\(\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx\) [490]
Optimal result
Integrand size = 30, antiderivative size = 121 \[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {3+n}{2}} a \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-n),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{3-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^{-1+n}}{d (3-n)}
\]
[Out]
I*2^(3/2+1/2*n)*a*hypergeom([3/2-1/2*n, -1/2-1/2*n],[5/2-1/2*n],1/2-1/2*I*tan(d*x+c))*(e*sec(d*x+c))^(3-n)*(1+
I*tan(d*x+c))^(-1/2-1/2*n)*(a+I*a*tan(d*x+c))^(-1+n)/d/(3-n)
Rubi [A] (verified)
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3586, 3604, 72, 71}
\[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\frac {i a 2^{\frac {n+3}{2}} (1+i \tan (c+d x))^{\frac {1}{2} (-n-1)} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{3-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-n-1),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d (3-n)}
\]
[In]
Int[(e*Sec[c + d*x])^(3 - n)*(a + I*a*Tan[c + d*x])^n,x]
[Out]
(I*2^((3 + n)/2)*a*Hypergeometric2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x]
)^(3 - n)*(1 + I*Tan[c + d*x])^((-1 - n)/2)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(3 - n))
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}& = \left ((e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-3+n)}\right ) \int (a-i a \tan (c+d x))^{\frac {3-n}{2}} (a+i a \tan (c+d x))^{\frac {3-n}{2}+n} \, dx \\ & = \frac {\left (a^2 (e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-3+n)}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {3-n}{2}} (a+i a x)^{-1+\frac {3-n}{2}+n} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{\frac {1}{2}+\frac {n}{2}} a^2 (e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac {1}{2}+\frac {1}{2} (-3+n)+\frac {n}{2}} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-\frac {n}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-1+\frac {3-n}{2}+n} (a-i a x)^{-1+\frac {3-n}{2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {i 2^{\frac {3+n}{2}} a \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-n),\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{3-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^{-1+n}}{d (3-n)} \\
\end{align*}
Mathematica [A] (verified)
Time = 17.56 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96
\[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\frac {8 e^3 \operatorname {Hypergeometric2F1}\left (3,\frac {3-n}{2},\frac {5-n}{2},-\cos (2 (c+d x))+i \sin (2 (c+d x))\right ) \sec (d x) (e \sec (c+d x))^{-n} (i+\tan (d x)) (a+i a \tan (c+d x))^n}{d (-3+n) (\cos (c)+i \sin (c))^3 (-i+\tan (d x))^2}
\]
[In]
Integrate[(e*Sec[c + d*x])^(3 - n)*(a + I*a*Tan[c + d*x])^n,x]
[Out]
(8*e^3*Hypergeometric2F1[3, (3 - n)/2, (5 - n)/2, -Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]]*Sec[d*x]*(I + Tan[d*
x])*(a + I*a*Tan[c + d*x])^n)/(d*(-3 + n)*(e*Sec[c + d*x])^n*(Cos[c] + I*Sin[c])^3*(-I + Tan[d*x])^2)
Maple [F]
\[\int \left (e \sec \left (d x +c \right )\right )^{3-n} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
[In]
int((e*sec(d*x+c))^(3-n)*(a+I*a*tan(d*x+c))^n,x)
[Out]
int((e*sec(d*x+c))^(3-n)*(a+I*a*tan(d*x+c))^n,x)
Fricas [F]
\[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-n + 3} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x }
\]
[In]
integrate((e*sec(d*x+c))^(3-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")
[Out]
1/8*(((-I*n - I)*e^(4*I*d*x + 4*I*c) - 2*I*n*e^(2*I*d*x + 2*I*c) - I*n + I)*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x +
2*I*c) + 1))^(-n + 3)*e^(I*d*n*x + I*c*n + n*log(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1)) + n*log(a/e))
+ 8*d*e^(2*I*d*x + 2*I*c)*integral(-1/8*(n^2 + (n^2 - 1)*e^(4*I*d*x + 4*I*c) + 2*(n^2 - 1)*e^(2*I*d*x + 2*I*c
) - 1)*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^(-n + 3)*e^(I*d*n*x + I*c*n - 2*I*d*x + n*log(2*e*e^(I*
d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1)) + n*log(a/e) - 2*I*c), x))*e^(-2*I*d*x - 2*I*c)/d
Sympy [F]
\[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{3 - n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx
\]
[In]
integrate((e*sec(d*x+c))**(3-n)*(a+I*a*tan(d*x+c))**n,x)
