∫ (e sec(c + dx))−3−n(a + ia tan(c + dx))n dx [484]

3.5.84 \(\int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx\) [484]

Optimal. Leaf size=205 \[ \frac {i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n}{d (3-n)}+\frac {3 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (3-4 n+n^2\right )}-\frac {6 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (3-n) \left (1-n^2\right )}+\frac {6 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d \left (9-10 n^2+n^4\right )} \]

[Out]

I*(e*sec(d*x+c))^(-3-n)*(a+I*a*tan(d*x+c))^n/d/(3-n)+3*I*(e*sec(d*x+c))^(-3-n)*(a+I*a*tan(d*x+c))^(1+n)/a/d/(n

^2-4*n+3)-6*I*(e*sec(d*x+c))^(-3-n)*(a+I*a*tan(d*x+c))^(2+n)/a^2/d/(3-n)/(-n^2+1)+6*I*(e*sec(d*x+c))^(-3-n)*(a

+I*a*tan(d*x+c))^(3+n)/a^3/d/(n^4-10*n^2+9)

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Rubi [A]
time = 0.23, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3585, 3569} \begin {gather*} \frac {6 i (a+i a \tan (c+d x))^{n+3} (e \sec (c+d x))^{-n-3}}{a^3 d \left (n^4-10 n^2+9\right )}-\frac {6 i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-3}}{a^2 d (3-n) \left (1-n^2\right )}+\frac {3 i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-3}}{a d \left (n^2-4 n+3\right )}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(I*(e*Sec[c + d*x])^(-3 - n)*(a + I*a*Tan[c + d*x])^n)/(d*(3 - n)) + ((3*I)*(e*Sec[c + d*x])^(-3 - n)*(a + I*a

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

))/(a^2*d*(3 - n)*(1 - n^2)) + ((6*I)*(e*Sec[c + d*x])^(-3 - n)*(a + I*a*Tan[c + d*x])^(3 + n))/(a^3*d*(9 - 10

*n^2 + n^4))

Rule 3569

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]

Rule 3585

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]

Rubi steps

\begin {align*} \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx &=\frac {i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n}{d (3-n)}+\frac {3 \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{1+n} \, dx}{a (3-n)}\\ &=\frac {i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n}{d (3-n)}+\frac {3 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (3-4 n+n^2\right )}+\frac {6 \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{2+n} \, dx}{a^2 (1-n) (3-n)}\\ &=\frac {i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n}{d (3-n)}+\frac {3 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (3-4 n+n^2\right )}-\frac {6 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (1-n) (3-n) (1+n)}-\frac {6 \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{3+n} \, dx}{a^3 (1-n) (3-n) (1+n)}\\ &=\frac {i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n}{d (3-n)}+\frac {3 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (3-4 n+n^2\right )}-\frac {6 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (1-n) (3-n) (1+n)}+\frac {6 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d \left (9-10 n^2+n^4\right )}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 119, normalized size = 0.58 \begin {gather*} \frac {(e \sec (c+d x))^{-n} \left (-3 i n \left (-9+n^2\right ) \cos (c+d x)-i n \left (-1+n^2\right ) \cos (3 (c+d x))-6 \left (-5+n^2+\left (-1+n^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right ) (a+i a \tan (c+d x))^n}{4 d e^3 (-3+n) (-1+n) (1+n) (3+n)} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(((-3*I)*n*(-9 + n^2)*Cos[c + d*x] - I*n*(-1 + n^2)*Cos[3*(c + d*x)] - 6*(-5 + n^2 + (-1 + n^2)*Cos[2*(c + d*x

)])*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^n)/(4*d*e^3*(-3 + n)*(-1 + n)*(1 + n)*(3 + n)*(e*Sec[c + d*x])^n)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.28, size = 4994, normalized size = 24.36


method	result	size

risch	\(\text {Expression too large to display}\)	\(4994\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(-3-n)*(a+I*a*tan(d*x+c))^n,x,method=_RETURNVERBOSE)

[Out]

