\(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [168]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {8 x}{a^8}-\frac {8 i \log (\cos (c+d x))}{a^8 d}+\frac {\tan (c+d x)}{a^8 d}+\frac {16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac {24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )} \] Output:

-8*x/a^8-8*I*ln(cos(d*x+c))/a^8/d+tan(d*x+c)/a^8/d+16/3*I/a^5/d/(a+I*a*tan 
(d*x+c))^3-16*I/d/(a^4+I*a^4*tan(d*x+c))^2+24*I/d/(a^8+I*a^8*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {i \left (-8 a \log (i-\tan (c+d x))+i a \tan (c+d x)+\frac {8 a \left (-5 i+12 \tan (c+d x)+9 i \tan ^2(c+d x)\right )}{3 (-i+\tan (c+d x))^3}\right )}{a^9 d} \] Input:

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^8,x]
 

Output:

((-I)*(-8*a*Log[I - Tan[c + d*x]] + I*a*Tan[c + d*x] + (8*a*(-5*I + 12*Tan 
[c + d*x] + (9*I)*Tan[c + d*x]^2))/(3*(-I + Tan[c + d*x])^3)))/(a^9*d)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{10}}{(a+i a \tan (c+d x))^8}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i \int \frac {(a-i a \tan (c+d x))^4}{(i \tan (c+d x) a+a)^4}d(i a \tan (c+d x))}{a^9 d}\)

\(\Big \downarrow \) 49

\(\displaystyle -\frac {i \int \left (\frac {16 a^4}{(i \tan (c+d x) a+a)^4}-\frac {32 a^3}{(i \tan (c+d x) a+a)^3}+\frac {24 a^2}{(i \tan (c+d x) a+a)^2}-\frac {8 a}{i \tan (c+d x) a+a}+1\right )d(i a \tan (c+d x))}{a^9 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \left (-\frac {16 a^4}{3 (a+i a \tan (c+d x))^3}+\frac {16 a^3}{(a+i a \tan (c+d x))^2}-\frac {24 a^2}{a+i a \tan (c+d x)}+i a \tan (c+d x)-8 a \log (a+i a \tan (c+d x))\right )}{a^9 d}\)

Input:

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^8,x]
 

Output:

((-I)*(-8*a*Log[a + I*a*Tan[c + d*x]] + I*a*Tan[c + d*x] - (16*a^4)/(3*(a 
+ I*a*Tan[c + d*x])^3) + (16*a^3)/(a + I*a*Tan[c + d*x])^2 - (24*a^2)/(a + 
 I*a*Tan[c + d*x])))/(a^9*d)
 

Defintions of rubi rules used

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[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]
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.59

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )+\frac {24}{-i+\tan \left (d x +c \right )}+8 i \ln \left (-i+\tan \left (d x +c \right )\right )-\frac {16}{3 \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {16 i}{\left (-i+\tan \left (d x +c \right )\right )^{2}}}{d \,a^{8}}\)	\(68\)
default	\(\frac {\tan \left (d x +c \right )+\frac {24}{-i+\tan \left (d x +c \right )}+8 i \ln \left (-i+\tan \left (d x +c \right )\right )-\frac {16}{3 \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {16 i}{\left (-i+\tan \left (d x +c \right )\right )^{2}}}{d \,a^{8}}\)	\(68\)
risch	\(\frac {6 i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{8} d}-\frac {2 i {\mathrm e}^{-4 i \left (d x +c \right )}}{a^{8} d}+\frac {2 i {\mathrm e}^{-6 i \left (d x +c \right )}}{3 a^{8} d}-\frac {16 x}{a^{8}}-\frac {16 c}{a^{8} d}+\frac {2 i}{d \,a^{8} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {8 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{8} d}\)	\(114\)

Input:

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)
 

Output:

1/d/a^8*(tan(d*x+c)+24/(-I+tan(d*x+c))+8*I*ln(-I+tan(d*x+c))-16/3/(-I+tan( 
d*x+c))^3+16*I/(-I+tan(d*x+c))^2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {2 \, {\left (24 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 12 \, {\left (2 \, d x - i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 12 \, {\left (i \, e^{\left (8 i \, d x + 8 i \, c\right )} + i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{3 \, {\left (a^{8} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{8} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \] Input:

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")
 

Output:

-2/3*(24*d*x*e^(8*I*d*x + 8*I*c) + 12*(2*d*x - I)*e^(6*I*d*x + 6*I*c) + 12 
*(I*e^(8*I*d*x + 8*I*c) + I*e^(6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2*I*c) + 
 1) - 6*I*e^(4*I*d*x + 4*I*c) + 2*I*e^(2*I*d*x + 2*I*c) - I)/(a^8*d*e^(8*I 
*d*x + 8*I*c) + a^8*d*e^(6*I*d*x + 6*I*c))
 

Sympy [F]

\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \] Input:

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**8,x)
 

Output:

Integral(sec(c + d*x)**10/(tan(c + d*x)**8 - 8*I*tan(c + d*x)**7 - 28*tan( 
c + d*x)**6 + 56*I*tan(c + d*x)**5 + 70*tan(c + d*x)**4 - 56*I*tan(c + d*x 
)**3 - 28*tan(c + d*x)**2 + 8*I*tan(c + d*x) + 1), x)/a**8
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.65 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\frac {8 \, {\left (9 \, \tan \left (d x + c\right )^{6} - 48 i \, \tan \left (d x + c\right )^{5} - 107 \, \tan \left (d x + c\right )^{4} + 128 i \, \tan \left (d x + c\right )^{3} + 87 \, \tan \left (d x + c\right )^{2} - 32 i \, \tan \left (d x + c\right ) - 5\right )}}{a^{8} \tan \left (d x + c\right )^{7} - 7 i \, a^{8} \tan \left (d x + c\right )^{6} - 21 \, a^{8} \tan \left (d x + c\right )^{5} + 35 i \, a^{8} \tan \left (d x + c\right )^{4} + 35 \, a^{8} \tan \left (d x + c\right )^{3} - 21 i \, a^{8} \tan \left (d x + c\right )^{2} - 7 \, a^{8} \tan \left (d x + c\right ) + i \, a^{8}} + \frac {24 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{8}} + \frac {3 \, \tan \left (d x + c\right )}{a^{8}}}{3 \, d} \] Input:

