\(\int (e \cos (c+d x))^m (a+i a \tan (c+d x)) \, dx\) [699]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x)) \, dx=-\frac {i 2^{1-\frac {m}{2}} a (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{d m} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x)) \, dx=-\frac {a (e \cos (c+d x))^m \left (i (1+m)+m \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{d m (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (c+d x)) (e \cos (c+d x))^mdx\)

\(\Big \downarrow \) 3998

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

I*a*(Integral(-I*(e*cos(c + d*x))**m, x) + Integral((e*cos(c + d*x))**m*ta 
n(c + d*x), x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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