\(\int \cot ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [501]

Optimal result

Integrand size = 31, antiderivative size = 293 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {(2 a B-A b (1-n)) \cot (c+d x) (a+b \tan (c+d x))^{1+n}}{2 a^2 d}-\frac {A \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{2 a d}+\frac {\left (2 a^2 A-2 a b B n+A b^2 (1-n) n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{2 a^3 d (1+n)}-\frac {(i A+B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (i a+b) d (1+n)}-\frac {(A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.78 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {\left (a^3 (a+i b) (A-i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \left (a^3 (A+i B) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )-2 (a+i b) \left (a^2 A \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right )+b \left (a B \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right )-A b \operatorname {Hypergeometric2F1}\left (3,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right )\right )\right )\right )\right ) (a+b \tan (c+d x))^{1+n}}{2 a^3 (a-i b) (a+i b) d (1+n)} \] Input:

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

Output:

-1/2*((a^3*(a + I*b)*(A - I*B)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*T 
an[c + d*x])/(a - I*b)] + (a - I*b)*(a^3*(A + I*B)*Hypergeometric2F1[1, 1 
+ n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)] - 2*(a + I*b)*(a^2*A*Hypergeom 
etric2F1[1, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a] + b*(a*B*Hypergeometric2 
F1[2, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a] - A*b*Hypergeometric2F1[3, 1 + 
 n, 2 + n, 1 + (b*Tan[c + d*x])/a]))))*(a + b*Tan[c + d*x])^(1 + n))/(a^3* 
(a - I*b)*(a + I*b)*d*(1 + n))
 

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {3042, 4092, 25, 3042, 4132, 3042, 4136, 27, 3042, 4022, 3042, 4020, 25, 78, 4117, 75}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4092

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 78

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

-1/2*(A*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n))/(a*d) + (-(((2*a*B - 
A*b*(1 - n))*Cot[c + d*x]*(a + b*Tan[c + d*x])^(1 + n))/(a*d)) - (-(((2*a^ 
2*A - 2*a*b*B*n + A*b^2*(1 - n)*n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + 
(b*Tan[c + d*x])/a]*(a + b*Tan[c + d*x])^(1 + n))/(a*d*(1 + n))) + 2*(((-1 
/2*I)*a^2*(I*A + B)*Hypergeometric2F1[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)* 
a^2*(I*A - B)*Hypergeometric2F1[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))))/a)/(2*a)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 75

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

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 4092

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*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1) 
/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^ 
2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b* 
B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 
)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n 
+ 2)*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] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] 
 || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4132

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[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[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] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4136

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate(cot(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*cot(c + d*x)**3, x)
 

Maxima [F]

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

integrate(cot(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*cot(d*x + c)^3, x)
 

Giac [F]

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

integrate(cot(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*cot(d*x + c)^3, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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