Integrand size = 26, antiderivative size = 307 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {3003 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16384 \sqrt {2} a^{7/2} d}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}+\frac {13 i \cos ^3(c+d x)}{160 a d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i \cos ^3(c+d x)}{1920 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {1001 i \cos (c+d x)}{8192 a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {429 i \cos ^3(c+d x)}{5120 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {3003 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16384 a^4 d}-\frac {1001 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{10240 a^4 d} \]
3003/32768*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/ 2))/a^(7/2)/d*2^(1/2)+1001/8192*I*cos(d*x+c)/a^3/d/(a+I*a*tan(d*x+c))^(1/2 )+429/5120*I*cos(d*x+c)^3/a^3/d/(a+I*a*tan(d*x+c))^(1/2)-3003/16384*I*cos( d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^4/d-1001/10240*I*cos(d*x+c)^3*(a+I*a*tan (d*x+c))^(1/2)/a^4/d+1/10*I*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(7/2)+13/160 *I*cos(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^(5/2)+143/1920*I*cos(d*x+c)^3/a^2/d /(a+I*a*tan(d*x+c))^(3/2)
Time = 2.97 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.57 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {\left (42140+20048 e^{-2 i (c+d x)}+71190 e^{2 i (c+d x)}+5856 e^{-4 i (c+d x)}-48640 e^{4 i (c+d x)}+768 e^{-6 i (c+d x)}-2560 e^{6 i (c+d x)}+\frac {90090 e^{4 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right ) \sec ^3(c+d x)}{491520 a^3 d (-i+\tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}} \]
-1/491520*((42140 + 20048/E^((2*I)*(c + d*x)) + 71190*E^((2*I)*(c + d*x)) + 5856/E^((4*I)*(c + d*x)) - 48640*E^((4*I)*(c + d*x)) + 768/E^((6*I)*(c + d*x)) - 2560*E^((6*I)*(c + d*x)) + (90090*E^((4*I)*(c + d*x))*ArcTanh[Sqr t[1 + E^((2*I)*(c + d*x))]])/Sqrt[1 + E^((2*I)*(c + d*x))])*Sec[c + d*x]^3 )/(a^3*d*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.49 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.654, Rules used = {3042, 3983, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^3 (a+i a \tan (c+d x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {13 \int \frac {\cos ^3(c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 \int \frac {1}{\sec (c+d x)^3 (i \tan (c+d x) a+a)^{5/2}}dx}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {13 \left (\frac {11 \int \frac {\cos ^3(c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 \left (\frac {11 \int \frac {1}{\sec (c+d x)^3 (i \tan (c+d x) a+a)^{3/2}}dx}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \int \frac {\cos ^3(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \int \frac {1}{\sec (c+d x)^3 \sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \int \cos ^3(c+d x) \sqrt {i \tan (c+d x) a+a}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)^3}dx}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3978 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3971 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3970 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {13 \left (\frac {11 \left (\frac {3 \left (\frac {7 \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )}{8 a}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {i \cos ^3(c+d x)}{6 d (a+i a \tan (c+d x))^{3/2}}\right )}{16 a}+\frac {i \cos ^3(c+d x)}{8 d (a+i a \tan (c+d x))^{5/2}}\right )}{20 a}+\frac {i \cos ^3(c+d x)}{10 d (a+i a \tan (c+d x))^{7/2}}\) |
((I/10)*Cos[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (13*(((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)))/(20*a)
3.4.93.3.1 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. 1056 vs. \(2 (252 ) = 504\).
Time = 11.08 (sec) , antiderivative size = 1057, normalized size of antiderivative = 3.44
1/491520/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))^3/a^3*(-256256*I*cos(d*x+c)*(-cos(d*x+c)/(c os(d*x+c)+1))^(1/2)-256256*I*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/ 2)+106496*sin(d*x+c)*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+36036 0*I*cos(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))+106496*sin(d*x+c)*cos(d*x+c)^2*(-cos(d*x+c) /(cos(d*x+c)+1))^(1/2)-90090*I*sec(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1))^( 1/2)+45045*I*sec(d*x+c)^3*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))+329472*(-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2)*cos(d*x+c)*sin(d*x+c)+492492*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2 )-90090*I*sec(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-360360*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)+329472*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-573 44*I*cos(d*x+c)^4*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+492492*I*sec(d*x+c)*( -cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-180180*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))-300300 *tan(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+180180*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))-57 344*I*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+180180*tan(d*x+c)*se c(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*...
Time = 0.28 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (-\frac {3003 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8192 \, a^{3} d}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (-\frac {3003 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8192 \, a^{3} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1280 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 25600 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 11275 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 56665 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 31094 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 12952 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3312 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 384 i\right )}\right )} e^{\left (-10 i \, d x - 10 i \, c\right )}}{491520 \, a^{4} d} \]
1/491520*(-45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(10*I*d*x + 10*I*c) *log(-3003/8192*(sqrt(2)*sqrt(1/2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d) *sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - I)*e^(-I*d*x - I*c) /(a^3*d)) + 45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(10*I*d*x + 10*I*c )*log(-3003/8192*(sqrt(2)*sqrt(1/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3* d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - I)*e^(-I*d*x - I* c)/(a^3*d)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1280*I*e^(14*I*d *x + 14*I*c) - 25600*I*e^(12*I*d*x + 12*I*c) + 11275*I*e^(10*I*d*x + 10*I* c) + 56665*I*e^(8*I*d*x + 8*I*c) + 31094*I*e^(6*I*d*x + 6*I*c) + 12952*I*e ^(4*I*d*x + 4*I*c) + 3312*I*e^(2*I*d*x + 2*I*c) + 384*I))*e^(-10*I*d*x - 1 0*I*c)/(a^4*d)
Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5821 vs. \(2 (236) = 472\).
Time = 0.62 (sec) , antiderivative size = 5821, normalized size of antiderivative = 18.96 \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Too large to display} \]
-1/1966080*(40*(cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + 2*cos(1/5* arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 1)^(3/4)*((79*(-I*sqrt( 2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*cos(1/5*arctan2(sin(10 *d*x + 10*c), cos(10*d*x + 10*c)))^2 + 79*(-I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))*sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d* x + 10*c)))^2 + 837*(-I*sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10 *d*x + 10*c)))^2 - I*sqrt(2)*sin(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d* x + 10*c)))^2 - 2*I*sqrt(2)*cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - I*sqrt(2))*cos(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 1 0*c))) + 158*(-I*sqrt(2)*cos(10*d*x + 10*c) - sqrt(2)*sin(10*d*x + 10*c))* cos(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 837*(sqrt(2)*co s(1/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + sqrt(2)*sin(1/5 *arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c)))^2 + 2*sqrt(2)*cos(1/5*ar ctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + sqrt(2))*sin(4/5*arctan2( sin(10*d*x + 10*c), cos(10*d*x + 10*c))) - 79*I*sqrt(2)*cos(10*d*x + 10*c) - 79*sqrt(2)*sin(10*d*x + 10*c))*cos(7/2*arctan2(sin(1/5*arctan2(sin(10*d *x + 10*c), cos(10*d*x + 10*c))), cos(1/5*arctan2(sin(10*d*x + 10*c), cos( 10*d*x + 10*c))) + 1)) + (-49*I*sqrt(2)*cos(10*d*x + 10*c) - 1155*I*sqrt(2 )*cos(4/5*arctan2(sin(10*d*x + 10*c), cos(10*d*x + 10*c))) + 3264*I*sqr...
\[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \]