3.5.80 \(\int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [480]

3.5.80.1 Optimal result

Integrand size = 31, antiderivative size = 267 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {b \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m)}+\frac {\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {b^2 (A b (3+m)+a B (5+m)) \tan ^{2+m}(c+d x)}{d (2+m) (3+m)}+\frac {\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (2+m)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m)} \]

output

b*(3*a*A*b*(3+m)-b^2*B*(3+m)+2*a^2*B*(4+m))*tan(d*x+c)^(1+m)/d/(1+m)/(3+m) 
+(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-t 
an(d*x+c)^2)*tan(d*x+c)^(1+m)/d/(1+m)+b^2*(A*b*(3+m)+a*B*(5+m))*tan(d*x+c) 
^(2+m)/d/(2+m)/(3+m)+(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)*hypergeom([1, 1+1/2 
*m],[2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(2+m)/d/(2+m)+b*B*tan(d*x+c)^(1+m) 
*(a+b*tan(d*x+c))^2/d/(3+m)
 

3.5.80.2 Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.87 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\tan ^{1+m}(c+d x) \left (b (2+m) \left (3 a A b (3+m)-b^2 B (3+m)+2 a^2 B (4+m)\right )+\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) (2+m) (3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )+b^2 (1+m) (A b (3+m)+a B (5+m)) \tan (c+d x)+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) (1+m) (3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)+b B (1+m) (2+m) (a+b \tan (c+d x))^2\right )}{d (1+m) (2+m) (3+m)} \]

input

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

output

(Tan[c + d*x]^(1 + m)*(b*(2 + m)*(3*a*A*b*(3 + m) - b^2*B*(3 + m) + 2*a^2* 
B*(4 + m)) + (a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*(2 + m)*(3 + m)*Hyper 
geometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2] + b^2*(1 + m)*(A*b* 
(3 + m) + a*B*(5 + m))*Tan[c + d*x] + (3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2 
*B)*(1 + m)*(3 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d* 
x]^2]*Tan[c + d*x] + b*B*(1 + m)*(2 + m)*(a + b*Tan[c + d*x])^2))/(d*(1 + 
m)*(2 + m)*(3 + m))
 

3.5.80.3 Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4090, 25, 3042, 4120, 25, 3042, 4113, 3042, 4021, 3042, 3957, 278}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^m (a+b \tan (c+d x))^3 (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4090

