\(\int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [115]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i (a-i a \tan (c+d x))^4}{2 a^6 d}-\frac {i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

[Out]

1/2*I*(a-I*a*tan(d*x+c))^4/a^6/d-1/5*I*(a-I*a*tan(d*x+c))^5/a^7/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i (a-i a \tan (c+d x))^4}{2 a^6 d}-\frac {i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((I/2)*(a - I*a*Tan[c + d*x])^4)/(a^6*d) - ((I/5)*(a - I*a*Tan[c + d*x])^5)/(a^7*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a (a-x)^3-(a-x)^4\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = \frac {i (a-i a \tan (c+d x))^4}{2 a^6 d}-\frac {i (a-i a \tan (c+d x))^5}{5 a^7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\tan (c+d x) \left (-10+10 i \tan (c+d x)+5 i \tan ^3(c+d x)+2 \tan ^4(c+d x)\right )}{10 a^2 d} \]

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-1/10*(Tan[c + d*x]*(-10 + (10*I)*Tan[c + d*x] + (5*I)*Tan[c + d*x]^3 + 2*Tan[c + d*x]^4))/(a^2*d)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {8 i \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{5 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(36\)
derivativedivides	\(\frac {\tan \left (d x +c \right )-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}-i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}\)	\(47\)
default	\(\frac {\tan \left (d x +c \right )-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}-i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}\)	\(47\)

[In]

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

8/5*I*(5*exp(2*I*(d*x+c))+1)/d/a^2/(exp(2*I*(d*x+c))+1)^5

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. 97 vs. \(2 (43) = 86\).

Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {8 \, {\left (-5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{5 \, {\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-8/5*(-5*I*e^(2*I*d*x + 2*I*c) - I)/(a^2*d*e^(10*I*d*x + 10*I*c) + 5*a^2*d*e^(8*I*d*x + 8*I*c) + 10*a^2*d*e^(6
*I*d*x + 6*I*c) + 10*a^2*d*e^(4*I*d*x + 4*I*c) + 5*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

Sympy [F]

\[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {\sec ^{8}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

[In]

integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**2,x)

[Out]

-Integral(sec(c + d*x)**8/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.73 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\sin \left (c+d\,x\right )\,\left (-10\,{\cos \left (c+d\,x\right )}^4+{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,10{}\mathrm {i}+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^3\,5{}\mathrm {i}+2\,{\sin \left (c+d\,x\right )}^4\right )}{10\,a^2\,d\,{\cos \left (c+d\,x\right )}^5} \]

[In]

int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

-(sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^3*5i + cos(c + d*x)^3*sin(c + d*x)*10i - 10*cos(c + d*x)^4 + 2*sin(c
+ d*x)^4))/(10*a^2*d*cos(c + d*x)^5)