1/8*exp(I*(d*x+c))^n*e^(-n)*a^n/e^3*exp(-1/2*I*(6*c-n*Pi*csgn(I/(exp(2*I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c)))*c

sgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+6*d*x-n*Pi*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^3+3*Pi*csgn(

I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*e)+n*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn

(I*e)+n*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+n*Pi*cs

gn(I/(exp(2*I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2+n*Pi*csgn(I*exp(I*(d*x+c)))*csgn(I*ex

p(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2-3*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))*csgn(I*e)*csgn(I*exp(I

*(d*x+c))/(exp(2*I*(d*x+c))+1))+3*Pi*csgn(I/(exp(2*I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^

2+3*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))-n*Pi*csgn(I

*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^3-3*Pi*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^3-3*Pi*csgn(I*e/(ex

p(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^3+Pi*csgn(I*a/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^3*n+Pi*csgn(I/(exp(2*I*

(d*x+c))+1)*exp(2*I*(d*x+c)))^3*n+Pi*csgn(I*exp(2*I*(d*x+c)))*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))*cs

gn(I/(exp(2*I*(d*x+c))+1))*n+3*Pi*csgn(I*exp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2+Pi*csgn

(I*exp(2*I*(d*x+c)))^3*n+Pi*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))*csgn(I*a/(exp(2*I*(d*x+c))+1)*exp(2*

I*(d*x+c)))*csgn(I*a)*n-Pi*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))*csgn(I*a/(exp(2*I*(d*x+c))+1)*exp(2*I

*(d*x+c)))^2*n-2*Pi*csgn(I*exp(2*I*(d*x+c)))^2*csgn(I*exp(I*(d*x+c)))*n-n*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp

(I*(d*x+c)))*csgn(I*e)*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))-3*Pi*csgn(I/(exp(2*I*(d*x+c))+1))*csgn(I*ex

p(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))-Pi*csgn(I*a/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^2

*csgn(I*a)*n-Pi*csgn(I*exp(2*I*(d*x+c)))*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^2*n+Pi*csgn(I*exp(2*I*(

d*x+c)))*csgn(I*exp(I*(d*x+c)))^2*n-Pi*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^2*csgn(I/(exp(2*I*(d*x+c)

)+1))*n))/(-3*I*d+I*n*d)+1/8*exp(I*(d*x+c))^n*e^(-n)*a^n/e^3*exp(1/2*I*(6*c+n*Pi*csgn(I/(exp(2*I*(d*x+c))+1))*

csgn(I*exp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+6*d*x+n*Pi*csgn(I*exp(I*(d*x+c))/(exp(2*I*(

d*x+c))+1))^3-3*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*e)-n*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1

)*exp(I*(d*x+c)))^2*csgn(I*e)-n*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*exp(I*(d*x+c))/(exp(

2*I*(d*x+c))+1))-n*Pi*csgn(I/(exp(2*I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2-n*Pi*csgn(I*e

xp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2+3*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c))

)*csgn(I*e)*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))-3*Pi*csgn(I/(exp(2*I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c)

)/(exp(2*I*(d*x+c))+1))^2-3*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*exp(I*(d*x+c))/(exp(2*I*

(d*x+c))+1))+n*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^3+3*Pi*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))

+1))^3+3*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^3-Pi*csgn(I*a/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))

^3*n-Pi*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^3*n-Pi*csgn(I*exp(2*I*(d*x+c)))*csgn(I/(exp(2*I*(d*x+c))

+1)*exp(2*I*(d*x+c)))*csgn(I/(exp(2*I*(d*x+c))+1))*n-3*Pi*csgn(I*exp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*

I*(d*x+c))+1))^2-Pi*csgn(I*exp(2*I*(d*x+c)))^3*n-Pi*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))*csgn(I*a/(ex

p(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))*csgn(I*a)*n+Pi*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))*csgn(I*a/(exp

(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^2*n+2*Pi*csgn(I*exp(2*I*(d*x+c)))^2*csgn(I*exp(I*(d*x+c)))*n+n*Pi*csgn(I*e/

(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))*csgn(I*e)*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+3*Pi*csgn(I/(exp(2*

I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+Pi*csgn(I*a/(exp(2*I*(d*x+c)

