\(\int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [347]

Optimal result

Integrand size = 26, antiderivative size = 117 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {16 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d}+\frac {24 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d}-\frac {4 i (a+i a \tan (c+d x))^{9/2}}{3 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{11/2}}{11 a^7 d} \] Output:

-16/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^4/d+24/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^5 
/d-4/3*I*(a+I*a*tan(d*x+c))^(9/2)/a^6/d+2/11*I*(a+I*a*tan(d*x+c))^(11/2)/a 
^7/d
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2 (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)} \left (-533 i-755 \tan (c+d x)+455 i \tan ^2(c+d x)+105 \tan ^3(c+d x)\right )}{1155 a^2 d} \] Input:

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

(-2*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]]*(-533*I - 755*Tan[c + 
 d*x] + (455*I)*Tan[c + d*x]^2 + 105*Tan[c + d*x]^3))/(1155*a^2*d)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3042, 3968, 53, 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.

\(\displaystyle \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^8}{(a+i a \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{3/2}d(i a \tan (c+d x))}{a^7 d}\)

\(\Big \downarrow \) 53

\(\displaystyle -\frac {i \int \left (-(i \tan (c+d x) a+a)^{9/2}+6 a (i \tan (c+d x) a+a)^{7/2}-12 a^2 (i \tan (c+d x) a+a)^{5/2}+8 a^3 (i \tan (c+d x) a+a)^{3/2}\right )d(i a \tan (c+d x))}{a^7 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \left (\frac {16}{5} a^3 (a+i a \tan (c+d x))^{5/2}-\frac {24}{7} a^2 (a+i a \tan (c+d x))^{7/2}-\frac {2}{11} (a+i a \tan (c+d x))^{11/2}+\frac {4}{3} a (a+i a \tan (c+d x))^{9/2}\right )}{a^7 d}\)

Input:

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-I)*((16*a^3*(a + I*a*Tan[c + d*x])^(5/2))/5 - (24*a^2*(a + I*a*Tan[c + 
d*x])^(7/2))/7 + (4*a*(a + I*a*Tan[c + d*x])^(9/2))/3 - (2*(a + I*a*Tan[c 
+ d*x])^(11/2))/11))/(a^7*d)
 

Defintions of rubi rules used

rule 53

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] && 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 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 = 1.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}+\frac {12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{7}}\)	\(82\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}+\frac {12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{7}}\)	\(82\)

Input:

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2*I/d/a^7*(1/11*(a+I*a*tan(d*x+c))^(11/2)-2/3*a*(a+I*a*tan(d*x+c))^(9/2)+1 
2/7*a^2*(a+I*a*tan(d*x+c))^(7/2)-8/5*a^3*(a+I*a*tan(d*x+c))^(5/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {64 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (16 i \, e^{\left (11 i \, d x + 11 i \, c\right )} + 88 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 198 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 231 i \, e^{\left (5 i \, d x + 5 i \, c\right )}\right )}}{1155 \, {\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \] Input:

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-64/1155*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(16*I*e^(11*I*d*x + 11* 
I*c) + 88*I*e^(9*I*d*x + 9*I*c) + 198*I*e^(7*I*d*x + 7*I*c) + 231*I*e^(5*I 
*d*x + 5*I*c))/(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]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 770 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 1980 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 1848 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3}\right )}}{1155 \, a^{7} d} \] Input:

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

2/1155*I*(105*(I*a*tan(d*x + c) + a)^(11/2) - 770*(I*a*tan(d*x + c) + a)^( 
9/2)*a + 1980*(I*a*tan(d*x + c) + a)^(7/2)*a^2 - 1848*(I*a*tan(d*x + c) + 
a)^(5/2)*a^3)/(a^7*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 5.13 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.16 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{1155\,a^2\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{1155\,a^2\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{385\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{231\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{33\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{11\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5} \] Input:

int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^(3/2)),x)
 

Output:

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)* 
256i)/(33*a^2*d*(exp(c*2i + d*x*2i) + 1)^4) - ((a - (a*(exp(c*2i + d*x*2i) 
*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(1155*a^2*d*(exp(c*2i 
+ d*x*2i) + 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d* 
x*2i) + 1))^(1/2)*128i)/(385*a^2*d*(exp(c*2i + d*x*2i) + 1)^2) - ((a - (a* 
(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(231 
*a^2*d*(exp(c*2i + d*x*2i) + 1)^3) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i) 
*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*1024i)/(1155*a^2*d) - ((a - (a*(exp(c 
*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(11*a^2*d* 
(exp(c*2i + d*x*2i) + 1)^5)
 

Reduce [F]

\[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8} i -13 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d -13 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d +11 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d i +11 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d i \right )}{a^{2} d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(sqrt(a)*( - 2*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*i - 13*int((sqrt(t 
an(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x)**2)/(tan(c + d*x)**3*i + t 
an(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d - 13*int((sqrt(t 
an(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x)**2)/(tan(c + d*x)**3*i + t 
an(c + d*x)**2 + tan(c + d*x)*i + 1),x)*d + 11*int((sqrt(tan(c + d*x)*i + 
1)*sec(c + d*x)**8*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + ta 
n(c + d*x)*i + 1),x)*tan(c + d*x)**2*d*i + 11*int((sqrt(tan(c + d*x)*i + 1 
)*sec(c + d*x)**8*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan 
(c + d*x)*i + 1),x)*d*i))/(a**2*d*(tan(c + d*x)**2 + 1))