\(\int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^3} \, dx\) [316]

Optimal result

Integrand size = 26, antiderivative size = 274 \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^3} \, dx=\frac {(1-2 n) (1-n) (3-n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{24 a^3 d f (1+n)}+\frac {i (5-2 n) (2-n) n \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{24 a^3 d^2 f (2+n)}+\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {(7-2 n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {(5-2 n) (2-n) (d \tan (e+f x))^{1+n}}{24 d f \left (a^3+i a^3 \tan (e+f x)\right )} \] Output:

1/24*(1-2*n)*(1-n)*(3-n)*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^3/d/f/(1+n)+1/24*I*(5-2*n)*(2-n)*n*hypergeom([1, 
 1+1/2*n],[2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e))^(2+n)/a^3/d^2/f/(2+n)+1/ 
6*(d*tan(f*x+e))^(1+n)/d/f/(a+I*a*tan(f*x+e))^3+1/24*(7-2*n)*(d*tan(f*x+e) 
)^(1+n)/a/d/f/(a+I*a*tan(f*x+e))^2+1/24*(5-2*n)*(2-n)*(d*tan(f*x+e))^(1+n) 
/d/f/(a^3+I*a^3*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 3.37 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.84 \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^3} \, dx=\frac {(d \tan (e+f x))^{1+n} \left (4 a^3 (1+n) (2+n)+(a+i a \tan (e+f x)) \left (-a^2 (1+n) (2+n) (-7+2 n)+(a+i a \tan (e+f x)) \left (a (1+n) (2+n) \left (10-9 n+2 n^2\right )+(a+i a \tan (e+f x)) \left (-\left ((-3+n) (-1+n) (2+n) (-1+2 n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )\right )+i (-2+n) n (1+n) (-5+2 n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )\right )\right )}{24 a^3 d f (1+n) (2+n) (a+i a \tan (e+f x))^3} \] Input:

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

Output:

((d*Tan[e + f*x])^(1 + n)*(4*a^3*(1 + n)*(2 + n) + (a + I*a*Tan[e + f*x])* 
(-(a^2*(1 + n)*(2 + n)*(-7 + 2*n)) + (a + I*a*Tan[e + f*x])*(a*(1 + n)*(2 
+ n)*(10 - 9*n + 2*n^2) + (a + I*a*Tan[e + f*x])*(-((-3 + n)*(-1 + n)*(2 + 
 n)*(-1 + 2*n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2] 
) + I*(-2 + n)*n*(1 + n)*(-5 + 2*n)*Hypergeometric2F1[1, (2 + n)/2, (4 + n 
)/2, -Tan[e + f*x]^2]*Tan[e + f*x])))))/(24*a^3*d*f*(1 + n)*(2 + n)*(a + I 
*a*Tan[e + f*x])^3)
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 4042, 3042, 4079, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {(d \tan (e+f x))^n \left (a^2 d^2 \left (2 n^2-9 n+13\right )-i a^2 d^2 (7-2 n) (1-n) \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(d \tan (e+f x))^n \left (a^2 d^2 \left (2 n^2-9 n+13\right )-i a^2 d^2 (7-2 n) (1-n) \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\frac {\int 2 (d \tan (e+f x))^n \left (a^3 (1-2 n) (1-n) (3-n) d^3+i a^3 (5-2 n) (2-n) n \tan (e+f x) d^3\right )dx}{2 a^2 d}+\frac {a^2 d (5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{f (a+i a \tan (e+f x))}}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int (d \tan (e+f x))^n \left (a^3 (1-2 n) (1-n) (3-n) d^3+i a^3 (5-2 n) (2-n) n \tan (e+f x) d^3\right )dx}{a^2 d}+\frac {a^2 d (5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{f (a+i a \tan (e+f x))}}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int (d \tan (e+f x))^n \left (a^3 (1-2 n) (1-n) (3-n) d^3+i a^3 (5-2 n) (2-n) n \tan (e+f x) d^3\right )dx}{a^2 d}+\frac {a^2 d (5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{f (a+i a \tan (e+f x))}}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {\frac {\frac {a^3 d^3 (1-2 n) (1-n) (3-n) \int (d \tan (e+f x))^ndx+i a^3 d^2 (5-2 n) (2-n) n \int (d \tan (e+f x))^{n+1}dx}{a^2 d}+\frac {a^2 d (5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{f (a+i a \tan (e+f x))}}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^3 d^3 (1-2 n) (1-n) (3-n) \int (d \tan (e+f x))^ndx+i a^3 d^2 (5-2 n) (2-n) n \int (d \tan (e+f x))^{n+1}dx}{a^2 d}+\frac {a^2 d (5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{f (a+i a \tan (e+f x))}}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {\frac {\frac {a^3 d^4 (1-2 n) (1-n) (3-n) \int \frac {(d \tan (e+f x))^n}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}+\frac {i a^3 d^3 (5-2 n) (2-n) n \int \frac {(d \tan (e+f x))^{n+1}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}}{a^2 d}+\frac {a^2 d (5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{f (a+i a \tan (e+f x))}}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {\frac {\frac {a^2 d (5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{f (a+i a \tan (e+f x))}+\frac {\frac {a^3 d^2 (1-2 n) (1-n) (3-n) (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{f (n+1)}+\frac {i a^3 d (5-2 n) (2-n) n (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{f (n+2)}}{a^2 d}}{4 a^2 d}+\frac {a (7-2 n) (d \tan (e+f x))^{n+1}}{4 f (a+i a \tan (e+f x))^2}}{6 a^2 d}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3}\)

