\(\int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [133]

Optimal result

Integrand size = 24, antiderivative size = 27 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i (a-i a \tan (c+d x))^4}{4 a^7 d} \]

[Out]

1/4*I*(a-I*a*tan(d*x+c))^4/a^7/d

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} \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i (a-i a \tan (c+d x))^4}{4 a^7 d} \]

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((I/4)*(a - I*a*Tan[c + d*x])^4)/(a^7*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)^3 \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = \frac {i (a-i a \tan (c+d x))^4}{4 a^7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\tan (c+d x) \left (4-6 i \tan (c+d x)-4 \tan ^2(c+d x)+i \tan ^3(c+d x)\right )}{4 a^3 d} \]

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Tan[c + d*x]*(4 - (6*I)*Tan[c + d*x] - 4*Tan[c + d*x]^2 + I*Tan[c + d*x]^3))/(4*a^3*d)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {i \left (\tan \left (d x +c \right )+i\right )^{4}}{4 a^{3} d}\)	\(21\)
default	\(\frac {i \left (\tan \left (d x +c \right )+i\right )^{4}}{4 a^{3} d}\)	\(21\)
risch	\(\frac {4 i}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}\)	\(23\)

[In]

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.56 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {4 i}{a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{8}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*Integral(sec(c + d*x)**8/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x)/a**3

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

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-i \, \tan \left (d x + c\right )^{4} + 4 \, \tan \left (d x + c\right )^{3} + 6 i \, \tan \left (d x + c\right )^{2} - 4 \, \tan \left (d x + c\right )}{4 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/4*(-I*tan(d*x + c)^4 + 4*tan(d*x + c)^3 + 6*I*tan(d*x + c)^2 - 4*tan(d*x + c))/(a^3*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. 47 vs. \(2 (21) = 42\).

Time = 0.60 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-i \, \tan \left (d x + c\right )^{4} + 4 \, \tan \left (d x + c\right )^{3} + 6 i \, \tan \left (d x + c\right )^{2} - 4 \, \tan \left (d x + c\right )}{4 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/4*(-I*tan(d*x + c)^4 + 4*tan(d*x + c)^3 + 6*I*tan(d*x + c)^2 - 4*tan(d*x + c))/(a^3*d)

Mupad [B] (verification not implemented)

Time = 4.45 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.85 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sin \left (c+d\,x\right )\,\left (-4\,{\cos \left (c+d\,x\right )}^3+{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,6{}\mathrm {i}+4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^2-{\sin \left (c+d\,x\right )}^3\,1{}\mathrm {i}\right )}{4\,a^3\,d\,{\cos \left (c+d\,x\right )}^4} \]

[In]

int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

-(sin(c + d*x)*(4*cos(c + d*x)*sin(c + d*x)^2 + cos(c + d*x)^2*sin(c + d*x)*6i - 4*cos(c + d*x)^3 - sin(c + d*
x)^3*1i))/(4*a^3*d*cos(c + d*x)^4)