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")
 

Output:

1/3*(8*(9*tan(d*x + c)^6 - 48*I*tan(d*x + c)^5 - 107*tan(d*x + c)^4 + 128* 
I*tan(d*x + c)^3 + 87*tan(d*x + c)^2 - 32*I*tan(d*x + c) - 5)/(a^8*tan(d*x 
 + c)^7 - 7*I*a^8*tan(d*x + c)^6 - 21*a^8*tan(d*x + c)^5 + 35*I*a^8*tan(d* 
x + c)^4 + 35*a^8*tan(d*x + c)^3 - 21*I*a^8*tan(d*x + c)^2 - 7*a^8*tan(d*x 
 + c) + I*a^8) + 24*I*log(I*tan(d*x + c) + 1)/a^8 + 3*tan(d*x + c)/a^8)/d
 

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.59 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {8 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{8} d} + \frac {\tan \left (d x + c\right )}{a^{8} d} + \frac {8 \, {\left (9 \, \tan \left (d x + c\right )^{2} - 12 i \, \tan \left (d x + c\right ) - 5\right )}}{3 \, a^{8} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \] Input:

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")
 

Output:

8*I*log(tan(d*x + c) - I)/(a^8*d) + tan(d*x + c)/(a^8*d) + 8/3*(9*tan(d*x 
+ c)^2 - 12*I*tan(d*x + c) - 5)/(a^8*d*(tan(d*x + c) - I)^3)
 

Mupad [B] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{a^8\,d}-\frac {\frac {32\,\mathrm {tan}\left (c+d\,x\right )}{a^8}-\frac {40{}\mathrm {i}}{3\,a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,24{}\mathrm {i}}{a^8}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,8{}\mathrm {i}}{a^8\,d} \] Input:

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^8),x)
 

Output:

(log(tan(c + d*x) - 1i)*8i)/(a^8*d) - ((32*tan(c + d*x))/a^8 - 40i/(3*a^8) 
 + (tan(c + d*x)^2*24i)/a^8)/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan( 
c + d*x)^3*1i + 1)) + tan(c + d*x)/(a^8*d)
 

Reduce [F]

\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\text {too large to display} \] Input:

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x)
 

Output:

(13653760*cos(c + d*x)*int(cos(c + d*x)/(128*cos(c + d*x)*sin(c + d*x)**9* 
i - 320*cos(c + d*x)*sin(c + d*x)**7*i + 272*cos(c + d*x)*sin(c + d*x)**5* 
i - 88*cos(c + d*x)*sin(c + d*x)**3*i + 8*cos(c + d*x)*sin(c + d*x)*i - 12 
8*sin(c + d*x)**10 + 384*sin(c + d*x)**8 - 416*sin(c + d*x)**6 + 192*sin(c 
 + d*x)**4 - 33*sin(c + d*x)**2 + 1),x)*d - 13655040*cos(c + d*x)*int(cos( 
c + d*x)/(128*cos(c + d*x)*sin(c + d*x)**9 - 320*cos(c + d*x)*sin(c + d*x) 
**7 + 272*cos(c + d*x)*sin(c + d*x)**5 - 88*cos(c + d*x)*sin(c + d*x)**3 + 
 8*cos(c + d*x)*sin(c + d*x) + 128*sin(c + d*x)**10*i - 384*sin(c + d*x)** 
8*i + 416*sin(c + d*x)**6*i - 192*sin(c + d*x)**4*i + 33*sin(c + d*x)**2*i 
 - i),x)*d*i - 161280*cos(c + d*x)*int(cos(c + d*x)/(128*cos(c + d*x)*sin( 
c + d*x)**8*i - 256*cos(c + d*x)*sin(c + d*x)**6*i + 160*cos(c + d*x)*sin( 
c + d*x)**4*i - 32*cos(c + d*x)*sin(c + d*x)**2*i + cos(c + d*x)*i - 128*s 
in(c + d*x)**9 + 320*sin(c + d*x)**7 - 272*sin(c + d*x)**5 + 88*sin(c + d* 
x)**3 - 8*sin(c + d*x)),x)*d*i - 774144000*cos(c + d*x)*int(sin(c + d*x)** 
10/(128*cos(c + d*x)*sin(c + d*x)**9*i - 320*cos(c + d*x)*sin(c + d*x)**7* 
i + 272*cos(c + d*x)*sin(c + d*x)**5*i - 88*cos(c + d*x)*sin(c + d*x)**3*i 
 + 8*cos(c + d*x)*sin(c + d*x)*i - 128*sin(c + d*x)**10 + 384*sin(c + d*x) 
**8 - 416*sin(c + d*x)**6 + 192*sin(c + d*x)**4 - 33*sin(c + d*x)**2 + 1), 
x)*d + 774144000*cos(c + d*x)*int(sin(c + d*x)**10/(128*cos(c + d*x)*sin(c 
 + d*x)**9 - 320*cos(c + d*x)*sin(c + d*x)**7 + 272*cos(c + d*x)*sin(c ...