\(\int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [136]

Optimal result

Integrand size = 24, antiderivative size = 27 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i}{2 a d (a+i a \tan (c+d x))^2} \]

[Out]

1/2*I/a/d/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, 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 = {3568, 32} \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i}{2 a d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(I/2)/(a*d*(a + I*a*Tan[c + d*x])^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int \frac {1}{(a+x)^3} \, dx,x,i a \tan (c+d x)\right )}{a d} \\ & = \frac {i}{2 a d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i}{2 a^3 d (-i+\tan (c+d x))^2} \]

[In]

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

[Out]

(-1/2*I)/(a^3*d*(-I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {i}{2 a d \left (a +i a \tan \left (d x +c \right )\right )^{2}}\)	\(24\)
default	\(\frac {i}{2 a d \left (a +i a \tan \left (d x +c \right )\right )^{2}}\)	\(24\)
risch	\(\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{3} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{8 a^{3} d}\)	\(38\)

[In]

int(sec(d*x+c)^2/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/2*I/a/d/(a+I*a*tan(d*x+c))^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{8 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^2/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/8*(2*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*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. 153 vs. \(2 (19) = 38\).

Time = 0.82 (sec) , antiderivative size = 153, normalized size of antiderivative = 5.67 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} - \frac {i \tan {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{8 a^{3} d \tan ^{3}{\left (c + d x \right )} - 24 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 24 a^{3} d \tan {\left (c + d x \right )} + 8 i a^{3} d} - \frac {3 \sec ^{2}{\left (c + d x \right )}}{8 a^{3} d \tan ^{3}{\left (c + d x \right )} - 24 i a^{3} d \tan ^{2}{\left (c + d x \right )} - 24 a^{3} d \tan {\left (c + d x \right )} + 8 i a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{2}{\left (c \right )}}{\left (i a \tan {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(sec(d*x+c)**2/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise((-I*tan(c + d*x)*sec(c + d*x)**2/(8*a**3*d*tan(c + d*x)**3 - 24*I*a**3*d*tan(c + d*x)**2 - 24*a**3*d
*tan(c + d*x) + 8*I*a**3*d) - 3*sec(c + d*x)**2/(8*a**3*d*tan(c + d*x)**3 - 24*I*a**3*d*tan(c + d*x)**2 - 24*a
**3*d*tan(c + d*x) + 8*I*a**3*d), Ne(d, 0)), (x*sec(c)**2/(I*a*tan(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i}{2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a d} \]

[In]

integrate(sec(d*x+c)^2/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*I/((I*a*tan(d*x + c) + a)^2*a*d)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).

Time = 0.55 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{4}} \]

[In]

integrate(sec(d*x+c)^2/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {1{}\mathrm {i}}{2\,a^3\,d\,{\left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}^2} \]

[In]

int(1/(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

-1i/(2*a^3*d*(tan(c + d*x) - 1i)^2)