\(\int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx\) [480]

Optimal result

Integrand size = 28, antiderivative size = 91 \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{\frac {3}{4}+n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n}{d \sqrt {e \sec (c+d x)}} \]

[Out]

-I*2^(3/4+n)*hypergeom([-1/4, 5/4-n],[3/4],1/2-1/2*I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/4-n)*(a+I*a*tan(d*x+c))^n
/d/(e*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3586, 3604, 72, 71} \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{n+\frac {3}{4}} (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{d \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

((-I)*2^(3/4 + n)*Hypergeometric2F1[-1/4, 5/4 - n, 3/4, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(1/4 - n)
*(a + I*a*Tan[c + d*x])^n)/(d*Sqrt[e*Sec[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 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}& = \frac {\left (\sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{-\frac {1}{4}+n}}{\sqrt [4]{a-i a \tan (c+d x)}} \, dx}{\sqrt {e \sec (c+d x)}} \\ & = \frac {\left (a^2 \sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-\frac {5}{4}+n}}{(a-i a x)^{5/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}} \\ & = \frac {\left (2^{-\frac {5}{4}+n} a \sqrt [4]{a-i a \tan (c+d x)} (a+i a \tan (c+d x))^n \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {1}{4}-n}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {5}{4}+n}}{(a-i a x)^{5/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}} \\ & = -\frac {i 2^{\frac {3}{4}+n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n}{d \sqrt {e \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.57 \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{\frac {1}{2}+n} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{-\frac {1}{2}+n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}+n,-\frac {1}{4}+n,\frac {3}{4}+n,-e^{2 i (c+d x)}\right ) \sec ^{\frac {1}{2}-n}(c+d x) (a+i a \tan (c+d x))^n}{d (-1+4 n) \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

((-I)*2^(1/2 + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(-1/2 + n)*(1 + E^((2*I)*(c + d*x)))^(-1/2 + n)*
Hypergeometric2F1[-1/2 + n, -1/4 + n, 3/4 + n, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(1/2 - n)*(a + I*a*Tan[c + d
*x])^n)/(d*(-1 + 4*n)*Sqrt[e*Sec[c + d*x]])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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