3.5.94 \(\int (e \sec (c+d x))^{3-2 n} (a+i a \tan (c+d x))^n \, dx\) [494]

3.5.94.1 Optimal result

Integrand size = 30, antiderivative size = 97 \[ \int (e \sec (c+d x))^{3-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{\frac {3}{2}-n} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-1+2 n),\frac {5}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (e \sec (c+d x))^{3-2 n} (1-i \tan (c+d x))^{-\frac {3}{2}+n} (a+i a \tan (c+d x))^n}{3 d} \]

output

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

3.5.94.2 Mathematica [A] (verified)

Time = 14.59 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.71 \[ \int (e \sec (c+d x))^{3-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{3-n} e^{3 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \left (1+e^{2 i (c+d x)}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3-n,\frac {5}{2},-e^{2 i (c+d x)}\right ) \sec ^{-3+n}(c+d x) (e \sec (c+d x))^{3-2 n} (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{3 d} \]

input

Integrate[(e*Sec[c + d*x])^(3 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
 

output

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

3.5.94.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.44, 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.

\(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{3-2 n} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

input

Int[(e*Sec[c + d*x])^(3 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
 

output

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

3.5.94.3.1 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]
 

3.5.94.4 Maple [F]

\[\int \left (e \sec \left (d x +c \right )\right )^{3-2 n} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]

input

int((e*sec(d*x+c))^(3-2*n)*(a+I*a*tan(d*x+c))^n,x)
 

output

int((e*sec(d*x+c))^(3-2*n)*(a+I*a*tan(d*x+c))^n,x)
 

3.5.94.5 Fricas [F]

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

input

integrate((e*sec(d*x+c))^(3-2*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 + 3)*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)
 

3.5.94.6 Sympy [F]

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

input

integrate((e*sec(d*x+c))**(3-2*n)*(a+I*a*tan(d*x+c))**n,x)
 

output

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

3.5.94.7 Maxima [F]

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

input

integrate((e*sec(d*x+c))^(3-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima 
")
 

output

integrate((e*sec(d*x + c))^(-2*n + 3)*(I*a*tan(d*x + c) + a)^n, x)
 

3.5.94.8 Giac [F]

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

input

integrate((e*sec(d*x+c))^(3-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")
 

output

integrate((e*sec(d*x + c))^(-2*n + 3)*(I*a*tan(d*x + c) + a)^n, x)
 

3.5.94.9 Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^{3-2 n} (a+i a \tan (c+d x))^n \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3-2\,n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

input

int((e/cos(c + d*x))^(3 - 2*n)*(a + a*tan(c + d*x)*1i)^n,x)
 

output

int((e/cos(c + d*x))^(3 - 2*n)*(a + a*tan(c + d*x)*1i)^n, x)