\(\int (d \tan (e+f x))^n (a+b \tan (e+f x))^3 \, dx\) [697]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 198 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^3 \, dx=\frac {a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac {a \left (a^2-3 b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {b \left (3 a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)} \] Output:

a*b^2*(5+2*n)*(d*tan(f*x+e))^(1+n)/d/f/(1+n)/(2+n)+a*(a^2-3*b^2)*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)+ 
b*(3*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))/d/f/(2+n)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.71 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^3 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (a \left (a^2-3 b^2\right ) (2+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )+b \left (\left (3 a^2-b^2\right ) (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)+b (3 a (2+n)+b (1+n) \tan (e+f x))\right )\right )}{f (1+n) (2+n)} \] Input:

Integrate[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x])^3,x]
 

Output:

(Tan[e + f*x]*(d*Tan[e + f*x])^n*(a*(a^2 - 3*b^2)*(2 + n)*Hypergeometric2F 
1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2] + b*((3*a^2 - b^2)*(1 + n)*Hyp 
ergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x] + b* 
(3*a*(2 + n) + b*(1 + n)*Tan[e + f*x]))))/(f*(1 + n)*(2 + n))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4049, 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.

\(\displaystyle \int (a+b \tan (e+f x))^3 (d \tan (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x))^3 (d \tan (e+f x))^ndx\)

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4021

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 278

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

Input:

Int[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x])^3,x]
 

Output:

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

\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{3}d x\]

Input:

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,x)
 

Output:

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,x)
 

Fricas [F]

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

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((d*tan(f*x+e))**n*(a+b*tan(f*x+e))**3,x)
 

Output:

Integral((d*tan(e + f*x))**n*(a + b*tan(e + f*x))**3, x)
 

Maxima [F]

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

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e) + a)^3*(d*tan(f*x + e))^n, x)
 

Giac [F]

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

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e) + a)^3*(d*tan(f*x + e))^n, x)
 

Mupad [F(-1)]

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

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x))^3,x)
 

Output:

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x))^3, x)
 

Reduce [F]

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

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^3,x)
 

Output:

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