\(\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [370]

Optimal result

Integrand size = 26, antiderivative size = 110 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {64 i a^3 \sec ^{11}(c+d x)}{2145 d (a+i a \tan (c+d x))^{11/2}}+\frac {16 i a^2 \sec ^{11}(c+d x)}{195 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}} \]

[Out]

64/2145*I*a^3*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(11/2)+16/195*I*a^2*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(9/2)+
2/15*I*a*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(7/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {64 i a^3 \sec ^{11}(c+d x)}{2145 d (a+i a \tan (c+d x))^{11/2}}+\frac {16 i a^2 \sec ^{11}(c+d x)}{195 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}} \]

[In]

Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((64*I)/2145)*a^3*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((16*I)/195)*a^2*Sec[c + d*x]^11)/(d*
(a + I*a*Tan[c + d*x])^(9/2)) + (((2*I)/15)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(7/2))

Rule 3574

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(
d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3575

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}+\frac {1}{15} (8 a) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx \\ & = \frac {16 i a^2 \sec ^{11}(c+d x)}{195 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}+\frac {1}{195} \left (32 a^2\right ) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx \\ & = \frac {64 i a^3 \sec ^{11}(c+d x)}{2145 d (a+i a \tan (c+d x))^{11/2}}+\frac {16 i a^2 \sec ^{11}(c+d x)}{195 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.74 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^{10}(c+d x) (60+203 \cos (2 (c+d x))+187 i \sin (2 (c+d x))) (-2 i \cos (3 (c+d x))-2 \sin (3 (c+d x)))}{2145 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(Sec[c + d*x]^10*(60 + 203*Cos[2*(c + d*x)] + (187*I)*Sin[2*(c + d*x)])*((-2*I)*Cos[3*(c + d*x)] - 2*Sin[3*(c
+ d*x)]))/(2145*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

Maple [F(-1)]

Timed out.

\[\int \frac {\sec ^{11}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]

[In]

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {256 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-195 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 60 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i\right )}}{2145 \, {\left (a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-256/2145*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-195*I*e^(4*I*d*x + 4*I*c) - 60*I*e^(2*I*d*x + 2*I*c) - 8
*I)/(a^3*d*e^(14*I*d*x + 14*I*c) + 7*a^3*d*e^(12*I*d*x + 12*I*c) + 21*a^3*d*e^(10*I*d*x + 10*I*c) + 35*a^3*d*e
^(8*I*d*x + 8*I*c) + 35*a^3*d*e^(6*I*d*x + 6*I*c) + 21*a^3*d*e^(4*I*d*x + 4*I*c) + 7*a^3*d*e^(2*I*d*x + 2*I*c)
+ a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**11/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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. 764 vs. \(2 (86) = 172\).

Time = 0.48 (sec) , antiderivative size = 764, normalized size of antiderivative = 6.95 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/2145*(-263*I*sqrt(a) - 830*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 760*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 - 4270*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1085*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 - 11576*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2000*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6
- 23000*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 2470*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 33
540*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 33540*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 2470*I
*sqrt(a)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 23000*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 2000*I*
sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 11576*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 + 1085*I*s
qrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 4270*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17 + 760*I*sqrt
(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 830*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 + 263*I*sqrt(a)*
sin(d*x + c)^20/(cos(d*x + c) + 1)^20)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(sin(d*x + c)/(cos(d*x + c)
+ 1) - 1)^(5/2)/((a^3 - 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)
^4 - 120*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 210*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 252*a^3*sin(d
*x + c)^10/(cos(d*x + c) + 1)^10 + 210*a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 120*a^3*sin(d*x + c)^14/(co
s(d*x + c) + 1)^14 + 45*a^3*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 10*a^3*sin(d*x + c)^18/(cos(d*x + c) + 1)^
18 + a^3*sin(d*x + c)^20/(cos(d*x + c) + 1)^20)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 - 1)^(5/2))

Giac [F]

\[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{11}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^11/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 9.81 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {256\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,60{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,195{}\mathrm {i}+8{}\mathrm {i}\right )}{2145\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7} \]

[In]

int(1/(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

(256*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*(exp(c*2i +
d*x*2i)*60i + exp(c*4i + d*x*4i)*195i + 8i))/(2145*a^3*d*(exp(c*2i + d*x*2i) + 1)^7)