\(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [17]

Optimal result

Integrand size = 34, antiderivative size = 182 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-4 a^3 (A-i B) x+\frac {4 a^3 (i A+B) \log (\cos (c+d x))}{d}+\frac {4 a^3 (A-i B) \tan (c+d x)}{d}+\frac {2 a^3 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d} \]

[Out]

-4*a^3*(A-I*B)*x+4*a^3*(I*A+B)*ln(cos(d*x+c))/d+4*a^3*(A-I*B)*tan(d*x+c)/d+2*a^3*(I*A+B)*tan(d*x+c)^2/d-1/60*a
^3*(45*A-47*I*B)*tan(d*x+c)^3/d+1/5*I*a*B*tan(d*x+c)^3*(a+I*a*tan(d*x+c))^2/d-1/20*(5*A-7*I*B)*tan(d*x+c)^3*(a
^3+I*a^3*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3675, 3673, 3609, 3606, 3556} \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}-\frac {(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {2 a^3 (B+i A) \tan ^2(c+d x)}{d}+\frac {4 a^3 (A-i B) \tan (c+d x)}{d}+\frac {4 a^3 (B+i A) \log (\cos (c+d x))}{d}-4 a^3 x (A-i B)+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

[In]

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

[Out]

-4*a^3*(A - I*B)*x + (4*a^3*(I*A + B)*Log[Cos[c + d*x]])/d + (4*a^3*(A - I*B)*Tan[c + d*x])/d + (2*a^3*(I*A +
B)*Tan[c + d*x]^2)/d - (a^3*(45*A - (47*I)*B)*Tan[c + d*x]^3)/(60*d) + ((I/5)*a*B*Tan[c + d*x]^3*(a + I*a*Tan[
c + d*x])^2)/d - ((5*A - (7*I)*B)*Tan[c + d*x]^3*(a^3 + I*a^3*Tan[c + d*x]))/(20*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 3606

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[b*d*(Tan[e + f*x]/f), x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3609

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[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 3675

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*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^2 (a (5 A-3 i B)+a (5 i A+7 B) \tan (c+d x)) \, dx \\ & = \frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \tan ^2(c+d x) (a+i a \tan (c+d x)) \left (a^2 (35 A-33 i B)+a^2 (45 i A+47 B) \tan (c+d x)\right ) \, dx \\ & = -\frac {a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \tan ^2(c+d x) \left (80 a^3 (A-i B)+80 a^3 (i A+B) \tan (c+d x)\right ) \, dx \\ & = \frac {2 a^3 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \tan (c+d x) \left (-80 a^3 (i A+B)+80 a^3 (A-i B) \tan (c+d x)\right ) \, dx \\ & = -4 a^3 (A-i B) x+\frac {4 a^3 (A-i B) \tan (c+d x)}{d}+\frac {2 a^3 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}-\left (4 a^3 (i A+B)\right ) \int \tan (c+d x) \, dx \\ & = -4 a^3 (A-i B) x+\frac {4 a^3 (i A+B) \log (\cos (c+d x))}{d}+\frac {4 a^3 (A-i B) \tan (c+d x)}{d}+\frac {2 a^3 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.68 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^3 \left (-15 i A-17 B-240 i (A-i B) \log (i+\tan (c+d x))+240 (A-i B) \tan (c+d x)+120 (i A+B) \tan ^2(c+d x)+(-60 A+80 i B) \tan ^3(c+d x)-15 i (A-3 i B) \tan ^4(c+d x)-12 i B \tan ^5(c+d x)\right )}{60 d} \]

[In]

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

[Out]

