\(\int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [333]

Optimal result

Integrand size = 26, antiderivative size = 117 \[ \int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {16 i (a+i a \tan (c+d x))^{7/2}}{7 a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{9/2}}{3 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{11/2}}{11 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^7 d} \] Output:

-16/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^4/d+8/3*I*(a+I*a*tan(d*x+c))^(9/2)/a^5/ 
d-12/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^6/d+2/13*I*(a+I*a*tan(d*x+c))^(13/2) 
/a^7/d
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 (-i+\tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)} \left (-835+1421 i \tan (c+d x)+945 \tan ^2(c+d x)-231 i \tan ^3(c+d x)\right )}{3003 a d} \] Input:

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

Output:

(2*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]]*(-835 + (1421*I)*Tan[c 
 + d*x] + 945*Tan[c + d*x]^2 - (231*I)*Tan[c + d*x]^3))/(3003*a*d)
 

Rubi [A] (verified)

Time = 0.27 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{5/2}d(i a \tan (c+d x))}{a^7 d}\)

\(\Big \downarrow \) 53

\(\displaystyle -\frac {i \int \left (-(i \tan (c+d x) a+a)^{11/2}+6 a (i \tan (c+d x) a+a)^{9/2}-12 a^2 (i \tan (c+d x) a+a)^{7/2}+8 a^3 (i \tan (c+d x) a+a)^{5/2}\right )d(i a \tan (c+d x))}{a^7 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \left (\frac {16}{7} a^3 (a+i a \tan (c+d x))^{7/2}-\frac {8}{3} a^2 (a+i a \tan (c+d x))^{9/2}-\frac {2}{13} (a+i a \tan (c+d x))^{13/2}+\frac {12}{11} a (a+i a \tan (c+d x))^{11/2}\right )}{a^7 d}\)

Input:

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

Output:

((-I)*((16*a^3*(a + I*a*Tan[c + d*x])^(7/2))/7 - (8*a^2*(a + I*a*Tan[c + d 
*x])^(9/2))/3 + (12*a*(a + I*a*Tan[c + d*x])^(11/2))/11 - (2*(a + I*a*Tan[ 
c + d*x])^(13/2))/13))/(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 = 1.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}\right )}{d \,a^{7}}\)	\(82\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}+\frac {4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}\right )}{d \,a^{7}}\)	\(82\)

Input:

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2*I/d/a^7*(1/13*(a+I*a*tan(d*x+c))^(13/2)-6/11*a*(a+I*a*tan(d*x+c))^(11/2) 
+4/3*a^2*(a+I*a*tan(d*x+c))^(9/2)-8/7*a^3*(a+I*a*tan(d*x+c))^(7/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.28 \[ \int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (16 i \, e^{\left (13 i \, d x + 13 i \, c\right )} + 104 i \, e^{\left (11 i \, d x + 11 i \, c\right )} + 286 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 429 i \, e^{\left (7 i \, d x + 7 i \, c\right )}\right )}}{3003 \, {\left (a d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \] Input:

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

-128/3003*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(16*I*e^(13*I*d*x + 13 
*I*c) + 104*I*e^(11*I*d*x + 11*I*c) + 286*I*e^(9*I*d*x + 9*I*c) + 429*I*e^ 
(7*I*d*x + 7*I*c))/(a*d*e^(12*I*d*x + 12*I*c) + 6*a*d*e^(10*I*d*x + 10*I*c 
) + 15*a*d*e^(8*I*d*x + 8*I*c) + 20*a*d*e^(6*I*d*x + 6*I*c) + 15*a*d*e^(4* 
I*d*x + 4*I*c) + 6*a*d*e^(2*I*d*x + 2*I*c) + a*d)
 

Sympy [F]

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

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

Output:

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

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. 297 vs. \(2 (85) = 170\).

Time = 0.04 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.54 \[ \int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 i \, {\left (15015 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} - \frac {3003 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{a^{2}} + \frac {143 \, {\left (35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 180 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 378 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}\right )}}{a^{4}} - \frac {5 \, {\left (231 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 1638 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{6}\right )}}{a^{6}}\right )}}{15015 \, a d} \] Input:

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

-2/15015*I*(15015*sqrt(I*a*tan(d*x + c) + a) - 3003*(3*(I*a*tan(d*x + c) + 
 a)^(5/2) - 10*(I*a*tan(d*x + c) + a)^(3/2)*a + 15*sqrt(I*a*tan(d*x + c) + 
 a)*a^2)/a^2 + 143*(35*(I*a*tan(d*x + c) + a)^(9/2) - 180*(I*a*tan(d*x + c 
) + a)^(7/2)*a + 378*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 420*(I*a*tan(d*x + 
 c) + a)^(3/2)*a^3 + 315*sqrt(I*a*tan(d*x + c) + a)*a^4)/a^4 - 5*(231*(I*a 
*tan(d*x + c) + a)^(13/2) - 1638*(I*a*tan(d*x + c) + a)^(11/2)*a + 5005*(I 
*a*tan(d*x + c) + a)^(9/2)*a^2 - 8580*(I*a*tan(d*x + c) + a)^(7/2)*a^3 + 9 
009*(I*a*tan(d*x + c) + a)^(5/2)*a^4 - 6006*(I*a*tan(d*x + c) + a)^(3/2)*a 
^5 + 3003*sqrt(I*a*tan(d*x + c) + a)*a^6)/a^6)/(a*d)
 

Giac [F(-2)]

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

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(1/2),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 = 6.25 (sec) , antiderivative size = 434, normalized size of antiderivative = 3.71 \[ \int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, 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}}\,2048{}\mathrm {i}}{3003\,a\,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}}\,1024{}\mathrm {i}}{3003\,a\,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}}\,256{}\mathrm {i}}{1001\,a\,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}}\,640{}\mathrm {i}}{3003\,a\,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}}\,6784{}\mathrm {i}}{429\,a\,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}}\,3456{}\mathrm {i}}{143\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}+\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}}{13\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6} \] Input:

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

Output:

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)* 
6784i)/(429*a*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)*1024i)/(3003*a*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)*256i)/(1001*a*d*(exp(c*2i + d*x*2i) + 1)^2) - ((a - (a*(e 
xp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*640i)/(3003 
*a*d*(exp(c*2i + d*x*2i) + 1)^3) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1 
i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*2048i)/(3003*a*d) - ((a - (a*(exp(c*2i 
+ d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*3456i)/(143*a*d*(ex 
p(c*2i + d*x*2i) + 1)^5) + ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp( 
c*2i + d*x*2i) + 1))^(1/2)*128i)/(13*a*d*(exp(c*2i + d*x*2i) + 1)^6)
 

Reduce [F]

\[ \int \frac {\sec ^8(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sqrt {a}\, i \left (-\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8}+6 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) \tan \left (d x +c \right )^{2} d +6 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{8} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) d \right )}{a d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(2*sqrt(a)*i*( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8 + 6*int((sqrt(ta 
n(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x))/(tan(c + d*x)**2 + 1),x)*t 
an(c + d*x)**2*d + 6*int((sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + 
 d*x))/(tan(c + d*x)**2 + 1),x)*d))/(a*d*(tan(c + d*x)**2 + 1))