\(\int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx\) [698]

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 = 140 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {\left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a b \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (\left (a^2-b^2\right ) (2+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )+b \left (b (2+n)+2 a (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{f (1+n) (2+n)} \] Input:

Integrate[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x])^2,x]
 

Output:

(Tan[e + f*x]*(d*Tan[e + f*x])^n*((a^2 - b^2)*(2 + n)*Hypergeometric2F1[1, 
 (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2] + b*(b*(2 + n) + 2*a*(1 + n)*Hyper 
geometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))/(f* 
(1 + n)*(2 + n))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4026, 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.

\(\displaystyle \int (a+b \tan (e+f x))^2 (d \tan (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x))^2 (d \tan (e+f x))^ndx\)

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4021

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 278

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

Input:

Int[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x])^2,x]
 

Output:

(b^2*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + ((a^2 - b^2)*Hypergeometric 
2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(d 
*f*(1 + n)) + (2*a*b*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f 
*x]^2]*(d*Tan[e + f*x])^(2 + n))/(d^2*f*(2 + n))
 

Defintions of rubi rules used

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 4026
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( 
m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* 
x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ 
[m, -1] &&  !(EqQ[m, 2] && EqQ[a, 0])
 
Maple [F]

\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{2}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((d*tan(f*x+e))**n*(a+b*tan(f*x+e))**2,x)
 

Output:

Integral((d*tan(e + f*x))**n*(a + b*tan(e + f*x))**2, x)
 

Maxima [F]

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

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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