[Out]
Integral((e*sec(c + d*x))**(3 - n)*(I*a*(tan(c + d*x) - I))**n, x)
Maxima [F]
\[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-n + 3} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x }
\]
[In]
integrate((e*sec(d*x+c))^(3-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")
[Out]
8*(6*a^n*e^3*cos(c*n + (d*n + d)*x + c) + 6*I*a^n*e^3*sin(c*n + (d*n + d)*x + c) - (a^n*e^3*n - 5*a^n*e^3)*cos
(c*n + (d*n + 3*d)*x + 3*c) - 6*((I*a^n*d*e^(n + 3)*n^3 - 7*I*a^n*d*e^(n + 3)*n^2 + 7*I*a^n*d*e^(n + 3)*n + 15
*I*a^n*d*e^(n + 3))*cos(c*n) + ((I*a^n*d*e^(n + 3)*n^3 - 7*I*a^n*d*e^(n + 3)*n^2 + 7*I*a^n*d*e^(n + 3)*n + 15*
I*a^n*d*e^(n + 3))*cos(c*n) - (a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^
(n + 3))*sin(c*n))*cos(6*d*x + 6*c) + 3*((I*a^n*d*e^(n + 3)*n^3 - 7*I*a^n*d*e^(n + 3)*n^2 + 7*I*a^n*d*e^(n + 3
)*n + 15*I*a^n*d*e^(n + 3))*cos(c*n) - (a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15
*a^n*d*e^(n + 3))*sin(c*n))*cos(4*d*x + 4*c) + 3*((I*a^n*d*e^(n + 3)*n^3 - 7*I*a^n*d*e^(n + 3)*n^2 + 7*I*a^n*d
*e^(n + 3)*n + 15*I*a^n*d*e^(n + 3))*cos(c*n) - (a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n +
3)*n + 15*a^n*d*e^(n + 3))*sin(c*n))*cos(2*d*x + 2*c) - (a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d
*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*sin(c*n) - ((a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n + 3
)*n + 15*a^n*d*e^(n + 3))*cos(c*n) - (-I*a^n*d*e^(n + 3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7*I*a^n*d*e^(n + 3)*n
- 15*I*a^n*d*e^(n + 3))*sin(c*n))*sin(6*d*x + 6*c) - 3*((a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*
d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*cos(c*n) - (-I*a^n*d*e^(n + 3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7*I*a^n*d*e
^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*sin(c*n))*sin(4*d*x + 4*c) - 3*((a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^
2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*cos(c*n) - (-I*a^n*d*e^(n + 3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7
*I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*sin(c*n))*sin(2*d*x + 2*c))*integrate(((cos(8*d*x + 8*c) + 4*cos(
6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos((d*n + d)*x + c) + (sin(8*d*x + 8*c) + 4*sin(6
*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*sin((d*n + d)*x + c))/(e^n*n^2 + (e^n*n^2 - 8*e^n*n + 1
5*e^n)*cos(8*d*x + 8*c)^2 + 16*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(6*d*x + 6*c)^2 + 36*(e^n*n^2 - 8*e^n*n + 15*e^
n)*cos(4*d*x + 4*c)^2 + 16*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(2*d*x + 2*c)^2 + (e^n*n^2 - 8*e^n*n + 15*e^n)*sin(
8*d*x + 8*c)^2 + 16*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(6*d*x + 6*c)^2 + 36*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(4*d*
x + 4*c)^2 + 48*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*(e^n*n^2 - 8*e^n*n + 15*e^
n)*sin(2*d*x + 2*c)^2 - 8*e^n*n + 2*(e^n*n^2 - 8*e^n*n + 4*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(6*d*x + 6*c) + 6*(
e^n*n^2 - 8*e^n*n + 15*e^n)*cos(4*d*x + 4*c) + 4*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(2*d*x + 2*c) + 15*e^n)*cos(8
*d*x + 8*c) + 8*(e^n*n^2 - 8*e^n*n + 6*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(4*d*x + 4*c) + 4*(e^n*n^2 - 8*e^n*n +
15*e^n)*cos(2*d*x + 2*c) + 15*e^n)*cos(6*d*x + 6*c) + 12*(e^n*n^2 - 8*e^n*n + 4*(e^n*n^2 - 8*e^n*n + 15*e^n)*c
os(2*d*x + 2*c) + 15*e^n)*cos(4*d*x + 4*c) + 8*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(2*d*x + 2*c) + 4*(2*(e^n*n^2 -
8*e^n*n + 15*e^n)*sin(6*d*x + 6*c) + 3*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(4*d*x + 4*c) + 2*(e^n*n^2 - 8*e^n*n +
15*e^n)*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(4*d*x + 4*c) + 2*(e^n*n^2
- 8*e^n*n + 15*e^n)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 15*e^n), x) + 6*((a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n
+ 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*cos(c*n) + ((a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2
+ 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*cos(c*n) - (-I*a^n*d*e^(n + 3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7*
