\(\int (e \tan (c+d x))^m (a-i a \tan (c+d x)) \, dx\) [309]

Optimal result

Integrand size = 24, antiderivative size = 43 \[ \int (e \tan (c+d x))^m (a-i a \tan (c+d x)) \, dx=\frac {a \operatorname {Hypergeometric2F1}(1,1+m,2+m,-i \tan (c+d x)) (e \tan (c+d x))^{1+m}}{d e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int (e \tan (c+d x))^m (a-i a \tan (c+d x)) \, dx=\frac {a \operatorname {Hypergeometric2F1}(1,1+m,2+m,-i \tan (c+d x)) \tan (c+d x) (e \tan (c+d x))^m}{d (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 4020, 25, 27, 74}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 74

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

rule 3042

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

rule 4020

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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