Integrand size = 30, antiderivative size = 167 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {16 i a^2}{45 d e^4 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{15 d e^2 (e \sec (c+d x))^{5/2}}-\frac {32 i a \sqrt {a+i a \tan (c+d x)}}{45 d e^4 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}} \]
16/45*I*a^2/d/e^4/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-4/15*I*a*( a+I*a*tan(d*x+c))^(1/2)/d/e^2/(e*sec(d*x+c))^(5/2)-32/45*I*a*(a+I*a*tan(d* x+c))^(1/2)/d/e^4/(e*sec(d*x+c))^(1/2)-2/9*I*(a+I*a*tan(d*x+c))^(3/2)/d/(e *sec(d*x+c))^(9/2)
Time = 1.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.68 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {a (\cos (d x)-i \sin (d x)) (-81 i \cos (c+d x)+5 i \cos (3 (c+d x))-54 \sin (c+d x)+10 \sin (3 (c+d x))) (\cos (c+2 d x)+i \sin (c+2 d x)) \sqrt {a+i a \tan (c+d x)}}{90 d e^4 \sqrt {e \sec (c+d x)}} \]
(a*(Cos[d*x] - I*Sin[d*x])*((-81*I)*Cos[c + d*x] + (5*I)*Cos[3*(c + d*x)] - 54*Sin[c + d*x] + 10*Sin[3*(c + d*x)])*(Cos[c + 2*d*x] + I*Sin[c + 2*d*x ])*Sqrt[a + I*a*Tan[c + d*x]])/(90*d*e^4*Sqrt[e*Sec[c + d*x]])
Time = 0.79 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3978, 3042, 3978, 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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3978 |
| \(\displaystyle \frac {2 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{5/2}}dx}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{5/2}}dx}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3978 |
| \(\displaystyle \frac {2 a \left (\frac {4 a \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\frac {4 a \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {2 a \left (\frac {4 a \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\frac {4 a \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
| \(\Big \downarrow \) 3969 |
| \(\displaystyle \frac {2 a \left (\frac {4 a \left (\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )}{3 e^2}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{9 d (e \sec (c+d x))^{9/2}}\) |
(((-2*I)/9)*(a + I*a*Tan[c + d*x])^(3/2))/(d*(e*Sec[c + d*x])^(9/2)) + (2* a*((((-2*I)/5)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(5/2)) + (4 *a*(((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]])))/(5*e^2)))/(3 *e^2)
3.5.7.3.1 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[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/( a*f*m)), x] + Simp[a*((m + n)/(m*d^2)) Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b ^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
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 = 9.89 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.57
| method | result | size |
| default | \(\frac {2 \cos \left (d x +c \right ) \left (\tan \left (d x +c \right )-i\right ) a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (10 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 \left (\cos ^{3}\left (d x +c \right )\right )-16 i \sin \left (d x +c \right )+24 \cos \left (d x +c \right )\right )}{45 d \sqrt {e \sec \left (d x +c \right )}\, e^{4}}\) | \(95\) |
| risch | \(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (5 \,{\mathrm e}^{4 i \left (d x +c \right )}+135+12 \cos \left (2 d x +2 c \right )+42 i \sin \left (2 d x +2 c \right )\right )}{180 e^{4} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(99\) |
2/45/d*cos(d*x+c)*(tan(d*x+c)-I)*a*(a*(1+I*tan(d*x+c)))^(1/2)*(10*I*cos(d* x+c)^2*sin(d*x+c)-5*cos(d*x+c)^3-16*I*sin(d*x+c)+24*cos(d*x+c))/(e*sec(d*x +c))^(1/2)/e^4
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.62 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {{\left (-5 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 32 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 162 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 120 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{180 \, d e^{5}} \]
1/180*(-5*I*a*e^(8*I*d*x + 8*I*c) - 32*I*a*e^(6*I*d*x + 6*I*c) - 162*I*a*e ^(4*I*d*x + 4*I*c) - 120*I*a*e^(2*I*d*x + 2*I*c) + 15*I*a)*sqrt(a/(e^(2*I* d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-3/2*I*d*x - 3/2*I *c)/(d*e^5)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx=\text {Timed out} \]
Time = 0.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.96 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx=\frac {{\left (-5 i \, a \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 15 i \, a \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 27 i \, a \cos \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 135 i \, a \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 5 \, a \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 15 \, a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 27 \, a \sin \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 135 \, a \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} \sqrt {a}}{180 \, d e^{\frac {9}{2}}} \]
1/180*(-5*I*a*cos(9/2*d*x + 9/2*c) + 15*I*a*cos(3/2*d*x + 3/2*c) - 27*I*a* cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 135*I*a*cos (1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 5*a*sin(9/2*d* x + 9/2*c) + 15*a*sin(3/2*d*x + 3/2*c) + 27*a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 135*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2 *c), cos(3/2*d*x + 3/2*c))))*sqrt(a)/(d*e^(9/2))
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]
Time = 6.54 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{9/2}} \, dx=-\frac {a\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-42\,\sin \left (c+d\,x\right )-47\,\sin \left (3\,c+3\,d\,x\right )-5\,\sin \left (5\,c+5\,d\,x\right )+\cos \left (c+d\,x\right )\,282{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,17{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,5{}\mathrm {i}\right )}{360\,d\,e^5} \]