\(\int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx\) [315]

Optimal result

Integrand size = 26, antiderivative size = 209 \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\frac {(1-n)^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{4 a^2 d f (1+n)}+\frac {(2-n) (d \tan (e+f x))^{1+n}}{4 a^2 d f (1+i \tan (e+f x))}+\frac {i (2-n) n \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{4 a^2 d^2 f (2+n)}+\frac {(d \tan (e+f x))^{1+n}}{4 d f (a+i a \tan (e+f x))^2} \] Output:

1/4*(1-n)^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)/a^2/d/f/(1+n)+1/4*(2-n)*(d*tan(f*x+e))^(1+n)/a^2/d/f/(1+I*tan(f 
*x+e))+1/4*I*(2-n)*n*hypergeom([1, 1+1/2*n],[2+1/2*n],-tan(f*x+e)^2)*(d*ta 
n(f*x+e))^(2+n)/a^2/d^2/f/(2+n)+1/4*(d*tan(f*x+e))^(1+n)/d/f/(a+I*a*tan(f* 
x+e))^2
 

Mathematica [A] (verified)

Time = 2.03 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.06 \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\frac {2 a^3 d (1+n) (2+n) (d \tan (e+f x))^{1+n}+(a+i a \tan (e+f x)) \left (-2 a^2 d (-2+n) (1+n) (2+n) (d \tan (e+f x))^{1+n}+2 a (a+i a \tan (e+f x)) \left (d (-1+n)^2 (2+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}+i (2-n) n (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}\right )\right )}{8 a^3 d^2 f (1+n) (2+n) (a+i a \tan (e+f x))^2} \] Input:

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

Output:

(2*a^3*d*(1 + n)*(2 + n)*(d*Tan[e + f*x])^(1 + n) + (a + I*a*Tan[e + f*x]) 
*(-2*a^2*d*(-2 + n)*(1 + n)*(2 + n)*(d*Tan[e + f*x])^(1 + n) + 2*a*(a + I* 
a*Tan[e + f*x])*(d*(-1 + n)^2*(2 + n)*Hypergeometric2F1[1, (1 + n)/2, (3 + 
 n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n) + I*(2 - n)*n*(1 + n)*Hyp 
ergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^ 
(2 + n))))/(8*a^3*d^2*f*(1 + n)*(2 + n)*(a + I*a*Tan[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4042, 3042, 4079, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4021

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

(d*Tan[e + f*x])^(1 + n)/(4*d*f*(a + I*a*Tan[e + f*x])^2) + (((2 - n)*(d*T 
an[e + f*x])^(1 + n))/(f*(1 + I*Tan[e + f*x])) + ((a^2*d*(1 - n)^2*Hyperge 
ometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + 
 n))/(f*(1 + n)) + (I*a^2*(2 - n)*n*Hypergeometric2F1[1, (2 + n)/2, (4 + n 
)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(f*(2 + n)))/(a^2*d))/(4*a 
^2*d)
 

Defintions of rubi rules used

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 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 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] 
 && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

Maple [F]

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

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]

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

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

Output:

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

Sympy [F]

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

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

Output:

-Integral((d*tan(e + f*x))**n/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) 
/a**2
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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