3.483 \(\int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n \, dx\)

Optimal. Leaf size=269 \[ \frac{24 i (a+i a \tan (c+d x))^{n+3} (e \sec (c+d x))^{-n-4}}{a^3 d (4-n) n \left (4-n^2\right )}-\frac{24 i (a+i a \tan (c+d x))^{n+4} (e \sec (c+d x))^{-n-4}}{a^4 d n \left (n^4-20 n^2+64\right )}-\frac{12 i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-4}}{a^2 d (2-n) (4-n) n}+\frac{4 i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-4}}{a d \left (n^2-6 n+8\right )}+\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-4}}{d (4-n)} \]

[Out]

(I*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^n)/(d*(4 - n)) + ((4*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a
*Tan[c + d*x])^(1 + n))/(a*d*(8 - 6*n + n^2)) - ((12*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(2 +
n))/(a^2*d*(2 - n)*(4 - n)*n) + ((24*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(3 + n))/(a^3*d*(4 -
n)*n*(4 - n^2)) - ((24*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(4 + n))/(a^4*d*n*(64 - 20*n^2 + n^
4))

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Rubi [A]  time = 0.406169, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3504, 3488} \[ \frac{24 i (a+i a \tan (c+d x))^{n+3} (e \sec (c+d x))^{-n-4}}{a^3 d (4-n) n \left (4-n^2\right )}-\frac{24 i (a+i a \tan (c+d x))^{n+4} (e \sec (c+d x))^{-n-4}}{a^4 d n \left (n^4-20 n^2+64\right )}-\frac{12 i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-4}}{a^2 d (2-n) (4-n) n}+\frac{4 i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-4}}{a d \left (n^2-6 n+8\right )}+\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-4}}{d (4-n)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

(I*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^n)/(d*(4 - n)) + ((4*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a
*Tan[c + d*x])^(1 + n))/(a*d*(8 - 6*n + n^2)) - ((12*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(2 +
n))/(a^2*d*(2 - n)*(4 - n)*n) + ((24*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(3 + n))/(a^3*d*(4 -
n)*n*(4 - n^2)) - ((24*I)*(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^(4 + n))/(a^4*d*n*(64 - 20*n^2 + n^
4))

Rule 3504

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), 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] && IL
tQ[Simplify[m + n], 0] && NeQ[m + 2*n, 0]

Rule 3488

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)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rubi steps

\begin{align*} \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n \, dx &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n} \, dx}{a (4-n)}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}+\frac{12 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n} \, dx}{a^2 (2-n) (4-n)}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac{12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}-\frac{24 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n} \, dx}{a^3 (2-n) (4-n) n}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac{12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}+\frac{24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d (2-n) (4-n) n (2+n)}+\frac{24 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{4+n} \, dx}{a^4 (2-n) (4-n) n (2+n)}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac{12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}+\frac{24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d (2-n) (4-n) n (2+n)}-\frac{24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{4+n}}{a^4 d n \left (64-20 n^2+n^4\right )}\\ \end{align*}

Mathematica [A]  time = 0.622421, size = 165, normalized size = 0.61 \[ -\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n} \left (-8 i n^3 \sin (2 (c+d x))-4 i n^3 \sin (4 (c+d x))+4 \left (n^2-16\right ) n^2 \cos (2 (c+d x))+\left (n^2-4\right ) n^2 \cos (4 (c+d x))+128 i n \sin (2 (c+d x))+16 i n \sin (4 (c+d x))+3 n^4-60 n^2+192\right )}{8 d e^4 (n-4) (n-2) n (n+2) (n+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(-4 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((-I/8)*(192 - 60*n^2 + 3*n^4 + 4*n^2*(-16 + n^2)*Cos[2*(c + d*x)] + n^2*(-4 + n^2)*Cos[4*(c + d*x)] + (128*I)
*n*Sin[2*(c + d*x)] - (8*I)*n^3*Sin[2*(c + d*x)] + (16*I)*n*Sin[4*(c + d*x)] - (4*I)*n^3*Sin[4*(c + d*x)])*(a
+ I*a*Tan[c + d*x])^n)/(d*e^4*(-4 + n)*(-2 + n)*n*(2 + n)*(4 + n)*(e*Sec[c + d*x])^n)

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Maple [C]  time = 1.361, size = 5866, normalized size = 21.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n,x)

[Out]

