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

Optimal. Leaf size=182 \[ -\frac{a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}+\frac{2 a^3 (B+i A) \tan ^2(c+d x)}{d}-\frac{(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac{4 a^3 (A-i B) \tan (c+d x)}{d}+\frac{4 a^3 (B+i A) \log (\cos (c+d x))}{d}-4 a^3 x (A-i B)+\frac{i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

[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)

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Rubi [A]  time = 0.423991, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3594, 3592, 3528, 3525, 3475} \[ -\frac{a^3 (45 A-47 i B) \tan ^3(c+d x)}{60 d}+\frac{2 a^3 (B+i A) \tan ^2(c+d x)}{d}-\frac{(5 A-7 i B) \tan ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac{4 a^3 (A-i B) \tan (c+d x)}{d}+\frac{4 a^3 (B+i A) \log (\cos (c+d x))}{d}-4 a^3 x (A-i B)+\frac{i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[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 3594

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]

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 3528

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 3525

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 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 ^2(c+d x) (a+i a \tan (c+d x))^3 (A+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{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 [B]  time = 8.20921, size = 847, normalized size = 4.65 \[ \frac{x \left (-2 A \cos ^3(c)+2 i B \cos ^3(c)+8 i A \sin (c) \cos ^2(c)+8 B \sin (c) \cos ^2(c)+12 A \sin ^2(c) \cos (c)-12 i B \sin ^2(c) \cos (c)+2 A \cos (c)-2 i B \cos (c)-8 i A \sin ^3(c)-8 B \sin ^3(c)-4 i A \sin (c)-4 B \sin (c)-2 A \sin ^3(c) \tan (c)+2 i B \sin ^3(c) \tan (c)-2 A \sin (c) \tan (c)+2 i B \sin (c) \tan (c)-i (A-i B) (4 \cos (3 c)-4 i \sin (3 c)) \tan (c)\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos ^4(c+d x)}{(\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{\left (i A \cos \left (\frac{3 c}{2}\right )+B \cos \left (\frac{3 c}{2}\right )+A \sin \left (\frac{3 c}{2}\right )-i B \sin \left (\frac{3 c}{2}\right )\right ) \left (2 \cos \left (\frac{3 c}{2}\right ) \log \left (\cos ^2(c+d x)\right )-2 i \log \left (\cos ^2(c+d x)\right ) \sin \left (\frac{3 c}{2}\right )\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \cos ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{\sec (c) \sec (c+d x) \left (\frac{1}{240} \cos (3 c)-\frac{1}{240} i \sin (3 c)\right ) (195 i A \cos (d x)+225 B \cos (d x)-300 A d x \cos (d x)+300 i B d x \cos (d x)+195 i A \cos (2 c+d x)+225 B \cos (2 c+d x)-300 A d x \cos (2 c+d x)+300 i B d x \cos (2 c+d x)+75 i A \cos (2 c+3 d x)+105 B \cos (2 c+3 d x)-150 A d x \cos (2 c+3 d x)+150 i B d x \cos (2 c+3 d x)+75 i A \cos (4 c+3 d x)+105 B \cos (4 c+3 d x)-150 A d x \cos (4 c+3 d x)+150 i B d x \cos (4 c+3 d x)-30 A d x \cos (4 c+5 d x)+30 i B d x \cos (4 c+5 d x)-30 A d x \cos (6 c+5 d x)+30 i B d x \cos (6 c+5 d x)+420 A \sin (d x)-470 i B \sin (d x)-330 A \sin (2 c+d x)+360 i B \sin (2 c+d x)+270 A \sin (2 c+3 d x)-280 i B \sin (2 c+3 d x)-105 A \sin (4 c+3 d x)+135 i B \sin (4 c+3 d x)+75 A \sin (4 c+5 d x)-83 i B \sin (4 c+5 d x)) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x))}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^4*(I*A*Cos[(3*c)/2] + B*Cos[(3*c)/2] + A*Sin[(3*c)/2] - I*B*Sin[(3*c)/2])*(2*Cos[(3*c)/2]*Log[Co
s[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*(Cos
[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Sec[c]*Sec[c + d*x]*(Cos[3*c]/240 - (I/240)*Sin[3*
c])*((195*I)*A*Cos[d*x] + 225*B*Cos[d*x] - 300*A*d*x*Cos[d*x] + (300*I)*B*d*x*Cos[d*x] + (195*I)*A*Cos[2*c + d
*x] + 225*B*Cos[2*c + d*x] - 300*A*d*x*Cos[2*c + d*x] + (300*I)*B*d*x*Cos[2*c + d*x] + (75*I)*A*Cos[2*c + 3*d*
x] + 105*B*Cos[2*c + 3*d*x] - 150*A*d*x*Cos[2*c + 3*d*x] + (150*I)*B*d*x*Cos[2*c + 3*d*x] + (75*I)*A*Cos[4*c +
3*d*x] + 105*B*Cos[4*c + 3*d*x] - 150*A*d*x*Cos[4*c + 3*d*x] + (150*I)*B*d*x*Cos[4*c + 3*d*x] - 30*A*d*x*Cos[
4*c + 5*d*x] + (30*I)*B*d*x*Cos[4*c + 5*d*x] - 30*A*d*x*Cos[6*c + 5*d*x] + (30*I)*B*d*x*Cos[6*c + 5*d*x] + 420
*A*Sin[d*x] - (470*I)*B*Sin[d*x] - 330*A*Sin[2*c + d*x] + (360*I)*B*Sin[2*c + d*x] + 270*A*Sin[2*c + 3*d*x] -
(280*I)*B*Sin[2*c + 3*d*x] - 105*A*Sin[4*c + 3*d*x] + (135*I)*B*Sin[4*c + 3*d*x] + 75*A*Sin[4*c + 5*d*x] - (83
*I)*B*Sin[4*c + 5*d*x])*(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])) + (x*Cos[c + d*x]^4*(2*A*Cos[c] - (2*I)*B*Cos[c] - 2*A*Cos[c]^3 + (2*I)*B*Cos[c]^3 -
(4*I)*A*Sin[c] - 4*B*Sin[c] + (8*I)*A*Cos[c]^2*Sin[c] + 8*B*Cos[c]^2*Sin[c] + 12*A*Cos[c]*Sin[c]^2 - (12*I)*B*
Cos[c]*Sin[c]^2 - (8*I)*A*Sin[c]^3 - 8*B*Sin[c]^3 - 2*A*Sin[c]*Tan[c] + (2*I)*B*Sin[c]*Tan[c] - 2*A*Sin[c]^3*T
an[c] + (2*I)*B*Sin[c]^3*Tan[c] - I*(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]))

