\(\int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx\) [696]

Optimal result

Integrand size = 23, antiderivative size = 261 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=-\frac {b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac {\left (a^4-6 a^2 b^2+b^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {4 a b \left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)} \] Output:

-b^2*(b^2*(3+n)-a^2*(17+5*n))*(d*tan(f*x+e))^(1+n)/d/f/(1+n)/(3+n)+(a^4-6* 
a^2*b^2+b^4)*hypergeom([1, 1/2+1/2*n],[3/2+1/2*n],-tan(f*x+e)^2)*(d*tan(f* 
x+e))^(1+n)/d/f/(1+n)+2*a*b^3*(4+n)*tan(f*x+e)*(d*tan(f*x+e))^(1+n)/d/f/(2 
+n)/(3+n)+4*a*b*(a^2-b^2)*hypergeom([1, 1+1/2*n],[2+1/2*n],-tan(f*x+e)^2)* 
(d*tan(f*x+e))^(2+n)/d^2/f/(2+n)+b^2*(d*tan(f*x+e))^(1+n)*(a+b*tan(f*x+e)) 
^2/d/f/(3+n)
 

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.73 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (\frac {-b^4 (3+n)+a^2 b^2 (17+5 n)}{1+n}+\frac {\left (a^4-6 a^2 b^2+b^4\right ) (3+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )}{1+n}+\frac {2 a b^3 (4+n) \tan (e+f x)}{2+n}+\frac {4 a (a-b) b (a+b) (3+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{2+n}+b^2 (a+b \tan (e+f x))^2\right )}{f (3+n)} \] Input:

Integrate[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x])^4,x]
 

Output:

(Tan[e + f*x]*(d*Tan[e + f*x])^n*((-(b^4*(3 + n)) + a^2*b^2*(17 + 5*n))/(1 
 + n) + ((a^4 - 6*a^2*b^2 + b^4)*(3 + n)*Hypergeometric2F1[1, (1 + n)/2, ( 
3 + n)/2, -Tan[e + f*x]^2])/(1 + n) + (2*a*b^3*(4 + n)*Tan[e + f*x])/(2 + 
n) + (4*a*(a - b)*b*(a + b)*(3 + n)*Hypergeometric2F1[1, (2 + n)/2, (4 + n 
)/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(2 + n) + b^2*(a + b*Tan[e + f*x])^2)) 
/(f*(3 + n))
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 4049, 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.

