\(\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [354]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

256/12155*I*a^4*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(11/2)+64/1105*I*a^3*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(9/
2)+8/85*I*a^2*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(7/2)+2/17*I*a*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((256*I)/12155)*a^4*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((64*I)/1105)*a^3*Sec[c + d*x]^11)/
(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((8*I)/85)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I)
/17)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(5/2))

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)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{17} (12 a) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx \\ & = \frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{85} \left (32 a^2\right ) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx \\ & = \frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}+\frac {\left (128 a^3\right ) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx}{1105} \\ & = \frac {256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sec ^9(c+d x) (i \cos (4 (c+d x))+\sin (4 (c+d x))) (475 i-2242 i \cos (2 (c+d x))+1089 \sec (c+d x) \sin (3 (c+d x))+374 \tan (c+d x))}{12155 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*Sec[c + d*x]^9*(I*Cos[4*(c + d*x)] + Sin[4*(c + d*x)])*(475*I - (2242*I)*Cos[2*(c + d*x)] + 1089*Sec[c + d*
x]*Sin[3*(c + d*x)] + 374*Tan[c + d*x]))/(12155*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

Maple [F(-1)]

Timed out.

\[\int \frac {\sec ^{11}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

[In]

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {512 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1105 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 510 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 136 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{12155 \, {\left (a^{2} d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a^{2} d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a^{2} d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-512/12155*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1105*I*e^(6*I*d*x + 6*I*c) - 510*I*e^(4*I*d*x + 4*I*c)
- 136*I*e^(2*I*d*x + 2*I*c) - 16*I)/(a^2*d*e^(16*I*d*x + 16*I*c) + 8*a^2*d*e^(14*I*d*x + 14*I*c) + 28*a^2*d*e^
(12*I*d*x + 12*I*c) + 56*a^2*d*e^(10*I*d*x + 10*I*c) + 70*a^2*d*e^(8*I*d*x + 8*I*c) + 56*a^2*d*e^(6*I*d*x + 6*
I*c) + 28*a^2*d*e^(4*I*d*x + 4*I*c) + 8*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**11/(a+I*a*tan(d*x+c))**(3/2),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. \(2 (115) = 230\).

Time = 0.48 (sec) , antiderivative size = 764, normalized size of antiderivative = 5.20 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/12155*(-1767*I*sqrt(a) - 6854*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 2088*I*sqrt(a)*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2 - 16438*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5661*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 - 56984*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 13328*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 - 129336*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7514*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 - 156468*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 156468*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11
- 7514*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 129336*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13
+ 13328*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 56984*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15
+ 5661*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 16438*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17
- 2088*I*sqrt(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 6854*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 +
1767*I*sqrt(a)*sin(d*x + c)^20/(cos(d*x + c) + 1)^20)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)^(3/2)/((a^2 - 10*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*a^2*sin(d*x + c)^4/(co
s(d*x + c) + 1)^4 - 120*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 210*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8
- 252*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 210*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 120*a^2*sin(
d*x + c)^14/(cos(d*x + c) + 1)^14 + 45*a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 10*a^2*sin(d*x + c)^18/(cos
(d*x + c) + 1)^18 + a^2*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)^(3/2))

Giac [F]

\[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{11}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^11/(I*a*tan(d*x + c) + a)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 11.26 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{11\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1536{}\mathrm {i}}{13\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{5\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{17\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8} \]

[In]

int(1/(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

(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)*512i)/(11*a^2*d
*(exp(c*2i + d*x*2i) + 1)^5) - (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)*1536i)/(13*a^2*d*(exp(c*2i + d*x*2i) + 1)^6) + (exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2
i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(5*a^2*d*(exp(c*2i + d*x*2i) + 1)^7) - (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)*512i)/(17*a^2*d*(exp(c*2i + d*x*
2i) + 1)^8)