I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*sin(c*n))*cos(6*d*x + 6*c) + 3*((a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(
n + 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*cos(c*n) - (-I*a^n*d*e^(n + 3)*n^3 + 7*I*a^n*d*e^(n + 3
)*n^2 - 7*I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*sin(c*n))*cos(4*d*x + 4*c) + 3*((a^n*d*e^(n + 3)*n^3 - 7
*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*cos(c*n) - (-I*a^n*d*e^(n + 3)*n^3 + 7*I*a^n*
d*e^(n + 3)*n^2 - 7*I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*sin(c*n))*cos(2*d*x + 2*c) - (-I*a^n*d*e^(n +
3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7*I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*sin(c*n) - ((-I*a^n*d*e^(n +
3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7*I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*cos(c*n) + (a^n*d*e^(n + 3)*n
^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*sin(c*n))*sin(6*d*x + 6*c) - 3*((-I*a^n
*d*e^(n + 3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7*I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*cos(c*n) + (a^n*d*e
^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*sin(c*n))*sin(4*d*x + 4*c) -
3*((-I*a^n*d*e^(n + 3)*n^3 + 7*I*a^n*d*e^(n + 3)*n^2 - 7*I*a^n*d*e^(n + 3)*n - 15*I*a^n*d*e^(n + 3))*cos(c*n)
+ (a^n*d*e^(n + 3)*n^3 - 7*a^n*d*e^(n + 3)*n^2 + 7*a^n*d*e^(n + 3)*n + 15*a^n*d*e^(n + 3))*sin(c*n))*sin(2*d*x
+ 2*c))*integrate(-((sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos((d*
n + d)*x + c) - (cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*sin((d*n
+ d)*x + c))/(e^n*n^2 + (e^n*n^2 - 8*e^n*n + 15*e^n)*cos(8*d*x + 8*c)^2 + 16*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos
(6*d*x + 6*c)^2 + 36*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(4*d*x + 4*c)^2 + 16*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(2*d
*x + 2*c)^2 + (e^n*n^2 - 8*e^n*n + 15*e^n)*sin(8*d*x + 8*c)^2 + 16*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(6*d*x + 6*
c)^2 + 36*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(4*d*x + 4*c)^2 + 48*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(4*d*x + 4*c)*s
in(2*d*x + 2*c) + 16*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(2*d*x + 2*c)^2 - 8*e^n*n + 2*(e^n*n^2 - 8*e^n*n + 4*(e^n
*n^2 - 8*e^n*n + 15*e^n)*cos(6*d*x + 6*c) + 6*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(4*d*x + 4*c) + 4*(e^n*n^2 - 8*e
^n*n + 15*e^n)*cos(2*d*x + 2*c) + 15*e^n)*cos(8*d*x + 8*c) + 8*(e^n*n^2 - 8*e^n*n + 6*(e^n*n^2 - 8*e^n*n + 15*
e^n)*cos(4*d*x + 4*c) + 4*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(2*d*x + 2*c) + 15*e^n)*cos(6*d*x + 6*c) + 12*(e^n*n
^2 - 8*e^n*n + 4*(e^n*n^2 - 8*e^n*n + 15*e^n)*cos(2*d*x + 2*c) + 15*e^n)*cos(4*d*x + 4*c) + 8*(e^n*n^2 - 8*e^n
*n + 15*e^n)*cos(2*d*x + 2*c) + 4*(2*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(6*d*x + 6*c) + 3*(e^n*n^2 - 8*e^n*n + 15
*e^n)*sin(4*d*x + 4*c) + 2*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*(e^n*n^2 -
8*e^n*n + 15*e^n)*sin(4*d*x + 4*c) + 2*(e^n*n^2 - 8*e^n*n + 15*e^n)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 15*e^
n), x) - (I*a^n*e^3*n - 5*I*a^n*e^3)*sin(c*n + (d*n + 3*d)*x + 3*c))/(-I*d*e^n*n^2 + 8*I*d*e^n*n - 15*I*d*e^n
+ (-I*d*e^n*n^2 + 8*I*d*e^n*n - 15*I*d*e^n)*cos(6*d*x + 6*c) - 3*(I*d*e^n*n^2 - 8*I*d*e^n*n + 15*I*d*e^n)*cos(
4*d*x + 4*c) - 3*(I*d*e^n*n^2 - 8*I*d*e^n*n + 15*I*d*e^n)*cos(2*d*x + 2*c) + (d*e^n*n^2 - 8*d*e^n*n + 15*d*e^n
)*sin(6*d*x + 6*c) + 3*(d*e^n*n^2 - 8*d*e^n*n + 15*d*e^n)*sin(4*d*x + 4*c) + 3*(d*e^n*n^2 - 8*d*e^n*n + 15*d*e
^n)*sin(2*d*x + 2*c))
Giac [F]
\[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-n + 3} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x }
\]
[In]
integrate((e*sec(d*x+c))^(3-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")
[Out]
integrate((e*sec(d*x + c))^(-n + 3)*(I*a*tan(d*x + c) + a)^n, x)
Mupad [F(-1)]
Timed out. \[
\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3-n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x
\]
[In]
int((e/cos(c + d*x))^(3 - n)*(a + a*tan(c + d*x)*1i)^n,x)
[Out]
int((e/cos(c + d*x))^(3 - n)*(a + a*tan(c + d*x)*1i)^n, x)