3.491 \(\int (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^n \, dx\)
Optimal. Leaf size=156 \[ \frac{8 i a^3 (a+i a \tan (c+d x))^{n-3} (e \sec (c+d x))^{6-2 n}}{d (5-n) \left (n^2-7 n+12\right )}+\frac{4 i a^2 (a+i a \tan (c+d x))^{n-2} (e \sec (c+d x))^{6-2 n}}{d \left (n^2-9 n+20\right )}+\frac{i a (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{6-2 n}}{d (5-n)} \]
[Out]
((8*I)*a^3*(e*Sec[c + d*x])^(6 - 2*n)*(a + I*a*Tan[c + d*x])^(-3 + n))/(d*(5 - n)*(12 - 7*n + n^2)) + ((4*I)*a
^2*(e*Sec[c + d*x])^(6 - 2*n)*(a + I*a*Tan[c + d*x])^(-2 + n))/(d*(20 - 9*n + n^2)) + (I*a*(e*Sec[c + d*x])^(6
- 2*n)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(5 - n))
________________________________________________________________________________________
Rubi [A] time = 0.225592, antiderivative size = 156, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used =
{3494, 3493} \[ \frac{8 i a^3 (a+i a \tan (c+d x))^{n-3} (e \sec (c+d x))^{6-2 n}}{d (5-n) \left (n^2-7 n+12\right )}+\frac{4 i a^2 (a+i a \tan (c+d x))^{n-2} (e \sec (c+d x))^{6-2 n}}{d \left (n^2-9 n+20\right )}+\frac{i a (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{6-2 n}}{d (5-n)} \]
Antiderivative was successfully verified.
[In]
Int[(e*Sec[c + d*x])^(6 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
[Out]
((8*I)*a^3*(e*Sec[c + d*x])^(6 - 2*n)*(a + I*a*Tan[c + d*x])^(-3 + n))/(d*(5 - n)*(12 - 7*n + n^2)) + ((4*I)*a
^2*(e*Sec[c + d*x])^(6 - 2*n)*(a + I*a*Tan[c + d*x])^(-2 + n))/(d*(20 - 9*n + n^2)) + (I*a*(e*Sec[c + d*x])^(6
- 2*n)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(5 - n))
Rule 3494
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] && !IntegerQ[n]
Rule 3493
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*
(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]
Rubi steps
\begin{align*} \int (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^n \, dx &=\frac{i a (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-1+n}}{d (5-n)}+\frac{(4 a) \int (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-1+n} \, dx}{5-n}\\ &=\frac{4 i a^2 (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-2+n}}{d \left (20-9 n+n^2\right )}+\frac{i a (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-1+n}}{d (5-n)}+\frac{\left (8 a^2\right ) \int (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-2+n} \, dx}{20-9 n+n^2}\\ &=\frac{8 i a^3 (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-3+n}}{d (3-n) \left (20-9 n+n^2\right )}+\frac{4 i a^2 (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-2+n}}{d \left (20-9 n+n^2\right )}+\frac{i a (e \sec (c+d x))^{6-2 n} (a+i a \tan (c+d x))^{-1+n}}{d (5-n)}\\ \end{align*}
Mathematica [A] time = 2.12678, size = 122, normalized size = 0.78 \[ -\frac{e^6 \sec ^5(c+d x) (\sin (3 (c+d x))+i \cos (3 (c+d x))) (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n} \left (i \left (n^2-9 n+18\right ) \sin (2 (c+d x))+\left (n^2-9 n+22\right ) \cos (2 (c+d x))-2 (n-5)\right )}{d (n-5) (n-4) (n-3)} \]
Antiderivative was successfully verified.
