\(\int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [694]

Optimal result

Integrand size = 28, antiderivative size = 104 \[ \int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i 2^{-\frac {1}{2}-\frac {m}{2}} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {3+m}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1+m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-I*2^(-1/2-1/2*m)*(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))^(1/2+1/2*m)/d/m/(a+I*a*tan(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3586, 3604, 72, 71} \[ \int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i 2^{-\frac {m}{2}-\frac {1}{2}} (1+i \tan (c+d x))^{\frac {m+1}{2}} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {m+3}{2},1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((-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))/(d*m*Sqrt[a + I*a*Tan[c + d*x]])

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 \frac {(e \sec (c+d x))^{-m}}{\sqrt {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))^{-\frac {1}{2}-\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)^{-\frac {3}{2}-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{-\frac {3}{2}-\frac {m}{2}} a (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 )^{\frac {1}{2}+\frac {m}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {3}{2}-\frac {m}{2}} (a-i a x)^{-1-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a+i a \tan (c+d x)}} \\ & = -\frac {i 2^{-\frac {1}{2}-\frac {m}{2}} (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 {1+m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.72 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.38 \[ \int \frac {(e \cos (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i 4^{-m} \left (1+e^{2 i (c+d x)}\right ) \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \left (e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \cos ^{-m}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {1-m}{2},-e^{2 i (c+d x)}\right )}{d (1+m) \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(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
*Sqrt[a + I*a*Tan[c + d*x]])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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