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

[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

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Rubi [A]  time = 0.42, antiderivative size = 182, normalized size of antiderivative = 1.00, 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 3475

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

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 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 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 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]

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*}

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Mathematica [B]  time = 8.60, 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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fricas [A]  time = 0.66, size = 282, normalized size = 1.55 \[ \frac {{\left (360 i \, A + 480 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (1050 i \, A + 1170 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (1230 i \, A + 1390 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (690 i \, A + 770 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (150 i \, A + 166 \, B\right )} a^{3} + {\left ({\left (60 i \, A + 60 \, B\right )} a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + {\left (300 i \, A + 300 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (600 i \, A + 600 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (600 i \, A + 600 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (300 i \, A + 300 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (60 i \, A + 60 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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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giac [B]  time = 1.01, size = 504, normalized size = 2.77 \[ \frac {60 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 ) + 60 \, 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 ) + 300 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 ) + 300 \, 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 ) + 600 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 ) + 600 \, 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 ) + 600 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 ) + 600 \, 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 ) + 300 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 ) + 300 \, 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 ) + 360 i \, A a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 480 \, B a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 1050 i \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 1170 \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 1230 i \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 1390 \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 690 i \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 770 \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 60 i \, A a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 60 \, B a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 150 i \, A a^{3} + 166 \, B a^{3}}{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 )}} \]

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)

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maple [A]  time = 0.02, size = 230, normalized size = 1.26 \[ -\frac {i a^{3} B \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {i a^{3} A \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {4 i a^{3} B \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {3 a^{3} B \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {2 i a^{3} A \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{3} A \left (\tan ^{3}\left (d x +c \right )\right )}{d}-\frac {4 i a^{3} B \tan \left (d x +c \right )}{d}+\frac {2 a^{3} B \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {4 A \,a^{3} \tan \left (d x +c \right )}{d}-\frac {2 i a^{3} A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a^{3} B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {4 i a^{3} B \arctan \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{3} A \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^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*a^3*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 = 0.87, size = 134, normalized size = 0.74 \[ -\frac {12 i \, B a^{3} \tan \left (d x + c\right )^{5} - {\left (-15 i \, A - 45 \, B\right )} a^{3} \tan \left (d x + c\right )^{4} + 20 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (d x + c\right )^{3} - {\left (120 i \, A + 120 \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 240 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{3} - 60 \, {\left (-2 i \, A - 2 \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 240 \, {\left (A - i \, B\right )} a^{3} \tan \left (d x + c\right )}{60 \, d} \]

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 - 45*B)*a^3*tan(d*x + c)^4 + 20*(3*A - 4*I*B)*a^3*tan(d*x + c)^3 -
(120*I*A + 120*B)*a^3*tan(d*x + c)^2 + 240*(d*x + c)*(A - I*B)*a^3 - 60*(-2*I*A - 2*B)*a^3*log(tan(d*x + c)^2
+ 1) - 240*(A - I*B)*a^3*tan(d*x + c))/d

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mupad [B]  time = 6.15, size = 230, normalized size = 1.26 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,a^3\,1{}\mathrm {i}}{3}-\frac {a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )}{3}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a^3-B\,a^3\,1{}\mathrm {i}+a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )-a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {B\,a^3}{4}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )}{4}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (4\,B\,a^3+A\,a^3\,4{}\mathrm {i}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A\,a^3\,1{}\mathrm {i}}{2}+\frac {B\,a^3}{2}+\frac {a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )}{2}\right )}{d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^5\,1{}\mathrm {i}}{5\,d} \]

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)^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)

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sympy [A]  time = 1.16, size = 303, normalized size = 1.66 \[ \frac {4 i a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 150 A a^{3} + 166 i B a^{3} + \left (- 690 A a^{3} e^{2 i c} + 770 i B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (- 1230 A a^{3} e^{4 i c} + 1390 i B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (- 1050 A a^{3} e^{6 i c} + 1170 i B a^{3} e^{6 i c}\right ) e^{6 i d x} + \left (- 360 A a^{3} e^{8 i c} + 480 i B a^{3} e^{8 i c}\right ) e^{8 i d x}}{15 i d e^{10 i c} e^{10 i d x} + 75 i d e^{8 i c} e^{8 i d x} + 150 i d e^{6 i c} e^{6 i d x} + 150 i d e^{4 i c} e^{4 i d x} + 75 i d e^{2 i c} e^{2 i d x} + 15 i d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**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*A*a**3 + 166*I*B*a**3 + (-690*A*a**3*exp(2*I*c) +
770*I*B*a**3*exp(2*I*c))*exp(2*I*d*x) + (-1230*A*a**3*exp(4*I*c) + 1390*I*B*a**3*exp(4*I*c))*exp(4*I*d*x) + (
-1050*A*a**3*exp(6*I*c) + 1170*I*B*a**3*exp(6*I*c))*exp(6*I*d*x) + (-360*A*a**3*exp(8*I*c) + 480*I*B*a**3*exp(
8*I*c))*exp(8*I*d*x))/(15*I*d*exp(10*I*c)*exp(10*I*d*x) + 75*I*d*exp(8*I*c)*exp(8*I*d*x) + 150*I*d*exp(6*I*c)*
exp(6*I*d*x) + 150*I*d*exp(4*I*c)*exp(4*I*d*x) + 75*I*d*exp(2*I*c)*exp(2*I*d*x) + 15*I*d)

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