Integrand size = 24, antiderivative size = 85 \[ \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {6 \sin (c+d x)}{7 a d}-\frac {4 \sin ^3(c+d x)}{7 a d}+\frac {6 \sin ^5(c+d x)}{35 a d}+\frac {i \cos ^5(c+d x)}{7 d (a+i a \tan (c+d x))} \]
6/7*sin(d*x+c)/a/d-4/7*sin(d*x+c)^3/a/d+6/35*sin(d*x+c)^5/a/d+1/7*I*cos(d* x+c)^5/d/(a+I*a*tan(d*x+c))
Time = 0.61 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {\sec (c+d x) (-350+175 \cos (2 (c+d x))+14 \cos (4 (c+d x))+\cos (6 (c+d x))+350 i \sin (2 (c+d x))+56 i \sin (4 (c+d x))+6 i \sin (6 (c+d x)))}{1120 a d (-i+\tan (c+d x))} \]
-1/1120*(Sec[c + d*x]*(-350 + 175*Cos[2*(c + d*x)] + 14*Cos[4*(c + d*x)] + Cos[6*(c + d*x)] + (350*I)*Sin[2*(c + d*x)] + (56*I)*Sin[4*(c + d*x)] + ( 6*I)*Sin[6*(c + d*x)]))/(a*d*(-I + Tan[c + d*x]))
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 3983, 3042, 3113, 2009}
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.
\(\displaystyle \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (c+d x)^5 (a+i a \tan (c+d x))}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {6 \int \cos ^5(c+d x)dx}{7 a}+\frac {i \cos ^5(c+d x)}{7 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6 \int \sin \left (c+d x+\frac {\pi }{2}\right )^5dx}{7 a}+\frac {i \cos ^5(c+d x)}{7 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle -\frac {6 \int \left (\sin ^4(c+d x)-2 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{7 a d}+\frac {i \cos ^5(c+d x)}{7 d (a+i a \tan (c+d x))}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{7 a d}+\frac {i \cos ^5(c+d x)}{7 d (a+i a \tan (c+d x))}\) |
(-6*(-Sin[c + d*x] + (2*Sin[c + d*x]^3)/3 - Sin[c + d*x]^5/5))/(7*a*d) + ( (I/7)*Cos[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x]))
3.2.13.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 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 = 0.58 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40
method | result | size |
risch | \(\frac {i {\mathrm e}^{-7 i \left (d x +c \right )}}{448 a d}+\frac {5 i \cos \left (d x +c \right )}{64 a d}+\frac {35 \sin \left (d x +c \right )}{64 a d}+\frac {i \cos \left (5 d x +5 c \right )}{64 a d}+\frac {7 \sin \left (5 d x +5 c \right )}{320 a d}+\frac {3 i \cos \left (3 d x +3 c \right )}{64 a d}+\frac {7 \sin \left (3 d x +3 c \right )}{64 a d}\) | \(119\) |
derivativedivides | \(\frac {-\frac {2}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {15 i}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {11 i}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {21}{10 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {11}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {21}{16 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {i}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{4}}-\frac {i}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}+\frac {1}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{5}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {11}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}}{a d}\) | \(207\) |
default | \(\frac {-\frac {2}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {15 i}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {11 i}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {21}{10 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {11}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {21}{16 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {i}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{4}}-\frac {i}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}+\frac {1}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{5}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {11}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}}{a d}\) | \(207\) |
1/448*I/a/d*exp(-7*I*(d*x+c))+5/64*I/a/d*cos(d*x+c)+35/64*sin(d*x+c)/a/d+1 /64*I/a/d*cos(5*d*x+5*c)+7/320/a/d*sin(5*d*x+5*c)+3/64*I/a/d*cos(3*d*x+3*c )+7/64/a/d*sin(3*d*x+3*c)
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {{\left (-7 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 70 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 525 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 700 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 175 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 42 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{2240 \, a d} \]
1/2240*(-7*I*e^(12*I*d*x + 12*I*c) - 70*I*e^(10*I*d*x + 10*I*c) - 525*I*e^ (8*I*d*x + 8*I*c) + 700*I*e^(6*I*d*x + 6*I*c) + 175*I*e^(4*I*d*x + 4*I*c) + 42*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-7*I*d*x - 7*I*c)/(a*d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (68) = 136\).
