Integrand size = 24, antiderivative size = 160 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {5 a^3 x}{32}-\frac {i a^7}{16 d (a-i a \tan (c+d x))^4}-\frac {i a^9}{12 d \left (a^2-i a^2 \tan (c+d x)\right )^3}-\frac {3 i a^9}{32 d \left (a^3-i a^3 \tan (c+d x)\right )^2}-\frac {i a^9}{8 d \left (a^6-i a^6 \tan (c+d x)\right )}+\frac {i a^9}{32 d \left (a^6+i a^6 \tan (c+d x)\right )} \] Output:
5/32*a^3*x-1/16*I*a^7/d/(a-I*a*tan(d*x+c))^4-1/12*I*a^9/d/(a^2-I*a^2*tan(d *x+c))^3-3/32*I*a^9/d/(a^3-I*a^3*tan(d*x+c))^2-1/8*I*a^9/d/(a^6-I*a^6*tan( d*x+c))+1/32*I*a^9/d/(a^6+I*a^6*tan(d*x+c))
Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.86 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i a^9 \left (\frac {5 i \arctan (\tan (c+d x))}{32 a^6}+\frac {1}{16 a^2 (a-i a \tan (c+d x))^4}+\frac {1}{12 a^3 (a-i a \tan (c+d x))^3}+\frac {3}{32 a^4 (a-i a \tan (c+d x))^2}+\frac {1}{8 a^5 (a-i a \tan (c+d x))}-\frac {1}{32 a^5 (a+i a \tan (c+d x))}\right )}{d} \] Input:
Integrate[Cos[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*a^9*((((5*I)/32)*ArcTan[Tan[c + d*x]])/a^6 + 1/(16*a^2*(a - I*a*Tan[ c + d*x])^4) + 1/(12*a^3*(a - I*a*Tan[c + d*x])^3) + 3/(32*a^4*(a - I*a*Ta n[c + d*x])^2) + 1/(8*a^5*(a - I*a*Tan[c + d*x])) - 1/(32*a^5*(a + I*a*Tan [c + d*x]))))/d
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 54, 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.
\(\displaystyle \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{\sec (c+d x)^8}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i a^9 \int \frac {1}{(a-i a \tan (c+d x))^5 (i \tan (c+d x) a+a)^2}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {i a^9 \int \left (\frac {1}{8 a^5 (a-i a \tan (c+d x))^2}+\frac {1}{32 a^5 (i \tan (c+d x) a+a)^2}+\frac {3}{16 a^4 (a-i a \tan (c+d x))^3}+\frac {1}{4 a^3 (a-i a \tan (c+d x))^4}+\frac {1}{4 a^2 (a-i a \tan (c+d x))^5}+\frac {5}{32 a^5 \left (\tan ^2(c+d x) a^2+a^2\right )}\right )d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i a^9 \left (\frac {5 i \arctan (\tan (c+d x))}{32 a^6}+\frac {1}{8 a^5 (a-i a \tan (c+d x))}-\frac {1}{32 a^5 (a+i a \tan (c+d x))}+\frac {3}{32 a^4 (a-i a \tan (c+d x))^2}+\frac {1}{12 a^3 (a-i a \tan (c+d x))^3}+\frac {1}{16 a^2 (a-i a \tan (c+d x))^4}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*a^9*((((5*I)/32)*ArcTan[Tan[c + d*x]])/a^6 + 1/(16*a^2*(a - I*a*Tan[ c + d*x])^4) + 1/(12*a^3*(a - I*a*Tan[c + d*x])^3) + 3/(32*a^4*(a - I*a*Ta n[c + d*x])^2) + 1/(8*a^5*(a - I*a*Tan[c + d*x])) - 1/(32*a^5*(a + I*a*Tan [c + d*x]))))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 128.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.61
method | result | size |
risch | \(\frac {5 a^{3} x}{32}-\frac {i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}}{256 d}-\frac {5 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}}{192 d}-\frac {5 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}}{64 d}-\frac {9 i a^{3} \cos \left (2 d x +2 c \right )}{64 d}+\frac {11 a^{3} \sin \left (2 d x +2 c \right )}{64 d}\) | \(97\) |
