\(\int (e \cos (c+d x))^m (a+i a \tan (c+d x)) \, dx\) [690]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

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

[Out]

-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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3596, 3586, 3604, 72, 71} \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x)) \, dx=-\frac {i a 2^{1-\frac {m}{2}} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m} \]

[In]

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

[Out]

((-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)

Rule 71

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 3586

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*S
ec[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 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[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 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[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]

Rubi steps \begin{align*} \text {integral}& = \left ((e \cos (c+d x))^m (e \sec (c+d x))^m\right ) \int (e \sec (c+d x))^{-m} (a+i a \tan (c+d x)) \, dx \\ & = \left ((e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \int (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{1-\frac {m}{2}} \, dx \\ & = \frac {\left (a^2 (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \text {Subst}\left (\int (a-i a x)^{-1-\frac {m}{2}} (a+i a x)^{-m/2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{-m/2} a^2 (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{m/2}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-m/2} (a-i a x)^{-1-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x)) \, dx=-\frac {a (e \cos (c+d x))^m \left (i (1+m)+m \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{d m (1+m)} \]

[In]

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

[Out]

-((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)))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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