\(\int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx\) [482]

Optimal result

Integrand size = 28, antiderivative size = 96 \[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {i 2^{-\frac {1}{4}+n} a \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {9}{4}-n,-\frac {1}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {9}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{5 d (e \sec (c+d x))^{5/2}} \] Output:

-1/5*I*2^(-1/4+n)*a*hypergeom([-5/4, 9/4-n],[-1/4],1/2-1/2*I*tan(d*x+c))*( 
1+I*tan(d*x+c))^(9/4-n)*(a+I*a*tan(d*x+c))^(-1+n)/d/(e*sec(d*x+c))^(5/2)
 

Mathematica [A] (verified)

Time = 8.03 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.64 \[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {i 2^{-\frac {3}{2}+n} e^{-3 i (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{\frac {1}{2}+n} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2}+n,-\frac {5}{4}+n,-\frac {1}{4}+n,-e^{2 i (c+d x)}\right ) \sec ^{\frac {1}{2}-n}(c+d x) (a+i a \tan (c+d x))^n}{d e^2 (-5+4 n) \sqrt {e \sec (c+d x)}} \] Input:

Integrate[(a + I*a*Tan[c + d*x])^n/(e*Sec[c + d*x])^(5/2),x]
 

Output:

((-I)*2^(-3/2 + 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[-5/2 + n, -5/4 + n, - 
1/4 + n, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(1/2 - n)*(a + I*a*Tan[c + d*x 
])^n)/(d*e^2*E^((3*I)*(c + d*x))*(-5 + 4*n)*Sqrt[e*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.02, 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.

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4} \int \frac {(i \tan (c+d x) a+a)^{n-\frac {9}{4}}}{(a-i a \tan (c+d x))^{9/4}}d\tan (c+d x)}{d (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {2^{n-\frac {9}{4}} (a-i a \tan (c+d x))^{5/4} (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^{n+1} \int \frac {\left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{n-\frac {9}{4}}}{(a-i a \tan (c+d x))^{9/4}}d\tan (c+d x)}{d (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {i 2^{n-\frac {1}{4}} (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {9}{4}-n,-\frac {1}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{5 a d (e \sec (c+d x))^{5/2}}\)

Input:

Int[(a + I*a*Tan[c + d*x])^n/(e*Sec[c + d*x])^(5/2),x]
 

Output:

((-1/5*I)*2^(-1/4 + n)*Hypergeometric2F1[-5/4, 9/4 - n, -1/4, (1 - I*Tan[c 
 + d*x])/2]*(1 + I*Tan[c + d*x])^(1/4 - n)*(a + I*a*Tan[c + d*x])^(1 + n)) 
/(a*d*(e*Sec[c + d*x])^(5/2))
 

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]

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

Input:

int((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(5/2),x)
 

Output:

int((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(5/2),x)
 

Fricas [F]

\[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

integral(1/8*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^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4 
*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1)*e^(-5/2*I*d*x - 5/2*I*c)/e^3, x)
 

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+I*a*tan(d*x+c))**n/(e*sec(d*x+c))**(5/2),x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**n/(e*sec(c + d*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((I*a*tan(d*x + c) + a)^n/(e*sec(d*x + c))^(5/2), x)
 

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate((I*a*tan(d*x + c) + a)^n/(e*sec(d*x + c))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^n/(e/cos(c + d*x))^(5/2),x)
 

Output:

int((a + a*tan(c + d*x)*1i)^n/(e/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \left (\tan \left (d x +c \right ) a i +a \right )^{n}}{\sec \left (d x +c \right )^{3}}d x \right )}{e^{3}} \] Input:

int((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(e)*int((sqrt(sec(c + d*x))*(tan(c + d*x)*a*i + a)**n)/sec(c + d*x)** 
3,x))/e**3