\(\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [386]
Optimal result
Integrand size = 26, antiderivative size = 73 \[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}
\]
[Out]
8/143*I*a^2*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(11/2)+2/13*I*a*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(9/2)
Rubi [A] (verified)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number
of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574}
\[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}
\]
[In]
Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(7/2),x]
[Out]
(((8*I)/143)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((2*I)/13)*a*Sec[c + d*x]^11)/(d*(a + I
*a*Tan[c + d*x])^(9/2))
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)}{13 d (a+i a \tan (c+d x))^{9/2}}+\frac {1}{13} (4 a) \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx \\ & = \frac {8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}} \\
\end{align*}
Mathematica [A] (verified)
Time = 1.66 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.12
\[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {2 i \sec ^9(c+d x) (\cos (2 (c+d x))-i \sin (2 (c+d x))) (-15 i+11 \tan (c+d x))}{143 a^3 d (-i+\tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}}
\]
[In]
Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(7/2),x]
[Out]
(((-2*I)/143)*Sec[c + d*x]^9*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(-15*I + 11*Tan[c + d*x]))/(a^3*d*(-I + T
an[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])
Maple [F(-1)]
Timed out.
\[\int \frac {\sec ^{11}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}d x\]
[In]
int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(7/2),x)
[Out]
int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(7/2),x)
Fricas [B] (verification not implemented)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf
count of optimal. 132 vs. \(2 (57) = 114\).
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.81
\[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-13 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )}}{143 \, {\left (a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}}
\]
[In]
integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
[Out]
-128/143*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-13*I*e^(2*I*d*x + 2*I*c) - 2*I)/(a^4*d*e^(12*I*d*x + 12*I
*c) + 6*a^4*d*e^(10*I*d*x + 10*I*c) + 15*a^4*d*e^(8*I*d*x + 8*I*c) + 20*a^4*d*e^(6*I*d*x + 6*I*c) + 15*a^4*d*e
^(4*I*d*x + 4*I*c) + 6*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)
Sympy [F(-1)]
Timed out. \[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out}
\]
[In]
integrate(sec(d*x+c)**11/(a+I*a*tan(d*x+c))**(7/2),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. \(2 (57) = 114\).
Time = 0.47 (sec) , antiderivative size = 764, normalized size of antiderivative = 10.47
\[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Too large to display}
\]
[In]
integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
[Out]
-2/143*(-15*I*sqrt(a) - 38*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 88*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 - 278*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 213*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- 920*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 272*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 1848*s
qrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 182*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2548*sqrt(a)*
sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 2548*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 182*I*sqrt(a)*sin(d
*x + c)^12/(cos(d*x + c) + 1)^12 - 1848*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 272*I*sqrt(a)*sin(d*x
+ c)^14/(cos(d*x + c) + 1)^14 - 920*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15 + 213*I*sqrt(a)*sin(d*x + c)
^16/(cos(d*x + c) + 1)^16 - 278*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17 + 88*I*sqrt(a)*sin(d*x + c)^18/(
cos(d*x + c) + 1)^18 - 38*sqrt(a)*sin(d*x + c)^19/(cos(d*x + c) + 1)^19 + 15*I*sqrt(a)*sin(d*x + c)^20/(cos(d*
x + c) + 1)^20)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(7/2)/((a^4
- 10*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 120*a^4*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 + 210*a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 252*a^4*sin(d*x + c)^10/(cos(d*x + c
) + 1)^10 + 210*a^4*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 120*a^4*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + 45
*a^4*sin(d*x + c)^16/(cos(d*x + c) + 1)^16 - 10*a^4*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 + a^4*sin(d*x + c)^2
0/(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)^(7
/2))
Giac [F]
\[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{11}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x }
\]
[In]
integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")
[Out]
integrate(sec(d*x + c)^11/(I*a*tan(d*x + c) + a)^(7/2), x)
Mupad [B] (verification not implemented)
Time = 7.62 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25
\[
\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {128\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,13{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{143\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}
\]
[In]
int(1/(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(7/2)),x)
[Out]
(128*exp(- c*1i - d*x*1i)*(exp(c*2i + d*x*2i)*13i + 2i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d
*x*2i) + 1))^(1/2))/(143*a^4*d*(exp(c*2i + d*x*2i) + 1)^6)