\(\int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [144]

Optimal result

Integrand size = 22, antiderivative size = 98 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac {2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac {2 i \sec (c+d x)}{15 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

1/5*I*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^3+2/15*I*sec(d*x+c)/a/d/(a+I*a*tan(d*x+c))^2+2/15*I*sec(d*x+c)/d/(a^3+I*
a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3583, 3569} \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 i \sec (c+d x)}{15 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3} \]

[In]

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

[Out]

((I/5)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^3) + (((2*I)/15)*Sec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^2) +
(((2*I)/15)*Sec[c + d*x])/(d*(a^3 + I*a^3*Tan[c + d*x]))

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 3583

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}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac {2 \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{5 a} \\ & = \frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac {2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac {2 \int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx}{15 a^2} \\ & = \frac {i \sec (c+d x)}{5 d (a+i a \tan (c+d x))^3}+\frac {2 i \sec (c+d x)}{15 a d (a+i a \tan (c+d x))^2}+\frac {2 i \sec (c+d x)}{15 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.55 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sec ^3(c+d x) (5+9 \cos (2 (c+d x))+6 i \sin (2 (c+d x)))}{30 a^3 d (-i+\tan (c+d x))^3} \]

[In]

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

[Out]

-1/30*(Sec[c + d*x]^3*(5 + 9*Cos[2*(c + d*x)] + (6*I)*Sin[2*(c + d*x)]))/(a^3*d*(-I + Tan[c + d*x])^3)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57

method	result	size
risch	\(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{4 a^{3} d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{6 a^{3} d}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{20 a^{3} d}\)	\(56\)
derivativedivides	\(\frac {\frac {8}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {16}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}}{a^{3} d}\)	\(90\)
default	\(\frac {\frac {8}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {16}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}}{a^{3} d}\)	\(90\)

[In]

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

[Out]

1/4*I/a^3/d*exp(-I*(d*x+c))+1/6*I/a^3/d*exp(-3*I*(d*x+c))+1/20*I/a^3/d*exp(-5*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.42 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{60 \, a^{3} d} \]

[In]

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

[Out]

1/60*(15*I*e^(4*I*d*x + 4*I*c) + 10*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d*x - 5*I*c)/(a^3*d)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (82) = 164\).

Time = 0.85 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.23 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {2 \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 a^{3} d \tan ^{3}{\left (c + d x \right )} - 45 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 45 a^{3} d \tan {\left (c + d x \right )} + 15 i a^{3} d} - \frac {6 i \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}}{15 a^{3} d \tan ^{3}{\left (c + d x \right )} - 45 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 45 a^{3} d \tan {\left (c + d x \right )} + 15 i a^{3} d} - \frac {7 \sec {\left (c + d x \right )}}{15 a^{3} d \tan ^{3}{\left (c + d x \right )} - 45 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 45 a^{3} d \tan {\left (c + d x \right )} + 15 i a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sec {\left (c \right )}}{\left (i a \tan {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise((2*tan(c + d*x)**2*sec(c + d*x)/(15*a**3*d*tan(c + d*x)**3 - 45*I*a**3*d*tan(c + d*x)**2 - 45*a**3*d
*tan(c + d*x) + 15*I*a**3*d) - 6*I*tan(c + d*x)*sec(c + d*x)/(15*a**3*d*tan(c + d*x)**3 - 45*I*a**3*d*tan(c +
d*x)**2 - 45*a**3*d*tan(c + d*x) + 15*I*a**3*d) - 7*sec(c + d*x)/(15*a**3*d*tan(c + d*x)**3 - 45*I*a**3*d*tan(
c + d*x)**2 - 45*a**3*d*tan(c + d*x) + 15*I*a**3*d), Ne(d, 0)), (x*sec(c)/(I*a*tan(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 10 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 10 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )}{60 \, a^{3} d} \]

[In]

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

[Out]

1/60*(3*I*cos(5*d*x + 5*c) + 10*I*cos(3*d*x + 3*c) + 15*I*cos(d*x + c) + 3*sin(5*d*x + 5*c) + 10*sin(3*d*x + 3
*c) + 15*sin(d*x + c))/(a^3*d)

Giac [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.74 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}} \]

[In]

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 30*I*tan(1/2*d*x + 1/2*c)^3 - 40*tan(1/2*d*x + 1/2*c)^2 + 20*I*tan(1/2*d*x +
1/2*c) + 7)/(a^3*d*(tan(1/2*d*x + 1/2*c) - I)^5)

Mupad [B] (verification not implemented)

Time = 4.74 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.36 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,15{}\mathrm {i}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,40{}\mathrm {i}-20\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+7{}\mathrm {i}\right )}{15\,a^3\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+1\right )} \]

[In]

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

[Out]

(2*(30*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^2*40i - 20*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^4*15i + 7i
))/(15*a^3*d*(tan(c/2 + (d*x)/2)*5i - 10*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*10i + 5*tan(c/2 + (d*x)/2
)^4 + tan(c/2 + (d*x)/2)^5*1i + 1))