\(\displaystyle \int (a+b \tan (e+f x))^4 (d \tan (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x))^4 (d \tan (e+f x))^ndx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {\int -(d \tan (e+f x))^n (a+b \tan (e+f x)) \left (-2 a b^2 d (n+4) \tan ^2(e+f x)-b \left (3 a^2-b^2\right ) d (n+3) \tan (e+f x)+a d \left (b^2 (n+1)-a^2 (n+3)\right )\right )dx}{d (n+3)}+\frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\int (d \tan (e+f x))^n (a+b \tan (e+f x)) \left (-2 a b^2 d (n+4) \tan ^2(e+f x)-b \left (3 a^2-b^2\right ) d (n+3) \tan (e+f x)+a d \left (b^2 (n+1)-a^2 (n+3)\right )\right )dx}{d (n+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\int (d \tan (e+f x))^n (a+b \tan (e+f x)) \left (-2 a b^2 d (n+4) \tan (e+f x)^2-b \left (3 a^2-b^2\right ) d (n+3) \tan (e+f x)+a d \left (b^2 (n+1)-a^2 (n+3)\right )\right )dx}{d (n+3)}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {-\frac {\int -(d \tan (e+f x))^n \left (b^2 (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) \tan ^2(e+f x) d^2+a^2 (n+2) \left (b^2 (n+1)-a^2 (n+3)\right ) d^2-4 a b \left (a^2-b^2\right ) (n+2) (n+3) \tan (e+f x) d^2\right )dx}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {\int (d \tan (e+f x))^n \left (b^2 (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) \tan ^2(e+f x) d^2+a^2 (n+2) \left (b^2 (n+1)-a^2 (n+3)\right ) d^2-4 a b \left (a^2-b^2\right ) (n+2) (n+3) \tan (e+f x) d^2\right )dx}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {\int (d \tan (e+f x))^n \left (b^2 (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) \tan (e+f x)^2 d^2+a^2 (n+2) \left (b^2 (n+1)-a^2 (n+3)\right ) d^2-4 a b \left (a^2-b^2\right ) (n+2) (n+3) \tan (e+f x) d^2\right )dx}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {\int (d \tan (e+f x))^n \left (-\left (\left (a^4-6 b^2 a^2+b^4\right ) (n+2) (n+3) d^2\right )-4 a b \left (a^2-b^2\right ) (n+2) (n+3) \tan (e+f x) d^2\right )dx+\frac {b^2 d (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{f (n+1)}}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {\int (d \tan (e+f x))^n \left (-\left (\left (a^4-6 b^2 a^2+b^4\right ) (n+2) (n+3) d^2\right )-4 a b \left (a^2-b^2\right ) (n+2) (n+3) \tan (e+f x) d^2\right )dx+\frac {b^2 d (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{f (n+1)}}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {-4 a b d (n+2) (n+3) \left (a^2-b^2\right ) \int (d \tan (e+f x))^{n+1}dx-d^2 (n+2) (n+3) \left (a^4-6 a^2 b^2+b^4\right ) \int (d \tan (e+f x))^ndx+\frac {b^2 d (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{f (n+1)}}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {-4 a b d (n+2) (n+3) \left (a^2-b^2\right ) \int (d \tan (e+f x))^{n+1}dx-d^2 (n+2) (n+3) \left (a^4-6 a^2 b^2+b^4\right ) \int (d \tan (e+f x))^ndx+\frac {b^2 d (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{f (n+1)}}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {-\frac {4 a b d^2 (n+2) (n+3) \left (a^2-b^2\right ) \int \frac {(d \tan (e+f x))^{n+1}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}-\frac {d^3 (n+2) (n+3) \left (a^4-6 a^2 b^2+b^4\right ) \int \frac {(d \tan (e+f x))^n}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}+\frac {b^2 d (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{f (n+1)}}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}-\frac {\frac {-\frac {4 a b (n+3) \left (a^2-b^2\right ) (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{f}+\frac {b^2 d (n+2) \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{f (n+1)}-\frac {d (n+2) (n+3) \left (a^4-6 a^2 b^2+b^4\right ) (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{f (n+1)}}{d (n+2)}-\frac {2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{f (n+2)}}{d (n+3)}\)

Input:

Int[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x])^4,x]
 

Output:

(b^2*(d*Tan[e + f*x])^(1 + n)*(a + b*Tan[e + f*x])^2)/(d*f*(3 + n)) - ((-2 
*a*b^3*(4 + n)*Tan[e + f*x]*(d*Tan[e + f*x])^(1 + n))/(f*(2 + n)) + ((b^2* 
d*(2 + n)*(b^2*(3 + n) - a^2*(17 + 5*n))*(d*Tan[e + f*x])^(1 + n))/(f*(1 + 
 n)) - ((a^4 - 6*a^2*b^2 + b^4)*d*(2 + n)*(3 + n)*Hypergeometric2F1[1, (1 
+ n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(f*(1 + n)) 
- (4*a*b*(a^2 - b^2)*(3 + n)*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -T 
an[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/f)/(d*(2 + n)))/(d*(3 + n))
 

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 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !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]
 

Maple [F]

\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{4}d x\]

Input:

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x)
 

Output:

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x)
 

Fricas [F]

\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x, algorithm="fricas")
 

Output:

integral((b^4*tan(f*x + e)^4 + 4*a*b^3*tan(f*x + e)^3 + 6*a^2*b^2*tan(f*x 
+ e)^2 + 4*a^3*b*tan(f*x + e) + a^4)*(d*tan(f*x + e))^n, x)
 

