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

Optimal result

Integrand size = 28, antiderivative size = 94 \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{\frac {3}{4}+n} a \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {5}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{d \sqrt {e \sec (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 7.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.52 \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{\frac {1}{2}+n} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{-\frac {1}{2}+n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}+n,-\frac {1}{4}+n,\frac {3}{4}+n,-e^{2 i (c+d x)}\right ) \sec ^{\frac {1}{2}-n}(c+d x) (a+i a \tan (c+d x))^n}{d (-1+4 n) \sqrt {e \sec (c+d x)}} \] Input:

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

Output:

((-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]])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

\(\displaystyle \frac {\sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)} \int \frac {(i \tan (c+d x) a+a)^{n-\frac {1}{4}}}{\sqrt [4]{a-i a \tan (c+d x)}}dx}{\sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)} \int \frac {(i \tan (c+d x) a+a)^{n-\frac {1}{4}}}{\sqrt [4]{a-i a \tan (c+d x)}}dx}{\sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 \sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)} \int \frac {(i \tan (c+d x) a+a)^{n-\frac {5}{4}}}{(a-i a \tan (c+d x))^{5/4}}d\tan (c+d x)}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a 2^{n-\frac {5}{4}} \sqrt [4]{a-i a \tan (c+d x)} (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n \int \frac {\left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{n-\frac {5}{4}}}{(a-i a \tan (c+d x))^{5/4}}d\tan (c+d x)}{d \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {i 2^{n+\frac {3}{4}} (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{d \sqrt {e \sec (c+d x)}}\)

Input:

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

Output:

((-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*S 
ec[c + d*x]])
 

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*int((sqrt(sec(c + d*x))*(tan(c + d*x)*a*i + a)**n)/sec(c + d*x),x 
))/e