Integrand size = 26, antiderivative size = 270 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {1155 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i \cos ^3(c+d x)}{96 a d (a+i a \tan (c+d x))^{3/2}}+\frac {385 i \cos (c+d x)}{2048 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {33 i \cos ^3(c+d x)}{256 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1155 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4096 a^3 d}-\frac {77 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{512 a^3 d} \]
1155/8192*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2 ))/a^(5/2)/d*2^(1/2)+385/2048*I*cos(d*x+c)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+ 33/256*I*cos(d*x+c)^3/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-1155/4096*I*cos(d*x+c )*(a+I*a*tan(d*x+c))^(1/2)/a^3/d-77/512*I*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^ (1/2)/a^3/d+1/8*I*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(5/2)+11/96*I*cos(d*x+ c)^3/a/d/(a+I*a*tan(d*x+c))^(3/2)
Time = 1.93 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \sec ^3(c+d x) \left (-3325-3465 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-1605 \cos (2 (c+d x))+1800 \cos (4 (c+d x))+80 \cos (6 (c+d x))+1111 i \sin (2 (c+d x))+2552 i \sin (4 (c+d x))+176 i \sin (6 (c+d x))\right )}{24576 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
((I/24576)*Sec[c + d*x]^3*(-3325 - 3465*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2 *I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] - 1605*Cos[2*(c + d *x)] + 1800*Cos[4*(c + d*x)] + 80*Cos[6*(c + d*x)] + (1111*I)*Sin[2*(c + d *x)] + (2552*I)*Sin[4*(c + d*x)] + (176*I)*Sin[6*(c + d*x)]))/(a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.27 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {3042, 3983, 3042, 3983, 3042, 3983, 3042, 3978, 3042, 3983, 3042, 3971, 3042, 3970, 219}
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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^3 (a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {11 \int \frac {\cos ^3(c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 \int \frac {1}{\sec (c+d x)^3 (i \tan (c+d x) a+a)^{3/2}}dx}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {11 \left (\frac {3 \int \frac {\cos ^3(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 \left (\frac {3 \int \frac {1}{\sec (c+d x)^3 \sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \int \cos ^3(c+d x) \sqrt {i \tan (c+d x) a+a}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)^3}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3978 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3971 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3970 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\) |
((I/8)*Cos[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (11*(((I/6)*Cos[ c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (3*(((I/4)*Cos[c + d*x]^3)/ (d*Sqrt[a + I*a*Tan[c + d*x]]) + (7*(((-1/3*I)*Cos[c + d*x]^3*Sqrt[a + I*a *Tan[c + d*x]])/d + (5*a*(((I/2)*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x ]]) + (3*((I*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a* Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]) /d))/(4*a)))/6))/(8*a)))/(4*a)))/(16*a)
3.4.77.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S ymbol] :> Simp[-2*(a/(b*f)) Subst[Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/ Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^2, 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/(2*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] && EqQ[m/2 + n, 0] && GtQ[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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 863 vs. \(2 (221 ) = 442\).
Time = 9.11 (sec) , antiderivative size = 864, normalized size of antiderivative = 3.20
1/24576/d/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(cos(d*x+c)+1)/(a*(1+I*tan(d* x+c)))^(1/2)/(1+I*tan(d*x+c))^2/a^2*(-10560*I*cos(d*x+c)*(-cos(d*x+c)/(cos (d*x+c)+1))^(1/2)+5632*sin(d*x+c)*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1) )^(1/2)-10560*I*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+5632*sin(d *x+c)*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+13860*I*cos(d*x+c)*a rctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x +c)+1))^(1/2))+14784*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x +c)+16170*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-2560*I*cos(d*x+c)^4*(-cos(d *x+c)/(cos(d*x+c)+1))^(1/2)+14784*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^ (1/2)-13860*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d* x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+6930*I*arctan(1/2*(I*sin(d*x+c)-cos (d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+16170*I*sec( d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-6930*tan(d*x+c)*(-cos(d*x+c)/(co s(d*x+c)+1))^(1/2)-6930*tan(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/ (cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2560*I*cos(d*x+c)^3*(-c os(d*x+c)/(cos(d*x+c)+1))^(1/2)-10395*I*sec(d*x+c)*arctan(1/2*(I*sin(d*x+c )-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-6930*ta n(d*x+c)*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+3465*tan(d*x+c)*sec (d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c) /(cos(d*x+c)+1))^(1/2))-3465*I*sec(d*x+c)^2*arctan(1/2*(I*sin(d*x+c)-co...
Time = 0.26 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {1155 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2048 \, a^{2} d}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (-\frac {1155 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2048 \, a^{2} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-128 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 2176 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 247 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3325 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1358 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 376 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i\right )}\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{24576 \, a^{3} d} \]
1/24576*(-3465*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(8*I*d*x + 8*I*c)*log (-1155/2048*(sqrt(2)*sqrt(1/2)*(I*a^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqr t(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^ 2*d)) + 3465*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(8*I*d*x + 8*I*c)*log(- 1155/2048*(sqrt(2)*sqrt(1/2)*(-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqrt (a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2 *d)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-128*I*e^(12*I*d*x + 12* I*c) - 2176*I*e^(10*I*d*x + 10*I*c) + 247*I*e^(8*I*d*x + 8*I*c) + 3325*I*e ^(6*I*d*x + 6*I*c) + 1358*I*e^(4*I*d*x + 4*I*c) + 376*I*e^(2*I*d*x + 2*I*c ) + 48*I))*e^(-8*I*d*x - 8*I*c)/(a^3*d)
\[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\cos ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3789 vs. \(2 (207) = 414\).
Time = 0.60 (sec) , antiderivative size = 3789, normalized size of antiderivative = 14.03 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/98304*(4*(cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + sin( 1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*cos(1/4*arctan2(sin (8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)^(3/4)*(15*((-I*sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d *x + 8*c)))^2 + (-I*sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*s in(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + 2*(-I*sqrt(2)*cos( 8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) - I*sqrt(2)*cos(8*d*x + 8*c) - sqrt(2)*sin(8*d*x + 8*c ))*cos(7/2*arctan2(sin(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))), c os(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 1)) + (55*I*sqrt(2)* cos(8*d*x + 8*c) + 960*I*sqrt(2)*cos(3/4*arctan2(sin(8*d*x + 8*c), cos(8*d *x + 8*c))) - 1296*I*sqrt(2)*cos(1/2*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 55*sqrt(2)*sin(8*d*x + 8*c) + 960*sqrt(2)*sin(3/4*arctan2(sin(8* d*x + 8*c), cos(8*d*x + 8*c))) - 1296*sqrt(2)*sin(1/2*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c))) + 128*I*sqrt(2))*cos(3/2*arctan2(sin(1/4*arctan2( sin(8*d*x + 8*c), cos(8*d*x + 8*c))), cos(1/4*arctan2(sin(8*d*x + 8*c), co s(8*d*x + 8*c))) + 1)) + 15*((sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d *x + 8*c))*cos(1/4*arctan2(sin(8*d*x + 8*c), cos(8*d*x + 8*c)))^2 + (sqrt( 2)*cos(8*d*x + 8*c) - I*sqrt(2)*sin(8*d*x + 8*c))*sin(1/4*arctan2(sin(8*d* x + 8*c), cos(8*d*x + 8*c)))^2 + 2*(sqrt(2)*cos(8*d*x + 8*c) - I*sqrt(2...
\[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]