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Maple [A]  time = 0.005, size = 230, normalized size = 1.3 \begin{align*}{\frac{-{\frac{i}{5}}{a}^{3}B \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{d}}-{\frac{{\frac{i}{4}}{a}^{3}A \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{{\frac{4\,i}{3}}{a}^{3}B \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{3\,{a}^{3}B \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{2\,i{a}^{3}A \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{a}^{3}A \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{4\,i{a}^{3}B\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{3}B \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+4\,{\frac{{a}^{3}A\tan \left ( dx+c \right ) }{d}}-{\frac{2\,i{a}^{3}A\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-2\,{\frac{{a}^{3}B\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+{\frac{4\,i{a}^{3}B\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}}-4\,{\frac{{a}^{3}A\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)^2*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.92663, size = 182, normalized size = 1. \begin{align*} -\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} +{\left (60 \, A - 80 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} + 60 \,{\left (d x + c\right )}{\left (4 \, A - 4 i \, B\right )} a^{3} + 120 \,{\left (i \, A + B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) -{\left (240 \, A - 240 i \, B\right )} a^{3} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 + (60*A - 80*I*B)*a^3*tan(d*x + c)^3 + 12
0*(-I*A - B)*a^3*tan(d*x + c)^2 + 60*(d*x + c)*(4*A - 4*I*B)*a^3 + 120*(I*A + B)*a^3*log(tan(d*x + c)^2 + 1) -
(240*A - 240*I*B)*a^3*tan(d*x + c))/d

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Fricas [A]  time = 1.41306, size = 836, normalized size = 4.59 \begin{align*} \frac{{\left (360 i \, A + 480 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (1050 i \, A + 1170 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (1230 i \, A + 1390 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (690 i \, A + 770 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (150 i \, A + 166 \, B\right )} a^{3} +{\left ({\left (60 i \, A + 60 \, B\right )} a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} +{\left (300 i \, A + 300 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (600 i \, A + 600 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (600 i \, A + 600 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (300 i \, A + 300 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (60 i \, A + 60 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

