\(\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^2 \, dx\) [312]

Optimal result

Integrand size = 26, antiderivative size = 75 \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^2 \, dx=-\frac {a^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a^2 \operatorname {Hypergeometric2F1}(1,1+n,2+n,i \tan (e+f x)) (d \tan (e+f x))^{1+n}}{d f (1+n)} \] Output:

-a^2*(d*tan(f*x+e))^(1+n)/d/f/(1+n)+2*a^2*hypergeom([1, 1+n],[2+n],I*tan(f 
*x+e))*(d*tan(f*x+e))^(1+n)/d/f/(1+n)
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.67 \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^2 \, dx=\frac {a^2 (-1+2 \operatorname {Hypergeometric2F1}(1,1+n,2+n,i \tan (e+f x))) \tan (e+f x) (d \tan (e+f x))^n}{f (1+n)} \] Input:

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

Output:

(a^2*(-1 + 2*Hypergeometric2F1[1, 1 + n, 2 + n, I*Tan[e + f*x]])*Tan[e + f 
*x]*(d*Tan[e + f*x])^n)/(f*(1 + n))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 4026, 3042, 4020, 25, 27, 74}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 74

\(\displaystyle -\frac {a^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac {2 a^2 (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}(1,n+1,n+2,i \tan (e+f x))}{d f (n+1)}\)

Input:

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

Output:

-((a^2*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n))) + (2*a^2*Hypergeometric2F1 
[1, 1 + n, 2 + n, I*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n))
 

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 74

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

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(4*a^2*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1) 
)^n*e^(4*I*f*x + 4*I*e)/(e^(4*I*f*x + 4*I*e) + 2*e^(2*I*f*x + 2*I*e) + 1), 
 x)
 

Sympy [F]

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

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

Output:

-a**2*(Integral(-(d*tan(e + f*x))**n, x) + Integral((d*tan(e + f*x))**n*ta 
n(e + f*x)**2, x) + Integral(-2*I*(d*tan(e + f*x))**n*tan(e + f*x), x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^2 \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \] Input:

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^2,x)
 

Output:

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^2, x)
 

Reduce [F]

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

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

Output:

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