\(\int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [703]

Optimal result

Integrand size = 28, antiderivative size = 105 \[ \int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i 2^{-\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {3+m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {3+m}{2}}}{d m (a+i a \tan (c+d x))^{3/2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.67 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.36 \[ \int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i 4^{-m} \left (1+e^{2 i (c+d x)}\right ) \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \left (e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \cos ^{-m}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {1-m}{2},-e^{2 i (c+d x)}\right )}{d (1+m) \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3998, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3998

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

((-I)*2^(-1/2 - m/2)*(e*Cos[c + d*x])^m*Hypergeometric2F1[-1/2*m, (3 + m)/ 
2, 1 - m/2, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^((1 + m)/2)*(a + 
I*a*Tan[c + d*x])^((-1 - m)/2 + m/2))/(d*m)
 

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 3998

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m   Int[( 
a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m 
, n}, x] &&  !IntegerQ[m]
 

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]

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

Input:

int((e*cos(d*x+c))^m/(a+I*a*tan(d*x+c))^(1/2),x)
 

Output:

int((e*cos(d*x+c))^m/(a+I*a*tan(d*x+c))^(1/2),x)
 

Fricas [F]

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

integrate((e*cos(d*x+c))^m/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((e*cos(d*x+c))**m/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((e*cos(d*x+c))^m/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

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

integrate((e*cos(d*x+c))^m/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 e^{m} \sqrt {a}\, i \left (-\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{m}-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{m} \sin \left (d x +c \right )}{\cos \left (d x +c \right ) \tan \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) \tan \left (d x +c \right )^{2} d m -\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{m} \sin \left (d x +c \right )}{\cos \left (d x +c \right ) \tan \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) d m -2 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{m} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) \tan \left (d x +c \right )^{2} d -2 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{m} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) d \right )}{a d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

int((e*cos(d*x+c))^m/(a+I*a*tan(d*x+c))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(2*e**m*sqrt(a)*i*( - sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**m - int((sqrt 
(tan(c + d*x)*i + 1)*cos(c + d*x)**m*sin(c + d*x))/(cos(c + d*x)*tan(c + d 
*x)**2 + cos(c + d*x)),x)*tan(c + d*x)**2*d*m - int((sqrt(tan(c + d*x)*i + 
 1)*cos(c + d*x)**m*sin(c + d*x))/(cos(c + d*x)*tan(c + d*x)**2 + cos(c + 
d*x)),x)*d*m - 2*int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**m*tan(c + d*x 
))/(tan(c + d*x)**2 + 1),x)*tan(c + d*x)**2*d - 2*int((sqrt(tan(c + d*x)*i 
 + 1)*cos(c + d*x)**m*tan(c + d*x))/(tan(c + d*x)**2 + 1),x)*d))/(a*d*(tan 
(c + d*x)**2 + 1))