1/15*((360*I*A + 480*B)*a^3*e^(8*I*d*x + 8*I*c) + (1050*I*A + 1170*B)*a^3*e^(6*I*d*x + 6*I*c) + (1230*I*A + 13
90*B)*a^3*e^(4*I*d*x + 4*I*c) + (690*I*A + 770*B)*a^3*e^(2*I*d*x + 2*I*c) + (150*I*A + 166*B)*a^3 + ((60*I*A +
60*B)*a^3*e^(10*I*d*x + 10*I*c) + (300*I*A + 300*B)*a^3*e^(8*I*d*x + 8*I*c) + (600*I*A + 600*B)*a^3*e^(6*I*d*
x + 6*I*c) + (600*I*A + 600*B)*a^3*e^(4*I*d*x + 4*I*c) + (300*I*A + 300*B)*a^3*e^(2*I*d*x + 2*I*c) + (60*I*A +
60*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)

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Sympy [A]  time = 54.8753, size = 272, normalized size = 1.49 \begin{align*} \frac{4 a^{3} \left (i A + B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{\left (24 i A a^{3} + 32 B a^{3}\right ) e^{- 2 i c} e^{8 i d x}}{d} + \frac{\left (70 i A a^{3} + 78 B a^{3}\right ) e^{- 4 i c} e^{6 i d x}}{d} + \frac{\left (138 i A a^{3} + 154 B a^{3}\right ) e^{- 8 i c} e^{2 i d x}}{3 d} + \frac{\left (150 i A a^{3} + 166 B a^{3}\right ) e^{- 10 i c}}{15 d} + \frac{\left (246 i A a^{3} + 278 B a^{3}\right ) e^{- 6 i c} e^{4 i d x}}{3 d}}{e^{10 i d x} + 5 e^{- 2 i c} e^{8 i d x} + 10 e^{- 4 i c} e^{6 i d x} + 10 e^{- 6 i c} e^{4 i d x} + 5 e^{- 8 i c} e^{2 i d x} + e^{- 10 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a**3*(I*A + B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + ((24*I*A*a**3 + 32*B*a**3)*exp(-2*I*c)*exp(8*I*d*x)/d + (
70*I*A*a**3 + 78*B*a**3)*exp(-4*I*c)*exp(6*I*d*x)/d + (138*I*A*a**3 + 154*B*a**3)*exp(-8*I*c)*exp(2*I*d*x)/(3*
d) + (150*I*A*a**3 + 166*B*a**3)*exp(-10*I*c)/(15*d) + (246*I*A*a**3 + 278*B*a**3)*exp(-6*I*c)*exp(4*I*d*x)/(3
*d))/(exp(10*I*d*x) + 5*exp(-2*I*c)*exp(8*I*d*x) + 10*exp(-4*I*c)*exp(6*I*d*x) + 10*exp(-6*I*c)*exp(4*I*d*x) +
5*exp(-8*I*c)*exp(2*I*d*x) + exp(-10*I*c))

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Giac [B]  time = 1.75459, size = 680, normalized size = 3.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

1/15*(60*I*A*a^3*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 60*B*a^3*e^(10*I*d*x + 10*I*c)*log(e^(2*
I*d*x + 2*I*c) + 1) + 300*I*A*a^3*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 300*B*a^3*e^(8*I*d*x + 8*
I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 600*I*A*a^3*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 600*B*a^3*e
^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 600*I*A*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1)
+ 600*B*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 300*I*A*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 300*B*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 360*I*A*a^3*e^(8*I*d*x + 8*I*c)
+ 480*B*a^3*e^(8*I*d*x + 8*I*c) + 1050*I*A*a^3*e^(6*I*d*x + 6*I*c) + 1170*B*a^3*e^(6*I*d*x + 6*I*c) + 1230*I*A
*a^3*e^(4*I*d*x + 4*I*c) + 1390*B*a^3*e^(4*I*d*x + 4*I*c) + 690*I*A*a^3*e^(2*I*d*x + 2*I*c) + 770*B*a^3*e^(2*I
*d*x + 2*I*c) + 60*I*A*a^3*log(e^(2*I*d*x + 2*I*c) + 1) + 60*B*a^3*log(e^(2*I*d*x + 2*I*c) + 1) + 150*I*A*a^3
+ 166*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)