Integrand size = 30, antiderivative size = 162 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i}{11 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {12 i}{77 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a^2 d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{77 a^3 d \sqrt {e \sec (c+d x)}} \]
16/77*I/a^2/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-32/77*I*(a+I*a *tan(d*x+c))^(1/2)/a^3/d/(e*sec(d*x+c))^(1/2)+2/11*I/d/(e*sec(d*x+c))^(1/2 )/(a+I*a*tan(d*x+c))^(5/2)+12/77*I/a/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x +c))^(3/2)
Time = 1.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \sec ^3(c+d x) (-55 \cos (c+d x)+35 \cos (3 (c+d x))-22 i \sin (c+d x)+42 i \sin (3 (c+d x)))}{154 a^2 d \sqrt {e \sec (c+d x)} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
((I/154)*Sec[c + d*x]^3*(-55*Cos[c + d*x] + 35*Cos[3*(c + d*x)] - (22*I)*S in[c + d*x] + (42*I)*Sin[3*(c + d*x)]))/(a^2*d*Sqrt[e*Sec[c + d*x]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.74 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3983, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}dx\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {6 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \int \frac {1}{\sqrt {e \sec (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {6 \left (\frac {4 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \left (\frac {4 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3969 |
| \(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}}\right )}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}\right )}{11 a}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}}\) |
((2*I)/11)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + (6*(((2 *I)/7)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + (4*(((2*I)/ 3)/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[e*Sec[c + d*x]])))/(7*a)))/(11*a)
3.5.35.3.1 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 = 10.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.52
| method | result | size |
| default | \(\frac {-\frac {10 i}{11}+\frac {12 \tan \left (d x +c \right )}{11}+\frac {80 i \left (\sec ^{2}\left (d x +c \right )\right )}{77}-\frac {32 \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{77}}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}\, \left (1+i \tan \left (d x +c \right )\right )^{2} a^{2}}\) | \(85\) |
2/77/d/(a*(1+I*tan(d*x+c)))^(1/2)/(e*sec(d*x+c))^(1/2)/(1+I*tan(d*x+c))^2/ a^2*(-35*I+42*tan(d*x+c)+40*I*sec(d*x+c)^2-16*sec(d*x+c)^2*tan(d*x+c))
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-77 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 110 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-\frac {11}{2} i \, d x - \frac {11}{2} i \, c\right )}}{308 \, a^{3} d e} \]
1/308*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))* (-77*I*e^(8*I*d*x + 8*I*c) + 110*I*e^(4*I*d*x + 4*I*c) + 40*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(-11/2*I*d*x - 11/2*I*c)/(a^3*d*e)
\[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{\sqrt {e \sec {\left (c + d x \right )}} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.77 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {7 i \, \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 i \, \cos \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 i \, \cos \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) - 77 i \, \cos \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 7 \, \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 \, \sin \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right )}{308 \, a^{\frac {5}{2}} d \sqrt {e}} \]
1/308*(7*I*cos(11/2*d*x + 11/2*c) + 33*I*cos(7/11*arctan2(sin(11/2*d*x + 1 1/2*c), cos(11/2*d*x + 11/2*c))) + 77*I*cos(3/11*arctan2(sin(11/2*d*x + 11 /2*c), cos(11/2*d*x + 11/2*c))) - 77*I*cos(1/11*arctan2(sin(11/2*d*x + 11/ 2*c), cos(11/2*d*x + 11/2*c))) + 7*sin(11/2*d*x + 11/2*c) + 33*sin(7/11*ar ctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 77*sin(3/11*arcta n2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 77*sin(1/11*arctan2( sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))/(a^(5/2)*d*sqrt(e))
\[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 5.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (154\,\sin \left (c+d\,x\right )+33\,\sin \left (3\,c+3\,d\,x\right )+7\,\sin \left (5\,c+5\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\,33{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,7{}\mathrm {i}\right )}{308\,a^2\,d\,e\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]