)+1)*exp(2*I*(d*x+c)))^2*csgn(I*a)*n+Pi*csgn(I*exp(2*I*(d*x+c)))*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))

^2*n-Pi*csgn(I*exp(2*I*(d*x+c)))*csgn(I*exp(I*(d*x+c)))^2*n+Pi*csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^2

*csgn(I/(exp(2*I*(d*x+c))+1))*n))/(I*n*d+3*I*d)+3/8*exp(I*(d*x+c))^n*e^(-n)*a^n/e^3*exp(-1/2*I*(2*c-n*Pi*csgn(

I/(exp(2*I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+2*d*x-n*Pi*csgn(I*e

xp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^3+3*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*e)+n*Pi*csgn

(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csgn(I*e)+n*Pi*csgn(I*e/(exp(2*I*(d*x+c))+1)*exp(I*(d*x+c)))^2*csg

n(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+n*Pi*csgn(I/(exp(2*I*(d*x+c))+1))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x

+c))+1))^2+n*Pi*csgn(I*exp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2-3*Pi*csgn(I*e/(exp(2*I*(d

*x+c))+1)*exp(I*(d*x+c)))*csgn(I*e)*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+3*Pi*csgn(I/(exp(2*I*(d*x+c))+

1))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))...

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Maxima [A]
time = 0.60, size = 341, normalized size = 1.66 \begin {gather*} \frac {{\left ({\left (-i \, a^{n} n^{3} + 3 i \, a^{n} n^{2} + i \, a^{n} n - 3 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n + 3\right )}\right ) - 3 \, {\left (i \, a^{n} n^{3} - i \, a^{n} n^{2} - 9 i \, a^{n} n + 9 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n + 1\right )}\right ) - 3 \, {\left (i \, a^{n} n^{3} + i \, a^{n} n^{2} - 9 i \, a^{n} n - 9 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n - 1\right )}\right ) + {\left (-i \, a^{n} n^{3} - 3 i \, a^{n} n^{2} + i \, a^{n} n + 3 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n - 3\right )}\right ) + {\left (a^{n} n^{3} - 3 \, a^{n} n^{2} - a^{n} n + 3 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n + 3\right )}\right ) + 3 \, {\left (a^{n} n^{3} - a^{n} n^{2} - 9 \, a^{n} n + 9 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n + 1\right )}\right ) + 3 \, {\left (a^{n} n^{3} + a^{n} n^{2} - 9 \, a^{n} n - 9 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n - 1\right )}\right ) + {\left (a^{n} n^{3} + 3 \, a^{n} n^{2} - a^{n} n - 3 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n - 3\right )}\right )\right )} e^{\left (-n\right )}}{8 \, {\left (n^{4} e^{3} - 10 \, n^{2} e^{3} + 9 \, e^{3}\right )} d} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((-I*a^n*n^3 + 3*I*a^n*n^2 + I*a^n*n - 3*I*a^n)*cos((d*x + c)*(n + 3)) - 3*(I*a^n*n^3 - I*a^n*n^2 - 9*I*a^

n*n + 9*I*a^n)*cos((d*x + c)*(n + 1)) - 3*(I*a^n*n^3 + I*a^n*n^2 - 9*I*a^n*n - 9*I*a^n)*cos((d*x + c)*(n - 1))

+ (-I*a^n*n^3 - 3*I*a^n*n^2 + I*a^n*n + 3*I*a^n)*cos((d*x + c)*(n - 3)) + (a^n*n^3 - 3*a^n*n^2 - a^n*n + 3*a^

n)*sin((d*x + c)*(n + 3)) + 3*(a^n*n^3 - a^n*n^2 - 9*a^n*n + 9*a^n)*sin((d*x + c)*(n + 1)) + 3*(a^n*n^3 + a^n*

n^2 - 9*a^n*n - 9*a^n)*sin((d*x + c)*(n - 1)) + (a^n*n^3 + 3*a^n*n^2 - a^n*n - 3*a^n)*sin((d*x + c)*(n - 3)))*

e^(-n)/((n^4*e^3 - 10*n^2*e^3 + 9*e^3)*d)

