3.211 \(\int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=386 \[ -\frac{\left (m^2-4 m+3\right ) \left (-A \left (m^2-4 m+1\right )+i B \left (1-m^2\right )\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{48 a^4 d (m+1)}+\frac{(2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{48 a^4 d (m+2)}-\frac{(2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{48 a^4 d (1+i \tan (c+d x))}-\frac{\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac{(A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4} \]

[Out]

-((3 - 4*m + m^2)*(I*B*(1 - m^2) - A*(1 - 4*m + m^2))*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]
^2]*Tan[c + d*x]^(1 + m))/(48*a^4*d*(1 + m)) - ((I*B*(1 + 3*m - m^2) - A*(13 - 7*m + m^2))*Tan[c + d*x]^(1 + m
))/(48*a^4*d*(1 + I*Tan[c + d*x])^2) - ((2 - m)*(I*B*(2 + 2*m - m^2) - A*(8 - 6*m + m^2))*Tan[c + d*x]^(1 + m)
)/(48*a^4*d*(1 + I*Tan[c + d*x])) + ((2 - m)*m*(B*(2 + 2*m - m^2) + I*A*(8 - 6*m + m^2))*Hypergeometric2F1[1,
(2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(48*a^4*d*(2 + m)) + ((A + I*B)*Tan[c + d*x]^(1 +
m))/(8*d*(a + I*a*Tan[c + d*x])^4) + ((I*B*(1 - m) + A*(5 - m))*Tan[c + d*x]^(1 + m))/(24*a*d*(a + I*a*Tan[c
+ d*x])^3)

________________________________________________________________________________________

Rubi [A]  time = 1.20719, antiderivative size = 386, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3596, 3538, 3476, 364} \[ -\frac{\left (m^2-4 m+3\right ) \left (-A \left (m^2-4 m+1\right )+i B \left (1-m^2\right )\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{48 a^4 d (m+1)}+\frac{(2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{48 a^4 d (m+2)}-\frac{(2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{48 a^4 d (1+i \tan (c+d x))}-\frac{\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac{(A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-((3 - 4*m + m^2)*(I*B*(1 - m^2) - A*(1 - 4*m + m^2))*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]
^2]*Tan[c + d*x]^(1 + m))/(48*a^4*d*(1 + m)) - ((I*B*(1 + 3*m - m^2) - A*(13 - 7*m + m^2))*Tan[c + d*x]^(1 + m
))/(48*a^4*d*(1 + I*Tan[c + d*x])^2) - ((2 - m)*(I*B*(2 + 2*m - m^2) - A*(8 - 6*m + m^2))*Tan[c + d*x]^(1 + m)
)/(48*a^4*d*(1 + I*Tan[c + d*x])) + ((2 - m)*m*(B*(2 + 2*m - m^2) + I*A*(8 - 6*m + m^2))*Hypergeometric2F1[1,
(2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(48*a^4*d*(2 + m)) + ((A + I*B)*Tan[c + d*x]^(1 +
m))/(8*d*(a + I*a*Tan[c + d*x])^4) + ((I*B*(1 - m) + A*(5 - m))*Tan[c + d*x]^(1 + m))/(24*a*d*(a + I*a*Tan[c
+ d*x])^3)