(a^3*((-15*I)*A - 17*B - (240*I)*(A - I*B)*Log[I + Tan[c + d*x]] + 240*(A - I*B)*Tan[c + d*x] + 120*(I*A + B)*
Tan[c + d*x]^2 + (-60*A + (80*I)*B)*Tan[c + d*x]^3 - (15*I)*(A - (3*I)*B)*Tan[c + d*x]^4 - (12*I)*B*Tan[c + d*
x]^5))/(60*d)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {i B \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {i A \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {4 i B \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {3 B \left (\tan ^{4}\left (d x +c \right )\right )}{4}+2 i A \left (\tan ^{2}\left (d x +c \right )\right )-A \left (\tan ^{3}\left (d x +c \right )\right )-4 i B \tan \left (d x +c \right )+2 B \left (\tan ^{2}\left (d x +c \right )\right )+4 A \tan \left (d x +c \right )+\frac {\left (-4 i A -4 B \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (4 i B -4 A \right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(146\)
default	\(\frac {a^{3} \left (-\frac {i B \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {i A \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {4 i B \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {3 B \left (\tan ^{4}\left (d x +c \right )\right )}{4}+2 i A \left (\tan ^{2}\left (d x +c \right )\right )-A \left (\tan ^{3}\left (d x +c \right )\right )-4 i B \tan \left (d x +c \right )+2 B \left (\tan ^{2}\left (d x +c \right )\right )+4 A \tan \left (d x +c \right )+\frac {\left (-4 i A -4 B \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (4 i B -4 A \right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(146\)
norman	\(\left (4 i B \,a^{3}-4 A \,a^{3}\right ) x -\frac {\left (i A \,a^{3}+3 B \,a^{3}\right ) \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {\left (-4 i B \,a^{3}+3 A \,a^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {4 \left (-i B \,a^{3}+A \,a^{3}\right ) \tan \left (d x +c \right )}{d}+\frac {2 \left (i A \,a^{3}+B \,a^{3}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {i B \,a^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {2 \left (i A \,a^{3}+B \,a^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(169\)
parallelrisch	\(-\frac {12 i B \,a^{3} \left (\tan ^{5}\left (d x +c \right )\right )+15 i A \left (\tan ^{4}\left (d x +c \right )\right ) a^{3}-80 i B \left (\tan ^{3}\left (d x +c \right )\right ) a^{3}+45 B \left (\tan ^{4}\left (d x +c \right )\right ) a^{3}-120 i A \left (\tan ^{2}\left (d x +c \right )\right ) a^{3}+60 A \left (\tan ^{3}\left (d x +c \right )\right ) a^{3}-240 i B x \,a^{3} d +120 i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3}+240 A x \,a^{3} d +240 i B \tan \left (d x +c \right ) a^{3}-120 B \left (\tan ^{2}\left (d x +c \right )\right ) a^{3}-240 A \tan \left (d x +c \right ) a^{3}+120 B \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3}}{60 d}\)	\(186\)
risch	\(-\frac {8 i a^{3} B c}{d}+\frac {8 a^{3} A c}{d}+\frac {2 a^{3} \left (180 i A \,{\mathrm e}^{8 i \left (d x +c \right )}+240 B \,{\mathrm e}^{8 i \left (d x +c \right )}+525 i A \,{\mathrm e}^{6 i \left (d x +c \right )}+585 B \,{\mathrm e}^{6 i \left (d x +c \right )}+615 i A \,{\mathrm e}^{4 i \left (d x +c \right )}+695 B \,{\mathrm e}^{4 i \left (d x +c \right )}+345 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+385 B \,{\mathrm e}^{2 i \left (d x +c \right )}+75 i A +83 B \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}+\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{d}\)	\(195\)
parts	\(\frac {\left (-i A \,a^{3}-3 B \,a^{3}\right ) \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 i B \,a^{3}-3 A \,a^{3}\right ) \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {A \,a^{3} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {i B \,a^{3} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(209\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.60 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (60 \, {\left (-3 i \, A - 4 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 15 \, {\left (-35 i \, A - 39 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (-123 i \, A - 139 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, {\left (-69 i \, A - 77 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-75 i \, A - 83 \, B\right )} a^{3} + 30 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, {\left (-i \, A - B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, {\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, {\left (-i \, A - B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, {\left (-i \, A - B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-2/15*(60*(-3*I*A - 4*B)*a^3*e^(8*I*d*x + 8*I*c) + 15*(-35*I*A - 39*B)*a^3*e^(6*I*d*x + 6*I*c) + 5*(-123*I*A -
139*B)*a^3*e^(4*I*d*x + 4*I*c) + 5*(-69*I*A - 77*B)*a^3*e^(2*I*d*x + 2*I*c) + (-75*I*A - 83*B)*a^3 + 30*((-I*
A - B)*a^3*e^(10*I*d*x + 10*I*c) + 5*(-I*A - B)*a^3*e^(8*I*d*x + 8*I*c) + 10*(-I*A - B)*a^3*e^(6*I*d*x + 6*I*c
) + 10*(-I*A - B)*a^3*e^(4*I*d*x + 4*I*c) + 5*(-I*A - B)*a^3*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^3)*log(e^(2*I*
d*x + 2*I*c) + 1))/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I
*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.60 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {4 i a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {150 i A a^{3} + 166 B a^{3} + \left (690 i A a^{3} e^{2 i c} + 770 B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (1230 i A a^{3} e^{4 i c} + 1390 B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (1050 i A a^{3} e^{6 i c} + 1170 B a^{3} e^{6 i c}\right ) e^{6 i d x} + \left (360 i A a^{3} e^{8 i c} + 480 B a^{3} e^{8 i c}\right ) e^{8 i d x}}{15 d e^{10 i c} e^{10 i d x} + 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} + 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} + 15 d} \]

[In]

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

[Out]

