\(\int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)} \, dx\) [454]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)} \, dx=\frac {i 2^{-1+\frac {m}{2}} \operatorname {Hypergeometric2F1}\left (2-\frac {m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-m/2}}{a d m} \] Output:

I*2^(-1+1/2*m)*hypergeom([1/2*m, 2-1/2*m],[1+1/2*m],1/2-1/2*I*tan(d*x+c))* 
(e*sec(d*x+c))^m/a/d/m/((1+I*tan(d*x+c))^(1/2*m))
 

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.77 \[ \int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)} \, dx=-\frac {i 2^{-1+m} e^{-i (c+2 d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (1+e^{2 i (c+d x)}\right )^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2+m),-1+m,\frac {m}{2},-e^{2 i (c+d x)}\right ) \sec ^{1-m}(c+d x) (e \sec (c+d x))^m (\cos (d x)+i \sin (d x))}{d (-2+m) (a+i a \tan (c+d x))} \] Input:

Integrate[(e*Sec[c + d*x])^m/(a + I*a*Tan[c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, 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.

\(\displaystyle \int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)}dx\)

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{\frac {m-2}{2}} (i \tan (c+d x) a+a)^{\frac {m-4}{2}}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {2^{\frac {m}{2}-2} (1+i \tan (c+d x))^{-m/2} (a-i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{\frac {m-4}{2}} (a-i a \tan (c+d x))^{\frac {m-2}{2}}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {i 2^{\frac {m}{2}-1} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {4-m}{2},\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{a d m}\)

Input:

Int[(e*Sec[c + d*x])^m/(a + I*a*Tan[c + d*x]),x]
 

Output:

(I*2^(-1 + m/2)*Hypergeometric2F1[(4 - m)/2, m/2, (2 + m)/2, (1 - I*Tan[c 
+ d*x])/2]*(e*Sec[c + d*x])^m)/(a*d*m*(1 + I*Tan[c + d*x])^(m/2))
 

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]

\[\int \frac {\left (e \sec \left (d x +c \right )\right )^{m}}{a +i a \tan \left (d x +c \right )}d x\]

Input:

int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),x)
 

Output:

int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),x)
 

Fricas [F]

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

integrate((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

integral(1/2*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*(e^(2*I*d*x 
 + 2*I*c) + 1)*e^(-2*I*d*x - 2*I*c)/a, x)
 

Sympy [F]

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

integrate((e*sec(d*x+c))**m/(a+I*a*tan(d*x+c)),x)
 

Output:

-I*Integral((e*sec(c + d*x))**m/(tan(c + d*x) - I), x)/a
 

Maxima [F(-2)]

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

integrate((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^m/(I*a*tan(d*x + c) + a), x)
 

Mupad [F(-1)]

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

int((e/cos(c + d*x))^m/(a + a*tan(c + d*x)*1i),x)
 

Output:

int((e/cos(c + d*x))^m/(a + a*tan(c + d*x)*1i), x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)} \, dx=\frac {e^{m} \left (\int \frac {\sec \left (d x +c \right )^{m}}{\tan \left (d x +c \right ) i +1}d x \right )}{a} \] Input:

int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),x)
 

Output:

(e**m*int(sec(c + d*x)**m/(tan(c + d*x)*i + 1),x))/a