185\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.64 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (12 \, {\left (2 \, A - 3 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (19 \, A - 23 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (23 \, A - 27 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (13 \, A - 15 i \, B\right )} a^{3} + 6 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[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 [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (114) = 228\).

Time = 0.47 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.70 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=- \frac {4 a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 26 A a^{3} + 30 i B a^{3} + \left (- 92 A a^{3} e^{2 i c} + 108 i B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (- 114 A a^{3} e^{4 i c} + 138 i B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (- 48 A a^{3} e^{6 i c} + 72 i B a^{3} e^{6 i c}\right ) e^{6 i d x}}{3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} + 3 d} \]

[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 + (-26*A*a**3 + 30*I*B*a**3 + (-92*A*a**3*exp(2*I*c) + 108
*I*B*a**3*exp(2*I*c))*exp(2*I*d*x) + (-114*A*a**3*exp(4*I*c) + 138*I*B*a**3*exp(4*I*c))*exp(4*I*d*x) + (-48*A*
a**3*exp(6*I*c) + 72*I*B*a**3*exp(6*I*c))*exp(6*I*d*x))/(3*d*exp(8*I*c)*exp(8*I*d*x) + 12*d*exp(6*I*c)*exp(6*I
*d*x) + 18*d*exp(4*I*c)*exp(4*I*d*x) + 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {3 i \, B a^{3} \tan \left (d x + c\right )^{4} + 4 \, {\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{3} + 6 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 48 \, {\left (d x + c\right )} {\left (i \, A + B\right )} a^{3} - 24 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 48 \, {\left (-i \, A - B\right )} a^{3} \tan \left (d x + c\right )}{12 \, d} \]

[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 + 6*(3*A - 4*I*B)*a^3*tan(d*x + c)^2 + 48*(
d*x + c)*(I*A + B)*a^3 - 24*(A - I*B)*a^3*log(tan(d*x + c)^2 + 1) + 48*(-I*A - B)*a^3*tan(d*x + c))/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (116) = 232\).

Time = 0.53 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.96 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (6 \, A a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, B a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 24 i \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 36 \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 36 i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 24 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 36 i \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 57 \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 69 i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 46 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 54 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 \, A a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, B a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 13 \, A a^{3} - 15 i \, B a^{3}\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-2/3*(6*A*a^3*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*B*a^3*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 24*A*a^3*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 24*I*B*a^3*e^(6*I*d*x + 6*I*c)*log
(e^(2*I*d*x + 2*I*c) + 1) + 36*A*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 36*I*B*a^3*e^(4*I*d*x
+ 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*A*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 24*I*B*a^3
*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*A*a^3*e^(6*I*d*x + 6*I*c) - 36*I*B*a^3*e^(6*I*d*x + 6*I
*c) + 57*A*a^3*e^(4*I*d*x + 4*I*c) - 69*I*B*a^3*e^(4*I*d*x + 4*I*c) + 46*A*a^3*e^(2*I*d*x + 2*I*c) - 54*I*B*a^
3*e^(2*I*d*x + 2*I*c) + 6*A*a^3*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*B*a^3*log(e^(2*I*d*x + 2*I*c) + 1) + 13*A*a
^3 - 15*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)

Mupad [B] (verification not implemented)

Time = 7.56 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.28 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B\,a^3\,1{}\mathrm {i}}{2}-\frac {a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )}{2}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a^3\,1{}\mathrm {i}+B\,a^3+a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,a^3}{3}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )}{3}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (4\,A\,a^3-B\,a^3\,4{}\mathrm {i}\right )}{d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{4\,d} \]

[In]

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

[Out]

(tan(c + d*x)^2*((B*a^3*1i)/2 - (a^3*(2*A - B*1i))/2 + (a^3*(A*1i + 2*B)*1i)/2))/d + (tan(c + d*x)*(A*a^3*1i +
B*a^3 + a^3*(2*A - B*1i)*1i + a^3*(A*1i + 2*B)))/d - (tan(c + d*x)^3*((B*a^3)/3 + (a^3*(A*1i + 2*B))/3))/d +
(log(tan(c + d*x) + 1i)*(4*A*a^3 - B*a^3*4i))/d - (B*a^3*tan(c + d*x)^4*1i)/(4*d)