\(\int (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n \, dx\) [487]

Optimal result

Integrand size = 28, antiderivative size = 37 \[ \int (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n \, dx=-\frac {i (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {3569} \[ \int (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n \, dx=-\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n}}{d n} \]

[In]

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

[Out]

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {i (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n \, dx=-\frac {i (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d n} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 5.52 (sec) , antiderivative size = 842, normalized size of antiderivative = 22.76

method	result	size
risch	\(\text {Expression too large to display}\)	\(842\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (33) = 66\).

Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.27 \[ \int (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n \, dx=-\frac {i \, e^{\left (i \, d n x + i \, c n + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right )\right )}}{d n \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.86 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.32 \[ \int (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n \, dx=\begin {cases} x & \text {for}\: d = 0 \wedge n = 0 \\x \left (e \sec {\left (c \right )}\right )^{- n} \left (i a \tan {\left (c \right )} + a\right )^{n} & \text {for}\: d = 0 \\x & \text {for}\: n = 0 \\- \frac {i \left (e \sec {\left (c + d x \right )}\right )^{- n} \left (i a \tan {\left (c + d x \right )} + a\right )^{n}}{d n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x, Eq(d, 0) & Eq(n, 0)), (x*(I*a*tan(c) + a)**n/(e*sec(c))**n, Eq(d, 0)), (x, Eq(n, 0)), (-I*(I*a*t
an(c + d*x) + a)**n/(d*n*(e*sec(c + d*x))**n), True))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (33) = 66\).

Time = 0.33 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.32 \[ \int (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n \, dx=-\frac {i \, a^{n} e^{\left (n \log \left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right ) - n \log \left (-\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )\right )}}{d e^{n} n} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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