\(\int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx\) [492]

Optimal result

Integrand size = 30, antiderivative size = 103 \[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{\frac {5}{2}-n} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (-3+2 n),\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (e \sec (c+d x))^{5-2 n} (1-i \tan (c+d x))^{-1+\frac {1}{2} (-3+2 n)} (a+i a \tan (c+d x))^n}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 10.66 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.61 \[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{5-n} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \left (1+e^{2 i (c+d x)}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},5-n,\frac {7}{2},-e^{2 i (c+d x)}\right ) \sec ^{-5+n}(c+d x) (e \sec (c+d x))^{5-2 n} (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{5 d} \] Input:

Integrate[(e*Sec[c + d*x])^(5 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
 

Output:

((-1/5*I)*2^(5 - n)*E^((5*I)*(c + d*x))*(E^(I*d*x))^n*Hypergeometric2F1[5/ 
2, 5 - n, 7/2, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(-5 + n)*(e*Sec[c + d*x] 
)^(5 - 2*n)*(a + I*a*Tan[c + d*x])^n)/(d*(E^(I*(c + d*x))/(1 + E^((2*I)*(c 
 + d*x))))^n*(1 + E^((2*I)*(c + d*x)))^n*(Cos[d*x] + I*Sin[d*x])^n)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.38, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3986, 3042, 4006, 80, 79}

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.

\(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{5-2 n} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{5-2 n}dx\)

\(\Big \downarrow \) 3986

\(\displaystyle (a-i a \tan (c+d x))^{\frac {1}{2} (2 n-5)} (a+i a \tan (c+d x))^{\frac {1}{2} (2 n-5)} (e \sec (c+d x))^{5-2 n} \int (a-i a \tan (c+d x))^{\frac {1}{2} (5-2 n)} (i \tan (c+d x) a+a)^{5/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (a-i a \tan (c+d x))^{\frac {1}{2} (2 n-5)} (a+i a \tan (c+d x))^{\frac {1}{2} (2 n-5)} (e \sec (c+d x))^{5-2 n} \int (a-i a \tan (c+d x))^{\frac {1}{2} (5-2 n)} (i \tan (c+d x) a+a)^{5/2}dx\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{\frac {1}{2} (2 n-5)} (a+i a \tan (c+d x))^{\frac {1}{2} (2 n-5)} (e \sec (c+d x))^{5-2 n} \int (a-i a \tan (c+d x))^{\frac {1}{2} (3-2 n)} (i \tan (c+d x) a+a)^{3/2}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a^3 2^{\frac {3}{2}-n} (1-i \tan (c+d x))^{n-\frac {1}{2}} (a-i a \tan (c+d x))^{-n+\frac {1}{2} (2 n-5)+\frac {1}{2}} (a+i a \tan (c+d x))^{\frac {1}{2} (2 n-5)} (e \sec (c+d x))^{5-2 n} \int \left (\frac {1}{2}-\frac {1}{2} i \tan (c+d x)\right )^{\frac {1}{2} (3-2 n)} (i \tan (c+d x) a+a)^{3/2}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 79

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

Input:

Int[(e*Sec[c + d*x])^(5 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
 

Output:

((-1/5*I)*2^(5/2 - n)*a^2*Hypergeometric2F1[5/2, (-3 + 2*n)/2, 7/2, (1 + I 
*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(5 - 2*n)*(1 - I*Tan[c + d*x])^(-1/2 + 
n)*(a - I*a*Tan[c + d*x])^(1/2 - n + (-5 + 2*n)/2)*(a + I*a*Tan[c + d*x])^ 
(5/2 + (-5 + 2*n)/2))/d
 

Defintions of rubi rules used

rule 79

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

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3986

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_.), x_Symbol] :> Simp[(d*Sec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/ 
2)*(a - b*Tan[e + f*x])^(m/2))   Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a - b* 
Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + 
 b^2, 0]
 

rule 4006

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

Maple [F]

\[\int \left (e \sec \left (d x +c \right )\right )^{5-2 n} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]

Input:

int((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x)
 

Output:

int((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x)
 

Fricas [F]

\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \] Input:

integrate((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas 
")
 

Output:

integral((2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^(-2*n + 5)*e^(I*d 
*n*x + I*c*n + n*log(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1)) + n*lo 
g(a/e)), x)
 

Sympy [F]

\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{5 - 2 n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \] Input:

integrate((e*sec(d*x+c))**(5-2*n)*(a+I*a*tan(d*x+c))**n,x)
 

Output:

Integral((e*sec(c + d*x))**(5 - 2*n)*(I*a*(tan(c + d*x) - I))**n, x)
 

Maxima [F]

\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \] Input:

integrate((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima 
")
 

Output:

integrate((e*sec(d*x + c))^(-2*n + 5)*(I*a*tan(d*x + c) + a)^n, x)
 

Giac [F]

\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \] Input:

integrate((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(-2*n + 5)*(I*a*tan(d*x + c) + a)^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5-2\,n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \] Input:

int((e/cos(c + d*x))^(5 - 2*n)*(a + a*tan(c + d*x)*1i)^n,x)
 

Output:

int((e/cos(c + d*x))^(5 - 2*n)*(a + a*tan(c + d*x)*1i)^n, x)
 

Reduce [F]

\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\frac {\left (\int \frac {\left (\tan \left (d x +c \right ) a i +a \right )^{n} \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{2 n}}d x \right ) e^{5}}{e^{2 n}} \] Input:

int((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x)
 

Output:

(int(((tan(c + d*x)*a*i + a)**n*sec(c + d*x)**5)/sec(c + d*x)**(2*n),x)*e* 
*5)/e**(2*n)