Time = 0.34 (sec) , antiderivative size = 264, normalized size of antiderivative = 3.11 \[ \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx=\begin {cases} \frac {\left (- 150323855360 i a^{6} d^{6} e^{21 i c} e^{5 i d x} - 1503238553600 i a^{6} d^{6} e^{19 i c} e^{3 i d x} - 11274289152000 i a^{6} d^{6} e^{17 i c} e^{i d x} + 15032385536000 i a^{6} d^{6} e^{15 i c} e^{- i d x} + 3758096384000 i a^{6} d^{6} e^{13 i c} e^{- 3 i d x} + 901943132160 i a^{6} d^{6} e^{11 i c} e^{- 5 i d x} + 107374182400 i a^{6} d^{6} e^{9 i c} e^{- 7 i d x}\right ) e^{- 16 i c}}{48103633715200 a^{7} d^{7}} & \text {for}\: a^{7} d^{7} e^{16 i c} \neq 0 \\\frac {x \left (e^{12 i c} + 6 e^{10 i c} + 15 e^{8 i c} + 20 e^{6 i c} + 15 e^{4 i c} + 6 e^{2 i c} + 1\right ) e^{- 7 i c}}{64 a} & \text {otherwise} \end {cases} \]
Piecewise(((-150323855360*I*a**6*d**6*exp(21*I*c)*exp(5*I*d*x) - 150323855 3600*I*a**6*d**6*exp(19*I*c)*exp(3*I*d*x) - 11274289152000*I*a**6*d**6*exp (17*I*c)*exp(I*d*x) + 15032385536000*I*a**6*d**6*exp(15*I*c)*exp(-I*d*x) + 3758096384000*I*a**6*d**6*exp(13*I*c)*exp(-3*I*d*x) + 901943132160*I*a**6 *d**6*exp(11*I*c)*exp(-5*I*d*x) + 107374182400*I*a**6*d**6*exp(9*I*c)*exp( -7*I*d*x))*exp(-16*I*c)/(48103633715200*a**7*d**7), Ne(a**7*d**7*exp(16*I* c), 0)), (x*(exp(12*I*c) + 6*exp(10*I*c) + 15*exp(8*I*c) + 20*exp(6*I*c) + 15*exp(4*I*c) + 6*exp(2*I*c) + 1)*exp(-7*I*c)/(64*a), True))
Exception generated. \[ \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (73) = 146\).
Time = 0.43 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.01 \[ \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\frac {7 \, {\left (55 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 180 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 250 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 160 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 43\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}^{5}} + \frac {735 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3360 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 7315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 8820 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6321 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2492 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 461}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}}}{560 \, d} \]
1/560*(7*(55*tan(1/2*d*x + 1/2*c)^4 + 180*I*tan(1/2*d*x + 1/2*c)^3 - 250*t an(1/2*d*x + 1/2*c)^2 - 160*I*tan(1/2*d*x + 1/2*c) + 43)/(a*(tan(1/2*d*x + 1/2*c) + I)^5) + (735*tan(1/2*d*x + 1/2*c)^6 - 3360*I*tan(1/2*d*x + 1/2*c )^5 - 7315*tan(1/2*d*x + 1/2*c)^4 + 8820*I*tan(1/2*d*x + 1/2*c)^3 + 6321*t an(1/2*d*x + 1/2*c)^2 - 2492*I*tan(1/2*d*x + 1/2*c) - 461)/(a*(tan(1/2*d*x + 1/2*c) - I)^7))/d
Time = 8.21 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.21 \[ \int \frac {\cos ^5(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\left (-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,35{}\mathrm {i}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,105{}\mathrm {i}-126\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,182{}\mathrm {i}+26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,130{}\mathrm {i}-15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,55{}\mathrm {i}+25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+5{}\mathrm {i}\right )\,2{}\mathrm {i}}{35\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^5\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^7} \]
((25*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2*55i - 15*tan(c/2 + (d*x)/2) ^3 + tan(c/2 + (d*x)/2)^4*130i + 26*tan(c/2 + (d*x)/2)^5 + tan(c/2 + (d*x) /2)^6*182i - 126*tan(c/2 + (d*x)/2)^7 + tan(c/2 + (d*x)/2)^8*105i - 35*tan (c/2 + (d*x)/2)^9 + tan(c/2 + (d*x)/2)^10*35i - 35*tan(c/2 + (d*x)/2)^11 + 5i)*2i)/(35*a*d*(tan(c/2 + (d*x)/2) + 1i)^5*(tan(c/2 + (d*x)/2)*1i + 1)^7 )