Input:

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

Output:

(d*Tan[e + f*x])^(1 + n)/(6*d*f*(a + I*a*Tan[e + f*x])^3) + ((a*(7 - 2*n)* 
(d*Tan[e + f*x])^(1 + n))/(4*f*(a + I*a*Tan[e + f*x])^2) + ((a^2*d*(5 - 2* 
n)*(2 - n)*(d*Tan[e + f*x])^(1 + n))/(f*(a + I*a*Tan[e + f*x])) + ((a^3*d^ 
2*(1 - 2*n)*(1 - n)*(3 - n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Ta 
n[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(f*(1 + n)) + (I*a^3*d*(5 - 2*n)*( 
2 - n)*n*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Ta 
n[e + f*x])^(2 + n))/(f*(2 + n)))/(a^2*d))/(4*a^2*d))/(6*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(1/8*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^ 
n*(e^(6*I*f*x + 6*I*e) + 3*e^(4*I*f*x + 4*I*e) + 3*e^(2*I*f*x + 2*I*e) + 1 
)*e^(-6*I*f*x - 6*I*e)/a^3, x)
 

Sympy [F]

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

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

Output:

I*Integral((d*tan(e + f*x))**n/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3* 
tan(e + f*x) + I), x)/a**3
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^3} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \] Input:

int((d*tan(e + f*x))^n/(a + a*tan(e + f*x)*1i)^3,x)
 

Output:

int((d*tan(e + f*x))^n/(a + a*tan(e + f*x)*1i)^3, x)
 

Reduce [F]

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

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

Output:

(d**n*( - tan(e + f*x)**n*i - int(tan(e + f*x)**n/(tan(e + f*x)**4 - 3*tan 
(e + f*x)**3*i - 3*tan(e + f*x)**2 + tan(e + f*x)*i),x)*f*n + int((tan(e + 
 f*x)**n*tan(e + f*x)**4)/(tan(e + f*x)**3 - 3*tan(e + f*x)**2*i - 3*tan(e 
 + f*x) + i),x)*f*i*n + 3*int((tan(e + f*x)**n*tan(e + f*x)**3)/(tan(e + f 
*x)**3 - 3*tan(e + f*x)**2*i - 3*tan(e + f*x) + i),x)*f*n - 2*int((tan(e + 
 f*x)**n*tan(e + f*x)**2)/(tan(e + f*x)**3 - 3*tan(e + f*x)**2*i - 3*tan(e 
 + f*x) + i),x)*f*i*n + 2*int((tan(e + f*x)**n*tan(e + f*x))/(tan(e + f*x) 
**3 - 3*tan(e + f*x)**2*i - 3*tan(e + f*x) + i),x)*f*n))/(3*a**3*f*n)