Integrand size = 30, antiderivative size = 206 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {256 i \sqrt {e \sec (c+d x)}}{315 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}} \]
2/9*I/d/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(1/2)+32/105*I/d/e^2/(e*se c(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2)+256/315*I*(e*sec(d*x+c))^(1/2)/d/ e^4/(a+I*a*tan(d*x+c))^(1/2)-16/63*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/(e*sec(d *x+c))^(7/2)-128/315*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/e^2/(e*sec(d*x+c))^(3/ 2)
Time = 1.40 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.42 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e \sec (c+d x)} (945 i-84 i \cos (2 (c+d x))-5 i \cos (4 (c+d x))+336 \sin (2 (c+d x))+40 \sin (4 (c+d x)))}{1260 d e^4 \sqrt {a+i a \tan (c+d x)}} \]
(Sqrt[e*Sec[c + d*x]]*(945*I - (84*I)*Cos[2*(c + d*x)] - (5*I)*Cos[4*(c + d*x)] + 336*Sin[2*(c + d*x)] + 40*Sin[4*(c + d*x)]))/(1260*d*e^4*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.94 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3983, 3042, 3978, 3042, 3983, 3042, 3978, 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.
\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {8 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{7/2}}dx}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{7/2}}dx}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle \frac {8 \left (\frac {6 a \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \left (\frac {6 a \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {8 \left (\frac {6 a \left (\frac {4 \left (\frac {4 i a \sqrt {e \sec (c+d x)}}{3 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{7 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{7 d (e \sec (c+d x))^{7/2}}\right )}{9 a}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}}\) |
((2*I)/9)/(d*(e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (8*((((- 2*I)/7)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(7/2)) + (6*a*(((2 *I)/5)/(d*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (4*((((4*I) /3)*a*Sqrt[e*Sec[c + d*x]])/(d*e^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/3 )*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(3/2))))/(5*a)))/(7*e^2) ))/(9*a)
3.5.22.3.1 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 = 10.60 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.43
method | result | size |
default | \(\frac {-\frac {2 i \left (\cos ^{3}\left (d x +c \right )\right )}{63}+\frac {16 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{63}-\frac {32 i \cos \left (d x +c \right )}{315}+\frac {128 \sin \left (d x +c \right )}{315}+\frac {256 i \sec \left (d x +c \right )}{315}}{d \sqrt {e \sec \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, e^{3}}\) | \(88\) |
2/315/d/(e*sec(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)/e^3*(-5*I*cos(d*x+ c)^3+40*cos(d*x+c)^2*sin(d*x+c)-16*I*cos(d*x+c)+64*sin(d*x+c)+128*I*sec(d* x+c))
Time = 0.24 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-45 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 465 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1470 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2142 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 287 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 35 i\right )} e^{\left (-\frac {9}{2} i \, d x - \frac {9}{2} i \, c\right )}}{2520 \, a d e^{4}} \]
1/2520*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^(10*I*d*x + 10*I*c) - 465*I*e^(8*I*d*x + 8*I*c) + 1470*I*e^(6*I* d*x + 6*I*c) + 2142*I*e^(4*I*d*x + 4*I*c) + 287*I*e^(2*I*d*x + 2*I*c) + 35 *I)*e^(-9/2*I*d*x - 9/2*I*c)/(a*d*e^4)
Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Timed out} \]
Time = 0.80 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {35 i \, \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) - 45 i \, \cos \left (\frac {7}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 252 i \, \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) - 420 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 1890 i \, \cos \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sin \left (\frac {7}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 252 \, \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 420 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 1890 \, \sin \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right )}{2520 \, \sqrt {a} d e^{\frac {7}{2}}} \]
1/2520*(35*I*cos(9/2*d*x + 9/2*c) - 45*I*cos(7/9*arctan2(sin(9/2*d*x + 9/2 *c), cos(9/2*d*x + 9/2*c))) + 252*I*cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 420*I*cos(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9 /2*d*x + 9/2*c))) + 1890*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d *x + 9/2*c))) + 35*sin(9/2*d*x + 9/2*c) + 45*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c) , cos(9/2*d*x + 9/2*c))) + 420*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9 /2*d*x + 9/2*c))) + 1890*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))/(sqrt(a)*d*e^(7/2))
\[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
Time = 4.94 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (336\,\sin \left (2\,c+2\,d\,x\right )-\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}-\cos \left (2\,c+2\,d\,x\right )\,84{}\mathrm {i}+40\,\sin \left (4\,c+4\,d\,x\right )+945{}\mathrm {i}\right )}{1260\,d\,e^4\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]