Integrand size = 26, antiderivative size = 110 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {64 i a^3 \sec ^{11}(c+d x)}{2145 d (a+i a \tan (c+d x))^{11/2}}+\frac {16 i a^2 \sec ^{11}(c+d x)}{195 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}} \] Output:
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)
Time = 1.68 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^{10}(c+d x) (60+203 \cos (2 (c+d x))+187 i \sin (2 (c+d x))) (-2 i \cos (3 (c+d x))-2 \sin (3 (c+d x)))}{2145 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(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]])
Time = 0.55 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3975, 3042, 3975, 3042, 3974}
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.
\(\displaystyle \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^{11}}{(a+i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{15} a \int \frac {\sec ^{11}(c+d x)}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{15} a \int \frac {\sec (c+d x)^{11}}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{15} a \left (\frac {4}{13} a \int \frac {\sec ^{11}(c+d x)}{(i \tan (c+d x) a+a)^{9/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{15} a \left (\frac {4}{13} a \int \frac {\sec (c+d x)^{11}}{(i \tan (c+d x) a+a)^{9/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {8}{15} a \left (\frac {8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\) |
Input:
Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(((2*I)/15)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (8*a*((( (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))))/15
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]
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] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[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] && !Inte gerQ[n]
Time = 6.64 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{7} \left (128 \cos \left (d x +c \right )^{4}+144 \cos \left (d x +c \right )^{2}+143\right )}{2145}+\frac {2 i \left (128 \sec \left (d x +c \right )^{3}+80 \sec \left (d x +c \right )^{5}+55 \sec \left (d x +c \right )^{7}\right )}{2145}}{d \left (2 i \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 \cos \left (d x +c \right )^{2}-1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}\) | \(125\) |
Input:
int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/2145/d/(2*I*sin(d*x+c)*cos(d*x+c)+2*cos(d*x+c)^2-1)/(a*(1+I*tan(d*x+c))) ^(1/2)/a^2*(tan(d*x+c)*sec(d*x+c)^7*(128*cos(d*x+c)^4+144*cos(d*x+c)^2+143 )+I*(128*sec(d*x+c)^3+80*sec(d*x+c)^5+55*sec(d*x+c)^7))
Time = 0.14 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {256 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-195 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 60 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i\right )}}{2145 \, {\left (a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \] Input:
integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-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)
Timed out. \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**11/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Timed out
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 = 0.39 (sec) , antiderivative size = 764, normalized size of antiderivative = 6.95 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
-2/2145*(-263*I*sqrt(a) - 830*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 76 0*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)*s in(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 - 33540* sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 33540*sqrt(a)*sin(d*x + c)^1 1/(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/(c os(d*x + c) + 1)^15 + 1085*I*sqrt(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/(cos(d*x + c) + 1)^14 + 45*a^3*sin(d*x + c)^16/(co...
Exception generated. \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),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
Time = 6.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {256\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,60{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,195{}\mathrm {i}+8{}\mathrm {i}\right )}{2145\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7} \] Input:
int(1/(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(5/2)),x)
Output:
(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)
\[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\sec \left (d x +c \right )^{11}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:
int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - int(sec(c + d*x)**11/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqr t(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt (a)*a**2)