3.5.95 \(\int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [495]

3.5.95.1 Optimal result

Integrand size = 31, antiderivative size = 291 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {\left (2 a^2 B-a A b (3+n)-b^2 B \left (6+5 n+n^2\right )\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n) (2+n) (3+n)}+\frac {(i A+B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a+b) d (1+n)}+\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)}-\frac {(2 a B-A b (3+n)) \tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}+\frac {B \tan ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{b d (3+n)} \]

output

(2*B*a^2-a*A*b*(3+n)-b^2*B*(n^2+5*n+6))*(a+b*tan(d*x+c))^(1+n)/b^3/d/(1+n) 
/(2+n)/(3+n)+1/2*(I*A+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)/(a+I*b)/d/(1+n)-(2*B* 
a-A*b*(3+n))*tan(d*x+c)*(a+b*tan(d*x+c))^(1+n)/b^2/d/(2+n)/(3+n)+B*tan(d*x 
+c)^2*(a+b*tan(d*x+c))^(1+n)/b/d/(3+n)
 

3.5.95.2 Mathematica [A] (verified)

Time = 2.61 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.97 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (2 (a-i b) (a+i b) \left (2 a^2 B-a A b (3+n)-b^2 B (2+n) (3+n)\right )+(a+i b) b^3 (A-i B) \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) b^3 (A+i B) \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )-2 (a-i b) (a+i b) b (1+n) (2 a B-A b (3+n)) \tan (c+d x)+2 (a-i b) (a+i b) b^2 B (1+n) (2+n) \tan ^2(c+d x)\right )}{2 (a-i b) (a+i b) b^3 d (1+n) (2+n) (3+n)} \]

input

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

output

((a + b*Tan[c + d*x])^(1 + n)*(2*(a - I*b)*(a + I*b)*(2*a^2*B - a*A*b*(3 + 
 n) - b^2*B*(2 + n)*(3 + n)) + (a + I*b)*b^3*(A - I*B)*(6 + 5*n + n^2)*Hyp 
ergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)] + (a - I*b 
)*b^3*(A + I*B)*(6 + 5*n + n^2)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b* 
Tan[c + d*x])/(a + I*b)] - 2*(a - I*b)*(a + I*b)*b*(1 + n)*(2*a*B - A*b*(3 
 + n))*Tan[c + d*x] + 2*(a - I*b)*(a + I*b)*b^2*B*(1 + n)*(2 + n)*Tan[c + 
d*x]^2))/(2*(a - I*b)*(a + I*b)*b^3*d*(1 + n)*(2 + n)*(3 + n))
 

3.5.95.3 Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4090, 25, 3042, 4130, 25, 3042, 4113, 3042, 4022, 3042, 4020, 25, 78}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 78

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

input

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

output

(B*Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n))/(b*d*(3 + n)) - (((2*a*B - 
 A*b*(3 + n))*Tan[c + d*x]*(a + b*Tan[c + d*x])^(1 + n))/(b*d*(2 + n)) - ( 
((2*a^2*B - a*A*b*(3 + n) - b^2*B*(2 + n)*(3 + n))*(a + b*Tan[c + d*x])^(1 
 + n))/(b*d*(1 + n)) - ((I/2)*b^2*(I*A + B)*(2 + n)*(3 + n)*Hypergeometric 
2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)]*(a + b*Tan[c + d*x])^ 
(1 + n))/((a - I*b)*d*(1 + n)) - ((I/2)*b^2*(I*A - B)*(2 + n)*(3 + n)*Hype 
rgeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)]*(a + b*Tan[ 
c + d*x])^(1 + n))/((a + I*b)*d*(1 + n)))/(b*(2 + n)))/(b*(3 + n))
 

3.5.95.3.1 Defintions of rubi rules used

rule 25

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

rule 78

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] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3042

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

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[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 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(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 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 4130

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] + Simp[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] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

3.5.95.4 Maple [F]

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

input

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

output

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

3.5.95.5 Fricas [F]

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

input

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

output

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

3.5.95.6 Sympy [F]

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

input

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

output

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

3.5.95.7 Maxima [F]

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

input

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

output

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

3.5.95.8 Giac [F]

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

input

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

output

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

3.5.95.9 Mupad [F(-1)]

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

input

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

output

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