Sympy [F]

\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{4}\, dx \] Input:

integrate((d*tan(f*x+e))**n*(a+b*tan(f*x+e))**4,x)
 

Output:

Integral((d*tan(e + f*x))**n*(a + b*tan(e + f*x))**4, x)
 

Maxima [F]

\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e) + a)^4*(d*tan(f*x + e))^n, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{4} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e) + a)^4*(d*tan(f*x + e))^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^4 \,d x \] Input:

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x))^4,x)
 

Output:

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x))^4, x)
 

Reduce [F]

\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx=\frac {d^{n} \left (4 \tan \left (f x +e \right )^{n} \tan \left (f x +e \right )^{2} a \,b^{3} n +4 \tan \left (f x +e \right )^{n} a^{3} b n +8 \tan \left (f x +e \right )^{n} a^{3} b -4 \tan \left (f x +e \right )^{n} a \,b^{3} n -8 \tan \left (f x +e \right )^{n} a \,b^{3}+\left (\int \tan \left (f x +e \right )^{n}d x \right ) a^{4} f \,n^{2}+2 \left (\int \tan \left (f x +e \right )^{n}d x \right ) a^{4} f n -4 \left (\int \frac {\tan \left (f x +e \right )^{n}}{\tan \left (f x +e \right )}d x \right ) a^{3} b f \,n^{2}-8 \left (\int \frac {\tan \left (f x +e \right )^{n}}{\tan \left (f x +e \right )}d x \right ) a^{3} b f n +4 \left (\int \frac {\tan \left (f x +e \right )^{n}}{\tan \left (f x +e \right )}d x \right ) a \,b^{3} f \,n^{2}+8 \left (\int \frac {\tan \left (f x +e \right )^{n}}{\tan \left (f x +e \right )}d x \right ) a \,b^{3} f n +\left (\int \tan \left (f x +e \right )^{n} \tan \left (f x +e \right )^{4}d x \right ) b^{4} f \,n^{2}+2 \left (\int \tan \left (f x +e \right )^{n} \tan \left (f x +e \right )^{4}d x \right ) b^{4} f n +6 \left (\int \tan \left (f x +e \right )^{n} \tan \left (f x +e \right )^{2}d x \right ) a^{2} b^{2} f \,n^{2}+12 \left (\int \tan \left (f x +e \right )^{n} \tan \left (f x +e \right )^{2}d x \right ) a^{2} b^{2} f n \right )}{f n \left (n +2\right )} \] Input:

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x)
 

Output:

(d**n*(4*tan(e + f*x)**n*tan(e + f*x)**2*a*b**3*n + 4*tan(e + f*x)**n*a**3 
*b*n + 8*tan(e + f*x)**n*a**3*b - 4*tan(e + f*x)**n*a*b**3*n - 8*tan(e + f 
*x)**n*a*b**3 + int(tan(e + f*x)**n,x)*a**4*f*n**2 + 2*int(tan(e + f*x)**n 
,x)*a**4*f*n - 4*int(tan(e + f*x)**n/tan(e + f*x),x)*a**3*b*f*n**2 - 8*int 
(tan(e + f*x)**n/tan(e + f*x),x)*a**3*b*f*n + 4*int(tan(e + f*x)**n/tan(e 
+ f*x),x)*a*b**3*f*n**2 + 8*int(tan(e + f*x)**n/tan(e + f*x),x)*a*b**3*f*n 
 + int(tan(e + f*x)**n*tan(e + f*x)**4,x)*b**4*f*n**2 + 2*int(tan(e + f*x) 
**n*tan(e + f*x)**4,x)*b**4*f*n + 6*int(tan(e + f*x)**n*tan(e + f*x)**2,x) 
*a**2*b**2*f*n**2 + 12*int(tan(e + f*x)**n*tan(e + f*x)**2,x)*a**2*b**2*f* 
n))/(f*n*(n + 2))