\(\int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [116]

Optimal result

Integrand size = 24, antiderivative size = 27 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i (a-i a \tan (c+d x))^3}{3 a^5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (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} \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i (a-i a \tan (c+d x))^3}{3 a^5 d} \]

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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 (a-x)^2 \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = \frac {i (a-i a \tan (c+d x))^3}{3 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\tan (c+d x)}{a^2 d}-\frac {i \tan ^2(c+d x)}{a^2 d}-\frac {\tan ^3(c+d x)}{3 a^2 d} \]

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(-\frac {\left (\tan \left (d x +c \right )+i\right )^{3}}{3 a^{2} d}\)	\(20\)
default	\(-\frac {\left (\tan \left (d x +c \right )+i\right )^{3}}{3 a^{2} d}\)	\(20\)
risch	\(\frac {8 i}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(23\)

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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. 54 vs. \(2 (21) = 42\).

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {8 i}{3 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

8/3*I/(a^2*d*e^(6*I*d*x + 6*I*c) + 3*a^2*d*e^(4*I*d*x + 4*I*c) + 3*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

Sympy [F]

\[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {\sec ^{6}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

[In]

integrate(sec(d*x+c)**6/(a+I*a*tan(d*x+c))**2,x)

[Out]

-Integral(sec(c + d*x)**6/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\tan \left (d x + c\right )^{3} + 3 i \, \tan \left (d x + c\right )^{2} - 3 \, \tan \left (d x + c\right )}{3 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\tan \left (d x + c\right )^{3} + 3 i \, \tan \left (d x + c\right )^{2} - 3 \, \tan \left (d x + c\right )}{3 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}-3\right )}{3\,a^2\,d} \]

[In]

int(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

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