\(\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [161]

Optimal result

Integrand size = 24, antiderivative size = 68 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}+\frac {i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3} \]

[Out]

1/5*I*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^4+1/15*I*sec(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^3

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3583, 3569} \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3}+\frac {i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4} \]

[In]

Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/5)*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^4) + ((I/15)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3)

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 ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{5 a} \\ & = \frac {i \sec ^3(c+d x)}{5 d (a+i a \tan (c+d x))^4}+\frac {i \sec ^3(c+d x)}{15 a d (a+i a \tan (c+d x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.59 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\sec ^3(c+d x) (-4 i+\tan (c+d x))}{15 a^4 d (-i+\tan (c+d x))^4} \]

[In]

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-1/15*(Sec[c + d*x]^3*(-4*I + Tan[c + d*x]))/(a^4*d*(-I + Tan[c + d*x])^4)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.56

method	result	size
risch	\(\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{6 a^{4} d}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{10 a^{4} d}\)	\(38\)
derivativedivides	\(\frac {-\frac {28}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {16}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}}{a^{4} d}\)	\(90\)
default	\(\frac {-\frac {28}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {8 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2}{-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {16}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}}{a^{4} d}\)	\(90\)

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/6*I/a^4/d*exp(-3*I*(d*x+c))+1/10*I/a^4/d*exp(-5*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.44 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{30 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/30*(5*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d*x - 5*I*c)/(a^4*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. 182 vs. \(2 (54) = 108\).

Time = 1.15 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.68 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} - \frac {\tan {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{15 a^{4} d \tan ^{4}{\left (c + d x \right )} - 60 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 90 a^{4} d \tan ^{2}{\left (c + d x \right )} + 60 i a^{4} d \tan {\left (c + d x \right )} + 15 a^{4} d} + \frac {4 i \sec ^{3}{\left (c + d x \right )}}{15 a^{4} d \tan ^{4}{\left (c + d x \right )} - 60 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 90 a^{4} d \tan ^{2}{\left (c + d x \right )} + 60 i a^{4} d \tan {\left (c + d x \right )} + 15 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{3}{\left (c \right )}}{\left (i a \tan {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise((-tan(c + d*x)*sec(c + d*x)**3/(15*a**4*d*tan(c + d*x)**4 - 60*I*a**4*d*tan(c + d*x)**3 - 90*a**4*d*
tan(c + d*x)**2 + 60*I*a**4*d*tan(c + d*x) + 15*a**4*d) + 4*I*sec(c + d*x)**3/(15*a**4*d*tan(c + d*x)**4 - 60*
I*a**4*d*tan(c + d*x)**3 - 90*a**4*d*tan(c + d*x)**2 + 60*I*a**4*d*tan(c + d*x) + 15*a**4*d), Ne(d, 0)), (x*se
c(c)**3/(I*a*tan(c) + a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 5 i \, \cos \left (3 \, d x + 3 \, c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 5 \, \sin \left (3 \, d x + 3 \, c\right )}{30 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/30*(3*I*cos(5*d*x + 5*c) + 5*I*cos(3*d*x + 3*c) + 3*sin(5*d*x + 5*c) + 5*sin(3*d*x + 3*c))/(a^4*d)

Giac [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 15 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{15 \, a^{4} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 15*I*tan(1/2*d*x + 1/2*c)^3 - 25*tan(1/2*d*x + 1/2*c)^2 + 5*I*tan(1/2*d*x +
1/2*c) + 4)/(a^4*d*(tan(1/2*d*x + 1/2*c) - I)^5)

Mupad [B] (verification not implemented)

Time = 4.41 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.96 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,15{}\mathrm {i}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,25{}\mathrm {i}-5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4{}\mathrm {i}\right )}{15\,a^4\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+1\right )} \]

[In]

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

(2*(15*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^2*25i - 5*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^4*15i + 4i)
)/(15*a^4*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))