\(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx\) [472]

Optimal result

Integrand size = 24, antiderivative size = 92 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {3}{2}+n} a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-n,\frac {5}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^3(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-1+n}}{3 d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 11.82 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.62 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{3+n} e^{3 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {5}{2}+n,-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d \left (1+e^{2 i (c+d x)}\right )^2 (3+2 n)} \] Input:

Integrate[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^n,x]
 

Output:

((-I)*2^(3 + n)*E^((3*I)*(c + d*x))*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^ 
((2*I)*(c + d*x))))^n*Hypergeometric2F1[-1/2, 1, 5/2 + n, -E^((2*I)*(c + d 
*x))]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + E^((2*I)*(c + d*x)))^2*(3 + 2*n)*S 
ec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.

\(\displaystyle \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^3 (a+i a \tan (c+d x))^ndx\)

\(\Big \downarrow \) 3986

\(\displaystyle \frac {\sec ^3(c+d x) \int (a-i a \tan (c+d x))^{3/2} (i \tan (c+d x) a+a)^{n+\frac {3}{2}}dx}{(a-i a \tan (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sec ^3(c+d x) \int (a-i a \tan (c+d x))^{3/2} (i \tan (c+d x) a+a)^{n+\frac {3}{2}}dx}{(a-i a \tan (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 \sec ^3(c+d x) \int \sqrt {a-i a \tan (c+d x)} (i \tan (c+d x) a+a)^{n+\frac {1}{2}}d\tan (c+d x)}{d (a-i a \tan (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a^2 2^{n+\frac {1}{2}} \sec ^3(c+d x) (1+i \tan (c+d x))^{-n-\frac {1}{2}} (a+i a \tan (c+d x))^{n-1} \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{n+\frac {1}{2}} \sqrt {a-i a \tan (c+d x)}d\tan (c+d x)}{d (a-i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 79

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

Input:

Int[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^n,x]
 

Output:

((I/3)*2^(3/2 + n)*a*Hypergeometric2F1[3/2, -1/2 - n, 5/2, (1 - I*Tan[c + 
d*x])/2]*Sec[c + d*x]^3*(1 + I*Tan[c + d*x])^(-1/2 - n)*(a + I*a*Tan[c + d 
*x])^(-1 + n))/d
 

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]

\[\int \sec \left (d x +c \right )^{3} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]

Input:

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x)
 

Output:

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x)
 

Fricas [F]

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

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")
 

Output:

integral(8*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*e^(3*I*d* 
x + 3*I*c)/(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), x)
 

Sympy [F]

\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)**3*(a+I*a*tan(d*x+c))**n,x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**n*sec(c + d*x)**3, x)
 

Maxima [F]

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

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")
 

Output:

integrate((I*a*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)
 

Giac [F]

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

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((I*a*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^n/cos(c + d*x)^3,x)
 

Output:

int((a + a*tan(c + d*x)*1i)^n/cos(c + d*x)^3, x)
 

Reduce [F]

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

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x)
 

Output:

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