Integrand size = 30, antiderivative size = 121 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i \sqrt {e \sec (c+d x)}}{45 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
16/45*I*(e*sec(d*x+c))^(1/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+2/9*I*(e*sec(d *x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(5/2)+8/45*I*(e*sec(d*x+c))^(1/2)/a/d/(a +I*a*tan(d*x+c))^(3/2)
Time = 1.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \sec ^2(c+d x) \sqrt {e \sec (c+d x)} (9+25 \cos (2 (c+d x))+20 i \sin (2 (c+d x)))}{45 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
((-1/45*I)*Sec[c + d*x]^2*Sqrt[e*Sec[c + d*x]]*(9 + 25*Cos[2*(c + d*x)] + (20*I)*Sin[2*(c + d*x)]))/(a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.55 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3983, 3042, 3983, 3042, 3969}
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.
\(\displaystyle \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {4 \int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^{3/2}}dx}{9 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^{3/2}}dx}{9 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{5 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{5 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {4 \left (\frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}\right )}{9 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
(((2*I)/9)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (4*((( (2*I)/5)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((4*I)/ 5)*Sqrt[e*Sec[c + d*x]])/(a*d*Sqrt[a + I*a*Tan[c + d*x]])))/(9*a)
3.5.34.3.1 Defintions of rubi rules used
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/ (a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ [Simplify[m + n], 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n)) Int[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x ] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* n]
Time = 13.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.57
method | result | size |
default | \(\frac {2 i \sqrt {e \sec \left (d x +c \right )}\, \left (20 i \tan \left (d x +c \right )+25-8 \left (\sec ^{2}\left (d x +c \right )\right )\right )}{45 d \left (1+i \tan \left (d x +c \right )\right )^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}\) | \(69\) |
2/45*I/d*(e*sec(d*x+c))^(1/2)/(1+I*tan(d*x+c))^2/(a*(1+I*tan(d*x+c)))^(1/2 )/a^2*(20*I*tan(d*x+c)+25-8*sec(d*x+c)^2)
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (45 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 63 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 23 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac {9}{2} i \, d x - \frac {9}{2} i \, c\right )}}{90 \, a^{3} d} \]
1/90*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*( 45*I*e^(6*I*d*x + 6*I*c) + 63*I*e^(4*I*d*x + 4*I*c) + 23*I*e^(2*I*d*x + 2* I*c) + 5*I)*e^(-9/2*I*d*x - 9/2*I*c)/(a^3*d)
\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.39 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {e} {\left (5 i \, \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 18 i \, \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 45 i \, \cos \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 18 \, \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 45 \, \sin \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right )\right )}}{90 \, a^{\frac {5}{2}} d} \]
1/90*sqrt(e)*(5*I*cos(9/2*d*x + 9/2*c) + 18*I*cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 45*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2* c), cos(9/2*d*x + 9/2*c))) + 5*sin(9/2*d*x + 9/2*c) + 18*sin(5/9*arctan2(s in(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 45*sin(1/9*arctan2(sin(9/2*d *x + 9/2*c), cos(9/2*d*x + 9/2*c))))/(a^(5/2)*d)
\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sqrt {e \sec \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 5.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,18{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}+18\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (4\,c+4\,d\,x\right )+45{}\mathrm {i}\right )}{90\,a^2\,d\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]