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Fricas [A]
time = 0.39, size = 264, normalized size = 1.29 \begin {gather*} \frac {{\left (-i \, n^{3} - 3 i \, n^{2} + {\left (-i \, n^{3} + 3 i \, n^{2} + i \, n - 3 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, {\left (i \, n^{3} - i \, n^{2} - 9 i \, n + 9 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (i \, n^{3} + i \, n^{2} - 9 i \, n - 9 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, n + 3 i\right )} \left (\frac {2 \, e^{\left (i \, d x + i \, c + 1\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-n - 3} e^{\left (i \, d n x + i \, c n + n \log \left (a e^{\left (-1\right )}\right ) + n \log \left (\frac {2 \, e^{\left (i \, d x + i \, c + 1\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )\right )}}{d n^{4} - 10 \, d n^{2} + {\left (d n^{4} - 10 \, d n^{2} + 9 \, d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (d n^{4} - 10 \, d n^{2} + 9 \, d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (d n^{4} - 10 \, d n^{2} + 9 \, d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 9 \, d} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*n^3 - 3*I*n^2 + (-I*n^3 + 3*I*n^2 + I*n - 3*I)*e^(6*I*d*x + 6*I*c) - 3*(I*n^3 - I*n^2 - 9*I*n + 9*I)*e^(4*

I*d*x + 4*I*c) - 3*(I*n^3 + I*n^2 - 9*I*n - 9*I)*e^(2*I*d*x + 2*I*c) + I*n + 3*I)*(2*e^(I*d*x + I*c + 1)/(e^(2

*I*d*x + 2*I*c) + 1))^(-n - 3)*e^(I*d*n*x + I*c*n + n*log(a*e^(-1)) + n*log(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x

+ 2*I*c) + 1)))/(d*n^4 - 10*d*n^2 + (d*n^4 - 10*d*n^2 + 9*d)*e^(6*I*d*x + 6*I*c) + 3*(d*n^4 - 10*d*n^2 + 9*d)*

e^(4*I*d*x + 4*I*c) + 3*(d*n^4 - 10*d*n^2 + 9*d)*e^(2*I*d*x + 2*I*c) + 9*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \sec {\left (c + d x \right )}\right )^{- n - 3} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 9.24, size = 425, normalized size = 2.07 \begin {gather*} -\frac {\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (2\,{\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2+\sin \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-n^3-3\,n^2+n+3\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}+\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (3\,c+3\,d\,x\right )}^2+\sin \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-n^3+3\,n^2+n-3\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}+\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-3\,n^3-3\,n^2+27\,n+27\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}+\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (2\,c+2\,d\,x\right )}^2+\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-3\,n^3+3\,n^2+27\,n-27\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}\right )}{8\,{\left (-\frac {e}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^{n+3}\,{\left ({\sin \left (c+d\,x\right )}^2-1\right )}^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*sin(c/2 + (d*x)/2)^2 - 1)*(sin(3*c + 3*d*x)*1i + 2*sin((3*c)/2 + (3*d*x)/2)^2 - 1)*(((a - (a*sin(c + d*x)

*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(n - 3*n^2 - n^3 + 3))/(d*(n^4*1i - n^2*10i + 9i)) + ((a - (a*sin(c + d*x

)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(6*c + 6*d*x)*1i - 2*sin(3*c + 3*d*x)^2 + 1)*(n + 3*n^2 - n^3 - 3))/

(d*(n^4*1i - n^2*10i + 9i)) + ((a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(2*c + 2*d*x)*1i -

2*sin(c + d*x)^2 + 1)*(27*n - 3*n^2 - 3*n^3 + 27))/(d*(n^4*1i - n^2*10i + 9i)) + ((a - (a*sin(c + d*x)*1i)/(2

*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(4*c + 4*d*x)*1i - 2*sin(2*c + 2*d*x)^2 + 1)*(27*n + 3*n^2 - 3*n^3 - 27))/(d

*(n^4*1i - n^2*10i + 9i))))/(8*(-e/(2*sin(c/2 + (d*x)/2)^2 - 1))^(n + 3)*(sin(c + d*x)^2 - 1)^2)

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