\(\int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [494]

Optimal result

Integrand size = 31, antiderivative size = 387 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {\left (A b^3 (2+n) (3+n) (4+n)-a \left (b^2 B (3+n) (4+n)-2 a (3 a B-A b (4+n))\right )\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}+\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a+b) d (1+n)}-\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a-b) d (1+n)}-\frac {\left (b^2 B (3+n) (4+n)-2 a (3 a B-A b (4+n))\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac {(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)} \]

[Out]

-(A*b^3*(2+n)*(3+n)*(4+n)-a*(b^2*B*(3+n)*(4+n)-2*a*(3*B*a-A*b*(4+n))))*(a+b*tan(d*x+c))^(1+n)/b^4/d/(1+n)/(2+n
)/(3+n)/(4+n)+1/2*(A-I*B)*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a-I*b))*(a+b*tan(d*x+c))^(1+n)/(I*a+b)/d/
(1+n)-1/2*(A+I*B)*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a+I*b))*(a+b*tan(d*x+c))^(1+n)/(I*a-b)/d/(1+n)-(b
^2*B*(3+n)*(4+n)-2*a*(3*B*a-A*b*(4+n)))*tan(d*x+c)*(a+b*tan(d*x+c))^(1+n)/b^3/d/(2+n)/(3+n)/(4+n)-(3*B*a-A*b*(
4+n))*tan(d*x+c)^2*(a+b*tan(d*x+c))^(1+n)/b^2/d/(3+n)/(4+n)+B*tan(d*x+c)^3*(a+b*tan(d*x+c))^(1+n)/b/d/(4+n)

Rubi [A] (verified)

Time = 1.59 (sec) , antiderivative size = 385, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3688, 3728, 3711, 3620, 3618, 70} \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {\tan (c+d x) \left (6 a^2 B-2 a A b (n+4)-b^2 B (n+3) (n+4)\right ) (a+b \tan (c+d x))^{n+1}}{b^3 d (n+2) (n+3) (n+4)}-\frac {\left (6 a^3 B-2 a^2 A b (n+4)-a b^2 B (n+3) (n+4)+A b^3 (n+2) (n+3) (n+4)\right ) (a+b \tan (c+d x))^{n+1}}{b^4 d (n+1) (n+2) (n+3) (n+4)}-\frac {\tan ^2(c+d x) (3 a B-A b (n+4)) (a+b \tan (c+d x))^{n+1}}{b^2 d (n+3) (n+4)}+\frac {(A-i B) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}-\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (-b+i a)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+4)} \]

[In]

Int[Tan[c + d*x]^4*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

-(((6*a^3*B - 2*a^2*A*b*(4 + n) - a*b^2*B*(3 + n)*(4 + n) + A*b^3*(2 + n)*(3 + n)*(4 + n))*(a + b*Tan[c + d*x]
)^(1 + n))/(b^4*d*(1 + n)*(2 + n)*(3 + n)*(4 + n))) + ((A - I*B)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan
[c + d*x])/(a - I*b)]*(a + b*Tan[c + d*x])^(1 + n))/(2*(I*a + b)*d*(1 + n)) - ((A + I*B)*Hypergeometric2F1[1,
1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)]*(a + b*Tan[c + d*x])^(1 + n))/(2*(I*a - b)*d*(1 + n)) + ((6*a^2*
B - 2*a*A*b*(4 + n) - b^2*B*(3 + n)*(4 + n))*Tan[c + d*x]*(a + b*Tan[c + d*x])^(1 + n))/(b^3*d*(2 + n)*(3 + n)
*(4 + n)) - ((3*a*B - A*b*(4 + n))*Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n))/(b^2*d*(3 + n)*(4 + n)) + (B*T
an[c + d*x]^3*(a + b*Tan[c + d*x])^(1 + n))/(b*d*(4 + n))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3620

