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)*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

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Rubi [A]  time = 0.13, antiderivative size = 138, normalized size of antiderivative = 1.00, 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 3475

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

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 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 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 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]

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*}

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Mathematica [B]  time = 8.10, size = 980, normalized size = 7.10 \[ \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])) + (((-1
/4*I)*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]))

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fricas [A]  time = 0.68, size = 227, normalized size = 1.64 \[ -\frac {2 \, {\left (12 \, {\left (2 \, A - 3 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (19 \, A - 23 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (23 \, A - 27 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (13 \, A - 15 i \, B\right )} a^{3} + 6 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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)

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giac [B]  time = 0.67, size = 408, normalized size = 2.96 \[ -\frac {12 \, A a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, B a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 48 i \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 72 \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 72 i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 48 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 \, A a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 72 i \, B a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 114 \, A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 138 i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 92 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 108 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 12 \, A a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, B a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 26 \, A a^{3} - 30 i \, B a^{3}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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)

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maple [A]  time = 0.02, size = 195, normalized size = 1.41 \[ -\frac {i a^{3} B \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {i a^{3} A \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 i a^{3} B \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{3} B \left (\tan ^{3}\left (d x +c \right )\right )}{d}+\frac {4 i a^{3} A \tan \left (d x +c \right )}{d}-\frac {3 a^{3} A \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {4 a^{3} B \tan \left (d x +c \right )}{d}-\frac {2 i a^{3} B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a^{3} A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {4 i a^{3} A \arctan \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{3} B \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)*(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))

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maxima [A]  time = 1.67, size = 116, normalized size = 0.84 \[ -\frac {3 i \, B a^{3} \tan \left (d x + c\right )^{4} - {\left (-4 i \, A - 12 \, B\right )} a^{3} \tan \left (d x + c\right )^{3} + 6 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} - 12 \, {\left (d x + c\right )} {\left (-4 i \, A - 4 \, B\right )} a^{3} - 24 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - {\left (48 i \, A + 48 \, B\right )} a^{3} \tan \left (d x + c\right )}{12 \, d} \]

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 - 12*B)*a^3*tan(d*x + c)^3 + 6*(3*A - 4*I*B)*a^3*tan(d*x + c)^2 - 12
*(d*x + c)*(-4*I*A - 4*B)*a^3 - 24*(A - I*B)*a^3*log(tan(d*x + c)^2 + 1) - (48*I*A + 48*B)*a^3*tan(d*x + c))/d

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mupad [B]  time = 6.04, size = 176, normalized size = 1.28 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B\,a^3\,1{}\mathrm {i}}{2}-\frac {a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )}{2}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a^3\,1{}\mathrm {i}+B\,a^3+a^3\,\left (2\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,a^3}{3}+\frac {a^3\,\left (2\,B+A\,1{}\mathrm {i}\right )}{3}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (4\,A\,a^3-B\,a^3\,4{}\mathrm {i}\right )}{d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [B]  time = 1.11, size = 236, normalized size = 1.71 \[ - \frac {4 a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {26 A a^{3} - 30 i B a^{3} + \left (92 A a^{3} e^{2 i c} - 108 i B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (114 A a^{3} e^{4 i c} - 138 i B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (48 A a^{3} e^{6 i c} - 72 i B a^{3} e^{6 i c}\right ) e^{6 i d x}}{- 3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

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 + (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)

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