Integrand size = 26, antiderivative size = 110 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {64 i a^3 \sec ^3(c+d x)}{105 d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i a^2 \sec ^3(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d} \]
16/35*I*a^2*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)+2/7*I*a*sec(d*x+c)^3*( a+I*a*tan(d*x+c))^(1/2)/d+64/105*I*a^3*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^( 3/2)
Time = 0.78 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 a \sec ^3(c+d x) (\cos (d x)-i \sin (d x)) (28+43 \cos (2 (c+d x))+27 i \sin (2 (c+d x))) (i \cos (2 c+d x)+\sin (2 c+d x)) \sqrt {a+i a \tan (c+d x)}}{105 d} \]
(2*a*Sec[c + d*x]^3*(Cos[d*x] - I*Sin[d*x])*(28 + 43*Cos[2*(c + d*x)] + (2 7*I)*Sin[2*(c + d*x)])*(I*Cos[2*c + d*x] + Sin[2*c + d*x])*Sqrt[a + I*a*Ta n[c + d*x]])/(105*d)
Time = 0.50 (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.
\(\displaystyle \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^3 (a+i a \tan (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{7} a \int \sec ^3(c+d x) \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{7} a \int \sec (c+d x)^3 \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{7} a \left (\frac {4}{5} a \int \frac {\sec ^3(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx+\frac {2 i a \sec ^3(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{7} a \left (\frac {4}{5} a \int \frac {\sec (c+d x)^3}{\sqrt {i \tan (c+d x) a+a}}dx+\frac {2 i a \sec ^3(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {8}{7} a \left (\frac {8 i a^2 \sec ^3(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^3(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
(((2*I)/7)*a*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (8*a*((((8*I)/ 15)*a^2*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((2*I)/5)*a*Se c[c + d*x]^3)/(d*Sqrt[a + I*a*Tan[c + d*x]])))/7
3.4.1.3.1 Defintions of rubi rules used
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 = 7.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {2 a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (128 i \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )+128 i \left (\cos ^{3}\left (d x +c \right )\right )+16 i \sin \left (d x +c \right ) \tan \left (d x +c \right )-48 i \cos \left (d x +c \right )+64 \sin \left (d x +c \right )+15 i \left (\tan ^{2}\left (d x +c \right )\right ) \sec \left (d x +c \right )-9 i \sec \left (d x +c \right )+24 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{105 d}\) | \(123\) |
2/105/d*a*(a*(1+I*tan(d*x+c)))^(1/2)*(128*I*cos(d*x+c)*sin(d*x+c)^2+128*I* cos(d*x+c)^3+16*I*sin(d*x+c)*tan(d*x+c)-48*I*cos(d*x+c)+64*sin(d*x+c)+15*I *tan(d*x+c)^2*sec(d*x+c)-9*I*sec(d*x+c)+24*sec(d*x+c)*tan(d*x+c))
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.81 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {16 \, \sqrt {2} {\left (-35 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 28 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-16/105*sqrt(2)*(-35*I*a*e^(4*I*d*x + 4*I*c) - 28*I*a*e^(2*I*d*x + 2*I*c) - 8*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^ (4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (86) = 172\).
Time = 0.61 (sec) , antiderivative size = 580, normalized size of antiderivative = 5.27 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {16 \, {\left (35 i \, \sqrt {2} a \cos \left (4 \, d x + 4 \, c\right ) + 28 i \, \sqrt {2} a \cos \left (2 \, d x + 2 \, c\right ) - 35 \, \sqrt {2} a \sin \left (4 \, d x + 4 \, c\right ) - 28 \, \sqrt {2} a \sin \left (2 \, d x + 2 \, c\right ) + 8 i \, \sqrt {2} a\right )} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sqrt {a}}{105 \, {\left ({\left (2 \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 5 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (2 i \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \cos \left (4 \, d x + 4 \, c\right ) + 5 i \, \cos \left (2 \, d x + 2 \, c\right )^{2} - {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + 4 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} d} \]
16/105*(35*I*sqrt(2)*a*cos(4*d*x + 4*c) + 28*I*sqrt(2)*a*cos(2*d*x + 2*c) - 35*sqrt(2)*a*sin(4*d*x + 4*c) - 28*sqrt(2)*a*sin(2*d*x + 2*c) + 8*I*sqrt (2)*a)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^ (1/4)*sqrt(a)/(((2*cos(2*d*x + 2*c)^3 + (2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 2*I*sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c )^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 5*cos(2*d*x + 2*c)^2 + (I *cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin (4*d*x + 4*c) + 2*(I*cos(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(2* d*x + 2*c) + 4*cos(2*d*x + 2*c) + 1)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c) + 1)) + (2*I*cos(2*d*x + 2*c)^3 + (2*I*cos(2*d*x + 2*c) + I) *sin(2*d*x + 2*c)^2 - 2*sin(2*d*x + 2*c)^3 + (I*cos(2*d*x + 2*c)^2 + I*sin (2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(4*d*x + 4*c) + 5*I*cos(2*d *x + 2*c)^2 - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c ) + 1)*sin(4*d*x + 4*c) - 2*(cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)* sin(2*d*x + 2*c) + 4*I*cos(2*d*x + 2*c) + I)*sin(3/2*arctan2(sin(2*d*x + 2 *c), cos(2*d*x + 2*c) + 1)))*d)
\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
Time = 6.53 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {16\,a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,28{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,35{}\mathrm {i}+8{}\mathrm {i}\right )}{105\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \]