Integrand size = 24, antiderivative size = 98 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {5 a^5 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 i a^5 \sec (c+d x)}{d}+\frac {10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d} \]
5*a^5*arctanh(sin(d*x+c))/d+5*I*a^5*sec(d*x+c)/d+10/3*I*a^3*cos(d*x+c)*(a+ I*a*tan(d*x+c))^2/d-2/3*I*a*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4/d
Time = 1.97 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.33 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 \cos ^4(c+d x) \left (30 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right ) \cos (c+d x) (i \cos (5 c)+\sin (5 c))-(\cos (3 c-2 d x)-i \sin (3 c-2 d x)) (10+13 \cos (2 (c+d x))-17 i \sin (2 (c+d x)))\right ) (-i+\tan (c+d x))^5}{3 d (\cos (d x)+i \sin (d x))^5} \]
(a^5*Cos[c + d*x]^4*(30*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]]*Cos[c + d*x] *(I*Cos[5*c] + Sin[5*c]) - (Cos[3*c - 2*d*x] - I*Sin[3*c - 2*d*x])*(10 + 1 3*Cos[2*(c + d*x)] - (17*I)*Sin[2*(c + d*x)]))*(-I + Tan[c + d*x])^5)/(3*d *(Cos[d*x] + I*Sin[d*x])^5)
Time = 0.53 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3977, 3042, 3977, 3042, 3967, 3042, 4257}
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.
| \(\displaystyle \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^5}{\sec (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle -\frac {5}{3} a^2 \int \cos (c+d x) (i \tan (c+d x) a+a)^3dx-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5}{3} a^2 \int \frac {(i \tan (c+d x) a+a)^3}{\sec (c+d x)}dx-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle -\frac {5}{3} a^2 \left (-3 a^2 \int \sec (c+d x) (i \tan (c+d x) a+a)dx-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^2}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5}{3} a^2 \left (-3 a^2 \int \sec (c+d x) (i \tan (c+d x) a+a)dx-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^2}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle -\frac {5}{3} a^2 \left (-3 a^2 \left (a \int \sec (c+d x)dx+\frac {i a \sec (c+d x)}{d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^2}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5}{3} a^2 \left (-3 a^2 \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {i a \sec (c+d x)}{d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^2}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {5}{3} a^2 \left (-3 a^2 \left (\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^2}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\) |
(((-2*I)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4)/d - (5*a^2*(-3*a^2* ((a*ArcTanh[Sin[c + d*x]])/d + (I*a*Sec[c + d*x])/d) - ((2*I)*a*Cos[c + d* x]*(a + I*a*Tan[c + d*x])^2)/d))/3
3.1.72.3.1 Defintions of rubi rules used
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 26.35 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.13
| method | result | size |
| risch | \(-\frac {4 i a^{5} {\mathrm e}^{3 i \left (d x +c \right )}}{3 d}+\frac {8 i a^{5} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {2 i a^{5} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {5 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {5 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(111\) |
| derivativedivides | \(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+5 a^{5} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {10 i a^{5} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-\frac {10 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {5 i a^{5} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{5} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(165\) |
| default | \(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+5 a^{5} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {10 i a^{5} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-\frac {10 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {5 i a^{5} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{5} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(165\) |
-4/3*I/d*a^5*exp(3*I*(d*x+c))+8*I/d*a^5*exp(I*(d*x+c))+2*I*a^5*exp(I*(d*x+ c))/d/(exp(2*I*(d*x+c))+1)+5/d*a^5*ln(exp(I*(d*x+c))+I)-5/d*a^5*ln(exp(I*( d*x+c))-I)
Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.24 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {-4 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 20 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 15 \, {\left (a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