\(\displaystyle \frac {\int -\tan ^m(c+d x) (a+b \tan (c+d x)) \left (-b (A b (m+3)+a B (m+5)) \tan ^2(c+d x)-\left (B a^2+2 A b a-b^2 B\right ) (m+3) \tan (c+d x)+a (b B (m+1)-a A (m+3))\right )dx}{m+3}+\frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\int \tan ^m(c+d x) (a+b \tan (c+d x)) \left (-b (A b (m+3)+a B (m+5)) \tan ^2(c+d x)-\left (B a^2+2 A b a-b^2 B\right ) (m+3) \tan (c+d x)+a (b B (m+1)-a A (m+3))\right )dx}{m+3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\int \tan (c+d x)^m (a+b \tan (c+d x)) \left (-b (A b (m+3)+a B (m+5)) \tan (c+d x)^2-\left (B a^2+2 A b a-b^2 B\right ) (m+3) \tan (c+d x)+a (b B (m+1)-a A (m+3))\right )dx}{m+3}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {-\frac {\int -\tan ^m(c+d x) \left ((m+2) (b B (m+1)-a A (m+3)) a^2-b (m+2) \left (2 B (m+4) a^2+3 A b (m+3) a-b^2 B (m+3)\right ) \tan ^2(c+d x)-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) (m+2) (m+3) \tan (c+d x)\right )dx}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {\int \tan ^m(c+d x) \left ((m+2) (b B (m+1)-a A (m+3)) a^2-b (m+2) \left (2 B (m+4) a^2+3 A b (m+3) a-b^2 B (m+3)\right ) \tan ^2(c+d x)-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) (m+2) (m+3) \tan (c+d x)\right )dx}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {\int \tan (c+d x)^m \left ((m+2) (b B (m+1)-a A (m+3)) a^2-b (m+2) \left (2 B (m+4) a^2+3 A b (m+3) a-b^2 B (m+3)\right ) \tan (c+d x)^2-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) (m+2) (m+3) \tan (c+d x)\right )dx}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {\int \tan ^m(c+d x) \left (-\left (\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) (m+2) (m+3)\right )-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) (m+2) \tan (c+d x) (m+3)\right )dx-\frac {b (m+2) \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {\int \tan (c+d x)^m \left (-\left (\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) (m+2) (m+3)\right )-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) (m+2) \tan (c+d x) (m+3)\right )dx-\frac {b (m+2) \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {-(m+2) (m+3) \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \int \tan ^{m+1}(c+d x)dx-(m+2) (m+3) \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \int \tan ^m(c+d x)dx-\frac {b (m+2) \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {-(m+2) (m+3) \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \int \tan (c+d x)^mdx-(m+2) (m+3) \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \int \tan (c+d x)^{m+1}dx-\frac {b (m+2) \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {-\frac {(m+2) (m+3) \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \int \frac {\tan ^{m+1}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {(m+2) (m+3) \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \int \frac {\tan ^m(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {b (m+2) \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}-\frac {\frac {-\frac {b (m+2) \left (2 a^2 B (m+4)+3 a A b (m+3)-b^2 B (m+3)\right ) \tan ^{m+1}(c+d x)}{d (m+1)}-\frac {(m+2) (m+3) \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}-\frac {(m+3) \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d}}{m+2}-\frac {b^2 (a B (m+5)+A b (m+3)) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}\)

input

Int[Tan[c + d*x]^m*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
 

output

(b*B*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^2)/(d*(3 + m)) - (-((b^2*(A 
*b*(3 + m) + a*B*(5 + m))*Tan[c + d*x]^(2 + m))/(d*(2 + m))) + (-((b*(2 + 
m)*(3*a*A*b*(3 + m) - b^2*B*(3 + m) + 2*a^2*B*(4 + m))*Tan[c + d*x]^(1 + m 
))/(d*(1 + m))) - ((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*(2 + m)*(3 + m) 
*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^ 
(1 + m))/(d*(1 + m)) - ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*(3 + m)*Hy 
pergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 
+ m))/d)/(2 + m))/(3 + m)
 

3.5.80.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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 4021

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Tan[e + f*x])^m, x], x] + Simp[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 4090

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
 n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta 
n[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 4113

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*Si 
mp[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 4120

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

3.5.80.4 Maple [F]

\[\int \tan \left (d x +c \right )^{m} \left (a +b \tan \left (d x +c \right )\right )^{3} \left (A +B \tan \left (d x +c \right )\right )d x\]

input

int(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)
 

output

int(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)
 

3.5.80.5 Fricas [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m} \,d x } \]

input

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="f 
ricas")
 

output

integral((B*b^3*tan(d*x + c)^4 + A*a^3 + (3*B*a*b^2 + A*b^3)*tan(d*x + c)^ 
3 + 3*(B*a^2*b + A*a*b^2)*tan(d*x + c)^2 + (B*a^3 + 3*A*a^2*b)*tan(d*x + c 
))*tan(d*x + c)^m, x)
 

3.5.80.6 Sympy [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3} \tan ^{m}{\left (c + d x \right )}\, dx \]

input

integrate(tan(d*x+c)**m*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)
 

output

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3*tan(c + d*x)**m, x)
 

3.5.80.7 Maxima [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m} \,d x } \]

input

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="m 
axima")
 

output

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^3*tan(d*x + c)^m, x)
 

3.5.80.8 Giac [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m} \,d x } \]

input

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="g 
iac")
 

output

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^3*tan(d*x + c)^m, x)
 

3.5.80.9 Mupad [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3 \,d x \]

input

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3,x)
 

output

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3, x)