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3538

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{\int \frac{\tan ^m(c+d x) (a (A (7-m)-i B (1+m))-a (i A-B) (3-m) \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\tan ^m(c+d x) \left (-2 a^2 \left (i B \left (4+3 m-m^2\right )-A \left (16-7 m+m^2\right )\right )+2 a^2 (B (1-m)-i A (5-m)) (2-m) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4}\\ &=-\frac{\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\tan ^m(c+d x) \left (4 a^3 \left (A \left (19-20 m+8 m^2-m^3\right )-i B \left (7+2 m-4 m^2+m^3\right )\right )-4 a^3 (1-m) \left (B \left (1+3 m-m^2\right )+i A \left (13-7 m+m^2\right )\right ) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6}\\ &=-\frac{\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\int \tan ^m(c+d x) \left (-8 a^4 \left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right )+8 a^4 (2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right ) \tan (c+d x)\right ) \, dx}{384 a^8}\\ &=-\frac{\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\left ((2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right )\right ) \int \tan ^{1+m}(c+d x) \, dx}{48 a^4}-\frac{\left (\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right )\right ) \int \tan ^m(c+d x) \, dx}{48 a^4}\\ &=-\frac{\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\left ((2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{48 a^4 d}-\frac{\left (\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{48 a^4 d}\\ &=-\frac{\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right ) \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+m)}-\frac{\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac{(2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right ) \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{48 a^4 d (2+m)}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 115.997, size = 921, normalized size = 2.39 \[ -\frac{i e^{-4 i c} \left (-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-m} \left (3 (A+i B) e^{-8 i d x} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{4-m}+e^{2 i (c-3 d x)} \left (-1+e^{2 i (c+d x)}\right )^{m+1} (A (7-2 m)-i (2 m B+B)) \left (1+e^{2 i (c+d x)}\right )^{4-m}-e^{4 i (c-d x)} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (i B \left (-2 m^2+2 m+3\right )+A \left (-2 m^2+10 m-9\right )\right ) \left (1+e^{2 i (c+d x)}\right )^{4-m}+\frac{e^{-2 i (d x-3 c)} \left (-1+e^{2 i (c+d x)}\right )^{m+1} m (m+1) \left (A \left (-4 m^3+26 m^2-44 m+13\right )-i B \left (4 m^3-10 m^2-4 m+7\right )\right ) \left (1+e^{2 i (c+d x)}\right )^{4-m}-2^{5-m} e^{8 i c} \left (-1+e^{2 i (c+d x)}\right )^{m+1} m \left (A \left (-4 m^3+26 m^2-44 m+13\right )-i B \left (4 m^3-10 m^2-4 m+7\right )\right ) \text{Hypergeometric2F1}\left (m-3,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+2^{2-m} e^{8 i c} \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (A \left (2 m^4-16 m^3+40 m^2-32 m+3\right )+i B \left (2 m^4-8 m^3+4 m^2+8 m-3\right )\right ) \left (\left (1+e^{2 i (c+d x)}\right )^m \left (4 \left (-1+e^{2 i (c+d x)}\right ) m \text{Hypergeometric2F1}\left (m-2,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+2 \left (-1+e^{2 i (c+d x)}\right ) m \text{Hypergeometric2F1}\left (m-1,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+m \text{Hypergeometric2F1}\left (m,m,m+1,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+\text{Hypergeometric2F1}\left (m,m,m+1,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+e^{2 i (c+d x)} m \text{Hypergeometric2F1}\left (m,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )-m \text{Hypergeometric2F1}\left (m,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right )-2^m (m+1) \text{Hypergeometric2F1}\left (1,m,m+1,\frac{1-e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right )}{m (m+1)}\right ) \sec ^3(c+d x) (\cos (d x)+i \sin (d x))^4 (A+B \tan (c+d x))}{384 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((-I/384)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*((3*(A + I*B)*(-1 + E^((2*I)*(c + d*
x)))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(4 - m))/E^((8*I)*d*x) + E^((2*I)*(c - 3*d*x))*(-1 + E^((2*I)*(c + d*x)
))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(4 - m)*(A*(7 - 2*m) - I*(B + 2*B*m)) - E^((4*I)*(c - d*x))*(-1 + E^((2*I
)*(c + d*x)))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(4 - m)*(I*B*(3 + 2*m - 2*m^2) + A*(-9 + 10*m - 2*m^2)) + (((-
1 + E^((2*I)*(c + d*x)))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(4 - m)*m*(1 + m)*(A*(13 - 44*m + 26*m^2 - 4*m^3) -
I*B*(7 - 4*m - 10*m^2 + 4*m^3)))/E^((2*I)*(-3*c + d*x)) - 2^(5 - m)*E^((8*I)*c)*(-1 + E^((2*I)*(c + d*x)))^(1
+ m)*m*(A*(13 - 44*m + 26*m^2 - 4*m^3) - I*B*(7 - 4*m - 10*m^2 + 4*m^3))*Hypergeometric2F1[-3 + m, 1 + m, 2 +
m, (1 - E^((2*I)*(c + d*x)))/2] + 2^(2 - m)*E^((8*I)*c)*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))
)^m*(A*(3 - 32*m + 40*m^2 - 16*m^3 + 2*m^4) + I*B*(-3 + 8*m + 4*m^2 - 8*m^3 + 2*m^4))*(-(2^m*(1 + m)*Hypergeom
etric2F1[1, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]) + (1 + E^((2*I)*(c + d*x)))^m*(4*(
-1 + E^((2*I)*(c + d*x)))*m*Hypergeometric2F1[-2 + m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2] + 2*(-1 + E^(
(2*I)*(c + d*x)))*m*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2] + Hypergeometric2F1[m
, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2] + m*Hypergeometric2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2] - m*H
ypergeometric2F1[m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2] + E^((2*I)*(c + d*x))*m*Hypergeometric2F1[m, 1
+ m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2])))/(m*(1 + m)))*Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[c
+ d*x]))/(d*E^((4*I)*c)*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(A*Cos[c + d*x] + B*Sin[c +
d*x])*(a + I*a*Tan[c + d*x])^4)

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Maple [F]  time = 0.901, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ) }{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x)

[Out]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (A - i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \,{\left (2 \, A - i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, A e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \,{\left (2 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} e^{\left (-8 i \, d x - 8 i \, c\right )}}{16 \, a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(1/16*((A - I*B)*e^(8*I*d*x + 8*I*c) + 2*(2*A - I*B)*e^(6*I*d*x + 6*I*c) + 6*A*e^(4*I*d*x + 4*I*c) + 2
*(2*A + I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*e^(-8*I
*d*x - 8*I*c)/a^4, x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**4,x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^m/(I*a*tan(d*x + c) + a)^4, x)