4*I*a**3*(A - I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (150*I*A*a**3 + 166*B*a**3 + (690*I*A*a**3*exp(2*I*c) +
770*B*a**3*exp(2*I*c))*exp(2*I*d*x) + (1230*I*A*a**3*exp(4*I*c) + 1390*B*a**3*exp(4*I*c))*exp(4*I*d*x) + (105
0*I*A*a**3*exp(6*I*c) + 1170*B*a**3*exp(6*I*c))*exp(6*I*d*x) + (360*I*A*a**3*exp(8*I*c) + 480*B*a**3*exp(8*I*c
))*exp(8*I*d*x))/(15*d*exp(10*I*c)*exp(10*I*d*x) + 75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6*I*d*x
) + 150*d*exp(4*I*c)*exp(4*I*d*x) + 75*d*exp(2*I*c)*exp(2*I*d*x) + 15*d)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.73 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {12 i \, B a^{3} \tan \left (d x + c\right )^{5} + 15 \, {\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{4} + 20 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (d x + c\right )^{3} + 120 \, {\left (-i \, A - B\right )} a^{3} \tan \left (d x + c\right )^{2} + 240 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{3} + 120 \, {\left (i \, A + B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 240 \, {\left (A - i \, B\right )} a^{3} \tan \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

-1/60*(12*I*B*a^3*tan(d*x + c)^5 + 15*(I*A + 3*B)*a^3*tan(d*x + c)^4 + 20*(3*A - 4*I*B)*a^3*tan(d*x + c)^3 + 1
20*(-I*A - B)*a^3*tan(d*x + c)^2 + 240*(d*x + c)*(A - I*B)*a^3 + 120*(I*A + B)*a^3*log(tan(d*x + c)^2 + 1) - 2
40*(A - I*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. 504 vs. \(2 (158) = 316\).

Time = 0.68 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.77 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (-30 i \, A a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 30 \, B a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 150 i \, 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 ) - 150 \, 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 ) - 300 i \, 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 ) - 300 \, 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 ) - 300 i \, 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 ) - 300 \, 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 ) - 150 i \, 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 ) - 150 \, 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 ) - 180 i \, A a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 240 \, B a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 525 i \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 585 \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 615 i \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 695 \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 345 i \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 385 \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 30 i \, A a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 30 \, B a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 75 i \, A a^{3} - 83 \, B a^{3}\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-2/15*(-30*I*A*a^3*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 30*B*a^3*e^(10*I*d*x + 10*I*c)*log(e^(
2*I*d*x + 2*I*c) + 1) - 150*I*A*a^3*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 150*B*a^3*e^(8*I*d*x +
8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*I*A*a^3*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*B*a^3
*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*I*A*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) +
1) - 300*B*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 150*I*A*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d
*x + 2*I*c) + 1) - 150*B*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 180*I*A*a^3*e^(8*I*d*x + 8*I*c
) - 240*B*a^3*e^(8*I*d*x + 8*I*c) - 525*I*A*a^3*e^(6*I*d*x + 6*I*c) - 585*B*a^3*e^(6*I*d*x + 6*I*c) - 615*I*A*
a^3*e^(4*I*d*x + 4*I*c) - 695*B*a^3*e^(4*I*d*x + 4*I*c) - 345*I*A*a^3*e^(2*I*d*x + 2*I*c) - 385*B*a^3*e^(2*I*d
*x + 2*I*c) - 30*I*A*a^3*log(e^(2*I*d*x + 2*I*c) + 1) - 30*B*a^3*log(e^(2*I*d*x + 2*I*c) + 1) - 75*I*A*a^3 - 8
3*B*a^3)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I
*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

Mupad [B] (verification not implemented)

Time = 7.27 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.26 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,a^3\,1{}\mathrm {i}}{3}-\frac {a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )}{3}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a^3-B\,a^3\,1{}\mathrm {i}+a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )-a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {B\,a^3}{4}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )}{4}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (4\,B\,a^3+A\,a^3\,4{}\mathrm {i}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A\,a^3\,1{}\mathrm {i}}{2}+\frac {B\,a^3}{2}+\frac {a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )}{2}\right )}{d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^5\,1{}\mathrm {i}}{5\,d} \]

[In]

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

[Out]

(tan(c + d*x)^3*((B*a^3*1i)/3 - (a^3*(2*A - B*1i))/3 + (a^3*(A*1i + 2*B)*1i)/3))/d + (tan(c + d*x)*(A*a^3 - B*
a^3*1i + a^3*(2*A - B*1i) - a^3*(A*1i + 2*B)*1i))/d - (tan(c + d*x)^4*((B*a^3)/4 + (a^3*(A*1i + 2*B))/4))/d -
(log(tan(c + d*x) + 1i)*(A*a^3*4i + 4*B*a^3))/d + (tan(c + d*x)^2*((A*a^3*1i)/2 + (B*a^3)/2 + (a^3*(2*A - B*1i
)*1i)/2 + (a^3*(A*1i + 2*B))/2))/d - (B*a^3*tan(c + d*x)^5*1i)/(5*d)