derivativedivides | \(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{6} \sin \left (d x +c \right )^{2}}{8}-\frac {\cos \left (d x +c \right )^{6}}{24}\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{7}}{8}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )-\frac {3 i a^{3} \cos \left (d x +c \right )^{8}}{8}+a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{7}+\frac {7 \cos \left (d x +c \right )^{5}}{6}+\frac {35 \cos \left (d x +c \right )^{3}}{24}+\frac {35 \cos \left (d x +c \right )}{16}\right ) \sin \left (d x +c \right )}{8}+\frac {35 d x}{128}+\frac {35 c}{128}\right )}{d}\) | \(176\) |
default | \(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{6} \sin \left (d x +c \right )^{2}}{8}-\frac {\cos \left (d x +c \right )^{6}}{24}\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{7}}{8}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )-\frac {3 i a^{3} \cos \left (d x +c \right )^{8}}{8}+a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{7}+\frac {7 \cos \left (d x +c \right )^{5}}{6}+\frac {35 \cos \left (d x +c \right )^{3}}{24}+\frac {35 \cos \left (d x +c \right )}{16}\right ) \sin \left (d x +c \right )}{8}+\frac {35 d x}{128}+\frac {35 c}{128}\right )}{d}\) | \(176\) |
Input:
int(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
5/32*a^3*x-1/256*I/d*a^3*exp(8*I*(d*x+c))-5/192*I/d*a^3*exp(6*I*(d*x+c))-5 /64*I/d*a^3*exp(4*I*(d*x+c))-9/64*I/d*a^3*cos(2*d*x+2*c)+11/64/d*a^3*sin(2 *d*x+2*c)
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.58 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {{\left (120 \, a^{3} d x e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 20 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 60 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 120 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 12 i \, a^{3}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{768 \, d} \] Input:
integrate(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/768*(120*a^3*d*x*e^(2*I*d*x + 2*I*c) - 3*I*a^3*e^(10*I*d*x + 10*I*c) - 2 0*I*a^3*e^(8*I*d*x + 8*I*c) - 60*I*a^3*e^(6*I*d*x + 6*I*c) - 120*I*a^3*e^( 4*I*d*x + 4*I*c) + 12*I*a^3)*e^(-2*I*d*x - 2*I*c)/d
Time = 0.35 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.41 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {5 a^{3} x}{32} + \begin {cases} \frac {\left (- 25165824 i a^{3} d^{4} e^{10 i c} e^{8 i d x} - 167772160 i a^{3} d^{4} e^{8 i c} e^{6 i d x} - 503316480 i a^{3} d^{4} e^{6 i c} e^{4 i d x} - 1006632960 i a^{3} d^{4} e^{4 i c} e^{2 i d x} + 100663296 i a^{3} d^{4} e^{- 2 i d x}\right ) e^{- 2 i c}}{6442450944 d^{5}} & \text {for}\: d^{5} e^{2 i c} \neq 0 \\x \left (- \frac {5 a^{3}}{32} + \frac {\left (a^{3} e^{10 i c} + 5 a^{3} e^{8 i c} + 10 a^{3} e^{6 i c} + 10 a^{3} e^{4 i c} + 5 a^{3} e^{2 i c} + a^{3}\right ) e^{- 2 i c}}{32}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**8*(a+I*a*tan(d*x+c))**3,x)
Output:
5*a**3*x/32 + Piecewise(((-25165824*I*a**3*d**4*exp(10*I*c)*exp(8*I*d*x) - 167772160*I*a**3*d**4*exp(8*I*c)*exp(6*I*d*x) - 503316480*I*a**3*d**4*exp (6*I*c)*exp(4*I*d*x) - 1006632960*I*a**3*d**4*exp(4*I*c)*exp(2*I*d*x) + 10 0663296*I*a**3*d**4*exp(-2*I*d*x))*exp(-2*I*c)/(6442450944*d**5), Ne(d**5* exp(2*I*c), 0)), (x*(-5*a**3/32 + (a**3*exp(10*I*c) + 5*a**3*exp(8*I*c) + 10*a**3*exp(6*I*c) + 10*a**3*exp(4*I*c) + 5*a**3*exp(2*I*c) + a**3)*exp(-2 *I*c)/32), True))
Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.80 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {15 \, {\left (d x + c\right )} a^{3} + \frac {15 \, a^{3} \tan \left (d x + c\right )^{7} + 55 \, a^{3} \tan \left (d x + c\right )^{5} + 73 \, a^{3} \tan \left (d x + c\right )^{3} + 16 i \, a^{3} \tan \left (d x + c\right )^{2} + 81 \, a^{3} \tan \left (d x + c\right ) - 32 i \, a^{3}}{\tan \left (d x + c\right )^{8} + 4 \, \tan \left (d x + c\right )^{6} + 6 \, \tan \left (d x + c\right )^{4} + 4 \, \tan \left (d x + c\right )^{2} + 1}}{96 \, d} \] Input:
integrate(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/96*(15*(d*x + c)*a^3 + (15*a^3*tan(d*x + c)^7 + 55*a^3*tan(d*x + c)^5 + 73*a^3*tan(d*x + c)^3 + 16*I*a^3*tan(d*x + c)^2 + 81*a^3*tan(d*x + c) - 32 *I*a^3)/(tan(d*x + c)^8 + 4*tan(d*x + c)^6 + 6*tan(d*x + c)^4 + 4*tan(d*x + c)^2 + 1))/d
Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.62 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {1}{192} i \, a^{3} {\left (\frac {15 \, \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {15 \, \log \left (\tan \left (d x + c\right ) - i\right )}{d} + \frac {2 \, {\left (-15 i \, \tan \left (d x + c\right )^{4} + 45 \, \tan \left (d x + c\right )^{3} + 35 i \, \tan \left (d x + c\right )^{2} + 15 \, \tan \left (d x + c\right ) + 32 i\right )}}{d {\left (\tan \left (d x + c\right ) + i\right )}^{4} {\left (\tan \left (d x + c\right ) - i\right )}}\right )} \] Input:
integrate(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/192*I*a^3*(15*log(tan(d*x + c) + I)/d - 15*log(tan(d*x + c) - I)/d + 2*( -15*I*tan(d*x + c)^4 + 45*tan(d*x + c)^3 + 35*I*tan(d*x + c)^2 + 15*tan(d* x + c) + 32*I)/(d*(tan(d*x + c) + I)^4*(tan(d*x + c) - I)))
Time = 0.74 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {5\,a^3\,x}{32}-\frac {\frac {5\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^4}{32}+\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3\,15{}\mathrm {i}}{32}-\frac {35\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{96}+\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}}{32}-\frac {a^3}{3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^5-{\mathrm {tan}\left (c+d\,x\right )}^4\,3{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,2{}\mathrm {i}+3\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \] Input:
int(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^3,x)
Output:
(5*a^3*x)/32 - ((a^3*tan(c + d*x)*5i)/32 - a^3/3 - (35*a^3*tan(c + d*x)^2) /96 + (a^3*tan(c + d*x)^3*15i)/32 + (5*a^3*tan(c + d*x)^4)/32)/(d*(3*tan(c + d*x) - tan(c + d*x)^2*2i + 2*tan(c + d*x)^3 - tan(c + d*x)^4*3i - tan(c + d*x)^5 + 1i))
Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.74 \[ \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} \left (-48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+152 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-170 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+81 \cos \left (d x +c \right ) \sin \left (d x +c \right )-48 \sin \left (d x +c \right )^{8} i +176 \sin \left (d x +c \right )^{6} i -240 \sin \left (d x +c \right )^{4} i +144 \sin \left (d x +c \right )^{2} i +15 d x \right )}{96 d} \] Input:
int(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x)
Output:
(a**3*( - 48*cos(c + d*x)*sin(c + d*x)**7 + 152*cos(c + d*x)*sin(c + d*x)* *5 - 170*cos(c + d*x)*sin(c + d*x)**3 + 81*cos(c + d*x)*sin(c + d*x) - 48* sin(c + d*x)**8*i + 176*sin(c + d*x)**6*i - 240*sin(c + d*x)**4*i + 144*si n(c + d*x)**2*i + 15*d*x))/(96*d)