1/3*(-4*I*a^5*e^(5*I*d*x + 5*I*c) + 20*I*a^5*e^(3*I*d*x + 3*I*c) + 30*I*a^ 5*e^(I*d*x + I*c) + 15*(a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) + I) - 15*(a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) - I))/(d*e^ (2*I*d*x + 2*I*c) + d)
Time = 0.32 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.51 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {2 i a^{5} e^{i c} e^{i d x}}{d e^{2 i c} e^{2 i d x} + d} + \frac {5 a^{5} \left (- \log {\left (e^{i d x} - i e^{- i c} \right )} + \log {\left (e^{i d x} + i e^{- i c} \right )}\right )}{d} + \begin {cases} \frac {- 4 i a^{5} d e^{3 i c} e^{3 i d x} + 24 i a^{5} d e^{i c} e^{i d x}}{3 d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (4 a^{5} e^{3 i c} - 8 a^{5} e^{i c}\right ) & \text {otherwise} \end {cases} \]
2*I*a**5*exp(I*c)*exp(I*d*x)/(d*exp(2*I*c)*exp(2*I*d*x) + d) + 5*a**5*(-lo g(exp(I*d*x) - I*exp(-I*c)) + log(exp(I*d*x) + I*exp(-I*c)))/d + Piecewise (((-4*I*a**5*d*exp(3*I*c)*exp(3*I*d*x) + 24*I*a**5*d*exp(I*c)*exp(I*d*x))/ (3*d**2), Ne(d**2, 0)), (x*(4*a**5*exp(3*I*c) - 8*a**5*exp(I*c)), True))
Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.57 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {10 i \, a^{5} \cos \left (d x + c\right )^{3} + 20 \, a^{5} \sin \left (d x + c\right )^{3} + 2 i \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{5} + 20 i \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{5} + 5 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{5} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5}}{6 \, d} \]
-1/6*(10*I*a^5*cos(d*x + c)^3 + 20*a^5*sin(d*x + c)^3 + 2*I*(cos(d*x + c)^ 3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^5 + 20*I*(cos(d*x + c)^3 - 3*cos(d* x + c))*a^5 + 5*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d* x + c) - 1) + 6*sin(d*x + c))*a^5 + 2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^ 5)/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1683 vs. \(2 (84) = 168\).
Time = 1.18 (sec) , antiderivative size = 1683, normalized size of antiderivative = 17.17 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\text {Too large to display} \]
-1/6144*(39225*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 31380 0*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1098300*a^5*e^(12* I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2196600*a^5*e^(10*I*d*x + 2*I* c)*log(I*e^(I*d*x + I*c) + 1) + 2196600*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I *d*x + I*c) + 1) + 1098300*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 313800*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2745750* a^5*e^(8*I*d*x)*log(I*e^(I*d*x + I*c) + 1) + 39225*a^5*e^(-8*I*c)*log(I*e^ (I*d*x + I*c) + 1) + 8520*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) - 1) + 68160*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 238560*a ^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 477120*a^5*e^(10*I*d* x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 477120*a^5*e^(6*I*d*x - 2*I*c)*log (I*e^(I*d*x + I*c) - 1) + 238560*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 68160*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 596 400*a^5*e^(8*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 8520*a^5*e^(-8*I*c)*log(I *e^(I*d*x + I*c) - 1) - 39225*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x + I *c) + 1) - 313800*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1 098300*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2196600*a^5* e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2196600*a^5*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1098300*a^5*e^(4*I*d*x - 4*I*c)*log (-I*e^(I*d*x + I*c) + 1) - 313800*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d...
Time = 6.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.65 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {10\,a^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,34{}\mathrm {i}-\frac {82\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,38{}\mathrm {i}+\frac {46\,a^5}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,3{}\mathrm {i}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,4{}\mathrm {i}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]
(10*a^5*atanh(tan(c/2 + (d*x)/2)))/d - (a^5*tan(c/2 + (d*x)/2)^3*34i - (82 *a^5*tan(c/2 + (d*x)/2)^2)/3 + 8*a^5*tan(c/2 + (d*x)/2)^4 + (46*a^5)/3 - a ^5*tan(c/2 + (d*x)/2)*38i)/(d*(3*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2 *4i - 4*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*3i + tan(c/2 + (d*x)/2 )^5 + 1i))