result too large to display

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Maxima [A]  time = 2.0563, size = 583, normalized size = 2.17 \begin{align*} \frac{{\left (-i \, a^{n} n^{4} + 4 i \, a^{n} n^{3} + 4 i \, a^{n} n^{2} - 16 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n + 4\right )}\right ) +{\left (-4 i \, a^{n} n^{4} + 8 i \, a^{n} n^{3} + 64 i \, a^{n} n^{2} - 128 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n + 2\right )}\right ) +{\left (-4 i \, a^{n} n^{4} - 8 i \, a^{n} n^{3} + 64 i \, a^{n} n^{2} + 128 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n - 2\right )}\right ) +{\left (-i \, a^{n} n^{4} - 4 i \, a^{n} n^{3} + 4 i \, a^{n} n^{2} + 16 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n - 4\right )}\right ) +{\left (-6 i \, a^{n} n^{4} + 120 i \, a^{n} n^{2} - 384 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} n\right ) +{\left (a^{n} n^{4} - 4 \, a^{n} n^{3} - 4 \, a^{n} n^{2} + 16 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n + 4\right )}\right ) + 4 \,{\left (a^{n} n^{4} - 2 \, a^{n} n^{3} - 16 \, a^{n} n^{2} + 32 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n + 2\right )}\right ) + 4 \,{\left (a^{n} n^{4} + 2 \, a^{n} n^{3} - 16 \, a^{n} n^{2} - 32 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n - 2\right )}\right ) +{\left (a^{n} n^{4} + 4 \, a^{n} n^{3} - 4 \, a^{n} n^{2} - 16 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n - 4\right )}\right ) + 6 \,{\left (a^{n} n^{4} - 20 \, a^{n} n^{2} + 64 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} n\right )}{16 \,{\left (e^{n + 4} n^{5} - 20 \, e^{n + 4} n^{3} + 64 \, e^{n + 4} n\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

1/16*((-I*a^n*n^4 + 4*I*a^n*n^3 + 4*I*a^n*n^2 - 16*I*a^n*n)*cos((d*x + c)*(n + 4)) + (-4*I*a^n*n^4 + 8*I*a^n*n
^3 + 64*I*a^n*n^2 - 128*I*a^n*n)*cos((d*x + c)*(n + 2)) + (-4*I*a^n*n^4 - 8*I*a^n*n^3 + 64*I*a^n*n^2 + 128*I*a
^n*n)*cos((d*x + c)*(n - 2)) + (-I*a^n*n^4 - 4*I*a^n*n^3 + 4*I*a^n*n^2 + 16*I*a^n*n)*cos((d*x + c)*(n - 4)) +
(-6*I*a^n*n^4 + 120*I*a^n*n^2 - 384*I*a^n)*cos((d*x + c)*n) + (a^n*n^4 - 4*a^n*n^3 - 4*a^n*n^2 + 16*a^n*n)*sin
((d*x + c)*(n + 4)) + 4*(a^n*n^4 - 2*a^n*n^3 - 16*a^n*n^2 + 32*a^n*n)*sin((d*x + c)*(n + 2)) + 4*(a^n*n^4 + 2*
a^n*n^3 - 16*a^n*n^2 - 32*a^n*n)*sin((d*x + c)*(n - 2)) + (a^n*n^4 + 4*a^n*n^3 - 4*a^n*n^2 - 16*a^n*n)*sin((d*
x + c)*(n - 4)) + 6*(a^n*n^4 - 20*a^n*n^2 + 64*a^n)*sin((d*x + c)*n))/((e^(n + 4)*n^5 - 20*e^(n + 4)*n^3 + 64*
e^(n + 4)*n)*d)

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Fricas [A]  time = 2.20184, size = 844, normalized size = 3.14 \begin{align*} \frac{{\left (-i \, n^{4} - 4 i \, n^{3} + 4 i \, n^{2} +{\left (-i \, n^{4} + 4 i \, n^{3} + 4 i \, n^{2} - 16 i \, n\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (-4 i \, n^{4} + 8 i \, n^{3} + 64 i \, n^{2} - 128 i \, n\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-6 i \, n^{4} + 120 i \, n^{2} - 384 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-4 i \, n^{4} - 8 i \, n^{3} + 64 i \, n^{2} + 128 i \, n\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, n\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 )^{-n - 4}}{d n^{5} - 20 \, d n^{3} + 64 \, d n +{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \,{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \,{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \,{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (2 i \, d x + 2 i \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(-4-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

(-I*n^4 - 4*I*n^3 + 4*I*n^2 + (-I*n^4 + 4*I*n^3 + 4*I*n^2 - 16*I*n)*e^(8*I*d*x + 8*I*c) + (-4*I*n^4 + 8*I*n^3
+ 64*I*n^2 - 128*I*n)*e^(6*I*d*x + 6*I*c) + (-6*I*n^4 + 120*I*n^2 - 384*I)*e^(4*I*d*x + 4*I*c) + (-4*I*n^4 - 8
*I*n^3 + 64*I*n^2 + 128*I*n)*e^(2*I*d*x + 2*I*c) + 16*I*n)*(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))^(-n - 4)/(d*n^5 - 20*d*n^3 + 64*d*n + (d*n^5 - 20*d*n^3 + 6
4*d*n)*e^(8*I*d*x + 8*I*c) + 4*(d*n^5 - 20*d*n^3 + 64*d*n)*e^(6*I*d*x + 6*I*c) + 6*(d*n^5 - 20*d*n^3 + 64*d*n)
*e^(4*I*d*x + 4*I*c) + 4*(d*n^5 - 20*d*n^3 + 64*d*n)*e^(2*I*d*x + 2*I*c))

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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))**(-4-n)*(a+I*a*tan(d*x+c))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \sec \left (d x + c\right )\right )^{-n - 4}{\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))^(-4-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(-n - 4)*(I*a*tan(d*x + c) + a)^n, x)