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

Rule 3688

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

Rule 3711

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((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*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac {\int \tan ^2(c+d x) (a+b \tan (c+d x))^n \left (-3 a B-b B (4+n) \tan (c+d x)-(3 a B-A b (4+n)) \tan ^2(c+d x)\right ) \, dx}{b (4+n)} \\ & = -\frac {(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac {\int \tan (c+d x) (a+b \tan (c+d x))^n \left (2 a (3 a B-A b (4+n))-A b^2 (3+n) (4+n) \tan (c+d x)+\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^2 (3+n) (4+n)} \\ & = \frac {\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac {(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac {\int (a+b \tan (c+d x))^n \left (-a \left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right )+b^3 B (2+n) (3+n) (4+n) \tan (c+d x)-\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) \tan ^2(c+d x)\right ) \, dx}{b^3 (2+n) (3+n) (4+n)} \\ & = -\frac {\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}+\frac {\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac {(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac {\int (a+b \tan (c+d x))^n \left (A b^3 (2+n) (3+n) (4+n)+b^3 B (2+n) (3+n) (4+n) \tan (c+d x)\right ) \, dx}{b^3 (2+n) (3+n) (4+n)} \\ & = -\frac {\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}+\frac {\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac {(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}+\frac {1}{2} (A-i B) \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx \\ & = -\frac {\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}+\frac {\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac {(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)}-\frac {(i A-B) \text {Subst}\left (\int \frac {(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}+\frac {(i A+B) \text {Subst}\left (\int \frac {(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d} \\ & = -\frac {\left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right ) (a+b \tan (c+d x))^{1+n}}{b^4 d (1+n) (2+n) (3+n) (4+n)}-\frac {(i A+B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}-\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a-b) d (1+n)}+\frac {\left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^3 d (2+n) (3+n) (4+n)}-\frac {(3 a B-A b (4+n)) \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (3+n) (4+n)}+\frac {B \tan ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (4+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.10 (sec) , antiderivative size = 384, normalized size of antiderivative = 0.99 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (i \left (2 i (a-i b) (a+i b) \left (6 a^3 B-2 a^2 A b (4+n)-a b^2 B (3+n) (4+n)+A b^3 (2+n) (3+n) (4+n)\right )-(a+i b) b^4 (A-i B) \left (24+26 n+9 n^2+n^3\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) b^4 (A+i B) \left (24+26 n+9 n^2+n^3\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )\right )+2 (a-i b) (a+i b) b (1+n) \left (6 a^2 B-2 a A b (4+n)-b^2 B (3+n) (4+n)\right ) \tan (c+d x)-2 (a-i b) (a+i b) b^2 (1+n) (2+n) (3 a B-A b (4+n)) \tan ^2(c+d x)+2 (a-i b) (a+i b) b^3 B (1+n) (2+n) (3+n) \tan ^3(c+d x)\right )}{2 (a-i b) (a+i b) b^4 d (1+n) (2+n) (3+n) (4+n)} \]

[In]

Integrate[Tan[c + d*x]^4*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

((a + b*Tan[c + d*x])^(1 + n)*(I*((2*I)*(a - I*b)*(a + I*b)*(6*a^3*B - 2*a^2*A*b*(4 + n) - a*b^2*B*(3 + n)*(4
+ n) + A*b^3*(2 + n)*(3 + n)*(4 + n)) - (a + I*b)*b^4*(A - I*B)*(24 + 26*n + 9*n^2 + n^3)*Hypergeometric2F1[1,
1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)] + (a - I*b)*b^4*(A + I*B)*(24 + 26*n + 9*n^2 + n^3)*Hypergeomet
ric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)]) + 2*(a - I*b)*(a + I*b)*b*(1 + n)*(6*a^2*B - 2*a*A*b*
(4 + n) - b^2*B*(3 + n)*(4 + n))*Tan[c + d*x] - 2*(a - I*b)*(a + I*b)*b^2*(1 + n)*(2 + n)*(3*a*B - A*b*(4 + n)
)*Tan[c + d*x]^2 + 2*(a - I*b)*(a + I*b)*b^3*B*(1 + n)*(2 + n)*(3 + n)*Tan[c + d*x]^3))/(2*(a - I*b)*(a + I*b)
*b^4*d*(1 + n)*(2 + n)*(3 + n)*(4 + n))

Maple [F]

\[\int \tan \left (d x +c \right )^{4} \left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]

[In]

int(tan(d*x+c)^4*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^4*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

Fricas [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*tan(d*x + c)^5 + A*tan(d*x + c)^4)*(b*tan(d*x + c) + a)^n, x)

Sympy [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(tan(d*x+c)**4*(a+b*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**n*tan(c + d*x)**4, x)

Maxima [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*tan(d*x + c)^4, x)

Giac [F]

\[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*tan(d*x + c)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^4\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

[In]

int(tan(c + d*x)^4*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^4*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n, x)