[In]
Integrate[(e*Sec[c + d*x])^(6 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
[Out]
-((e^6*Sec[c + d*x]^5*(-2*(-5 + n) + (22 - 9*n + n^2)*Cos[2*(c + d*x)] + I*(18 - 9*n + n^2)*Sin[2*(c + d*x)])*
(I*Cos[3*(c + d*x)] + Sin[3*(c + d*x)])*(a + I*a*Tan[c + d*x])^n)/(d*(-5 + n)*(-4 + n)*(-3 + n)*(e*Sec[c + d*x
])^(2*n)))
________________________________________________________________________________________
Maple [F] time = 0.758, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{6-2\,n} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*sec(d*x+c))^(6-2*n)*(a+I*a*tan(d*x+c))^n,x)
[Out]
int((e*sec(d*x+c))^(6-2*n)*(a+I*a*tan(d*x+c))^n,x)
________________________________________________________________________________________
Maxima [B] time = 12.563, size = 1434, normalized size = 9.19 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(6-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")
[Out]
-(64*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/2*n)*a^n*e^6*cos(n*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c) + 1)) + 64*I*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/
2*n)*a^n*e^6*sin(n*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 32*(a^n*e^6*n^2 - 9*a^n*e^6*n + 20*a^n*e
^6)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/2*n)*cos(4*d*x + n*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1) + 4*c) - 64*(a^n*e^6*n - 5*a^n*e^6)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)^(1/2*n)*cos(2*d*x + n*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) + 2*c) + (32*I*a
^n*e^6*n^2 - 288*I*a^n*e^6*n + 640*I*a^n*e^6)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/2*n)*sin(4*d*x + n*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) + 4*c) + (-64*I*a^n*e^6*n + 320*I*a^n
*e^6)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/2*n)*sin(2*d*x + n*arctan2(sin(2*d
*x + 2*c), cos(2*d*x + 2*c) + 1) + 2*c))/(((-I*e^(2*n)*n^3 + 12*I*e^(2*n)*n^2 - 47*I*e^(2*n)*n + 60*I*e^(2*n))
*2^n*cos(10*d*x + 10*c) + (-5*I*e^(2*n)*n^3 + 60*I*e^(2*n)*n^2 - 235*I*e^(2*n)*n + 300*I*e^(2*n))*2^n*cos(8*d*
x + 8*c) + (-10*I*e^(2*n)*n^3 + 120*I*e^(2*n)*n^2 - 470*I*e^(2*n)*n + 600*I*e^(2*n))*2^n*cos(6*d*x + 6*c) + (-
10*I*e^(2*n)*n^3 + 120*I*e^(2*n)*n^2 - 470*I*e^(2*n)*n + 600*I*e^(2*n))*2^n*cos(4*d*x + 4*c) + (-5*I*e^(2*n)*n
^3 + 60*I*e^(2*n)*n^2 - 235*I*e^(2*n)*n + 300*I*e^(2*n))*2^n*cos(2*d*x + 2*c) + (e^(2*n)*n^3 - 12*e^(2*n)*n^2
+ 47*e^(2*n)*n - 60*e^(2*n))*2^n*sin(10*d*x + 10*c) + 5*(e^(2*n)*n^3 - 12*e^(2*n)*n^2 + 47*e^(2*n)*n - 60*e^(2
*n))*2^n*sin(8*d*x + 8*c) + 10*(e^(2*n)*n^3 - 12*e^(2*n)*n^2 + 47*e^(2*n)*n - 60*e^(2*n))*2^n*sin(6*d*x + 6*c)
+ 10*(e^(2*n)*n^3 - 12*e^(2*n)*n^2 + 47*e^(2*n)*n - 60*e^(2*n))*2^n*sin(4*d*x + 4*c) + 5*(e^(2*n)*n^3 - 12*e^
(2*n)*n^2 + 47*e^(2*n)*n - 60*e^(2*n))*2^n*sin(2*d*x + 2*c) + (-I*e^(2*n)*n^3 + 12*I*e^(2*n)*n^2 - 47*I*e^(2*n
)*n + 60*I*e^(2*n))*2^n)*d)
________________________________________________________________________________________
Fricas [A] time = 2.07946, size = 419, normalized size = 2.69 \begin{align*} \frac{{\left ({\left (-i \, n^{2} + 9 i \, n - 20 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-i \, n^{2} + 11 i \, n - 30 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (2 i \, n - 12 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} \left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \left (\frac{2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-2 \, n + 6} e^{\left (-6 i \, d x - 6 i \, c\right )}}{2 \,{\left (d n^{3} - 12 \, d n^{2} + 47 \, d n - 60 \, d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(6-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")
[Out]
1/2*((-I*n^2 + 9*I*n - 20*I)*e^(6*I*d*x + 6*I*c) + (-I*n^2 + 11*I*n - 30*I)*e^(4*I*d*x + 4*I*c) + (2*I*n - 12*
I)*e^(2*I*d*x + 2*I*c) - 2*I)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*(2*e*e^(I*d*x + I*c)/(e^(2
*I*d*x + 2*I*c) + 1))^(-2*n + 6)*e^(-6*I*d*x - 6*I*c)/(d*n^3 - 12*d*n^2 + 47*d*n - 60*d)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))**(6-2*n)*(a+I*a*tan(d*x+c))**n,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 6}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(6-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")
[Out]
integrate((e*sec(d*x + c))^(-2*n + 6)*(I*a*tan(d*x + c) + a)^n, x)