Integrand size = 24, antiderivative size = 124 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=80 a^8 x-\frac {80 i a^8 \log (\cos (c+d x))}{d}-\frac {31 a^8 \tan (c+d x)}{d}-\frac {4 i a^8 \tan ^2(c+d x)}{d}+\frac {a^8 \tan ^3(c+d x)}{3 d}-\frac {16 i a^{10}}{d (a-i a \tan (c+d x))^2}+\frac {80 i a^9}{d (a-i a \tan (c+d x))} \] Output:
80*a^8*x-80*I*a^8*ln(cos(d*x+c))/d-31*a^8*tan(d*x+c)/d-4*I*a^8*tan(d*x+c)^ 2/d+1/3*a^8*tan(d*x+c)^3/d-16*I*a^10/d/(a-I*a*tan(d*x+c))^2+80*I*a^9/d/(a- I*a*tan(d*x+c))
Time = 1.38 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.69 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {i a^8 \left (-93 i \tan (c+d x)+12 \tan ^2(c+d x)+i \tan ^3(c+d x)+48 \left (-5 \log (i+\tan (c+d x))+\frac {4-5 i \tan (c+d x)}{(i+\tan (c+d x))^2}\right )\right )}{3 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]
Output:
((-1/3*I)*a^8*((-93*I)*Tan[c + d*x] + 12*Tan[c + d*x]^2 + I*Tan[c + d*x]^3 + 48*(-5*Log[I + Tan[c + d*x]] + (4 - (5*I)*Tan[c + d*x])/(I + Tan[c + d* x])^2)))/d
Time = 0.30 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 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.
\(\displaystyle \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^8}{\sec (c+d x)^4}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i a^5 \int \frac {(i \tan (c+d x) a+a)^5}{(a-i a \tan (c+d x))^3}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i a^5 \int \left (\frac {32 a^5}{(a-i a \tan (c+d x))^3}-\frac {80 a^4}{(a-i a \tan (c+d x))^2}+\frac {80 a^3}{a-i a \tan (c+d x)}+\tan ^2(c+d x) a^2-8 i \tan (c+d x) a^2-31 a^2\right )d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i a^5 \left (\frac {16 a^5}{(a-i a \tan (c+d x))^2}-\frac {80 a^4}{a-i a \tan (c+d x)}+\frac {1}{3} i a^3 \tan ^3(c+d x)+4 a^3 \tan ^2(c+d x)-31 i a^3 \tan (c+d x)-80 a^3 \log (a-i a \tan (c+d x))\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]
Output:
((-I)*a^5*(-80*a^3*Log[a - I*a*Tan[c + d*x]] - (31*I)*a^3*Tan[c + d*x] + 4 *a^3*Tan[c + d*x]^2 + (I/3)*a^3*Tan[c + d*x]^3 + (16*a^5)/(a - I*a*Tan[c + d*x])^2 - (80*a^4)/(a - I*a*Tan[c + d*x])))/d
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[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 = 193.98 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {4 i a^{8} {\mathrm e}^{4 i \left (d x +c \right )}}{d}+\frac {32 i a^{8} {\mathrm e}^{2 i \left (d x +c \right )}}{d}-\frac {160 a^{8} c}{d}-\frac {4 i a^{8} \left (60 \,{\mathrm e}^{4 i \left (d x +c \right )}+105 \,{\mathrm e}^{2 i \left (d x +c \right )}+47\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {80 i a^{8} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(114\) |
derivativedivides | \(\frac {a^{8} \left (\frac {\sin \left (d x +c \right )^{9}}{3 \cos \left (d x +c \right )^{3}}-\frac {2 \sin \left (d x +c \right )^{9}}{\cos \left (d x +c \right )}-2 \left (\sin \left (d x +c \right )^{7}+\frac {7 \sin \left (d x +c \right )^{5}}{6}+\frac {35 \sin \left (d x +c \right )^{3}}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )+\frac {35 d x}{8}+\frac {35 c}{8}\right )-2 i a^{8} \cos \left (d x +c \right )^{4}-28 a^{8} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )-8 i a^{8} \left (\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{6}}{2}+\frac {3 \sin \left (d x +c \right )^{4}}{4}+\frac {3 \sin \left (d x +c \right )^{2}}{2}+3 \ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+56 i a^{8} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-28 a^{8} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-14 i a^{8} \sin \left (d x +c \right )^{4}+a^{8} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(409\) |
default | \(\frac {a^{8} \left (\frac {\sin \left (d x +c \right )^{9}}{3 \cos \left (d x +c \right )^{3}}-\frac {2 \sin \left (d x +c \right )^{9}}{\cos \left (d x +c \right )}-2 \left (\sin \left (d x +c \right )^{7}+\frac {7 \sin \left (d x +c \right )^{5}}{6}+\frac {35 \sin \left (d x +c \right )^{3}}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )+\frac {35 d x}{8}+\frac {35 c}{8}\right )-2 i a^{8} \cos \left (d x +c \right )^{4}-28 a^{8} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )-8 i a^{8} \left (\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{6}}{2}+\frac {3 \sin \left (d x +c \right )^{4}}{4}+\frac {3 \sin \left (d x +c \right )^{2}}{2}+3 \ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+56 i a^{8} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-28 a^{8} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-14 i a^{8} \sin \left (d x +c \right )^{4}+a^{8} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(409\) |
Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)
Output:
-4*I/d*a^8*exp(4*I*(d*x+c))+32*I/d*a^8*exp(2*I*(d*x+c))-160/d*a^8*c-4/3*I* a^8*(60*exp(4*I*(d*x+c))+105*exp(2*I*(d*x+c))+47)/d/(exp(2*I*(d*x+c))+1)^3 -80*I/d*a^8*ln(exp(2*I*(d*x+c))+1)
Time = 0.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.44 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {4 \, {\left (3 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} - 15 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 63 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} - 9 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 81 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 47 i \, a^{8} + 60 \, {\left (i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{8}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")
Output:
-4/3*(3*I*a^8*e^(10*I*d*x + 10*I*c) - 15*I*a^8*e^(8*I*d*x + 8*I*c) - 63*I* a^8*e^(6*I*d*x + 6*I*c) - 9*I*a^8*e^(4*I*d*x + 4*I*c) + 81*I*a^8*e^(2*I*d* x + 2*I*c) + 47*I*a^8 + 60*(I*a^8*e^(6*I*d*x + 6*I*c) + 3*I*a^8*e^(4*I*d*x + 4*I*c) + 3*I*a^8*e^(2*I*d*x + 2*I*c) + I*a^8)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2 *I*c) + d)
Time = 0.39 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.74 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=- \frac {80 i a^{8} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 240 i a^{8} e^{4 i c} e^{4 i d x} - 420 i a^{8} e^{2 i c} e^{2 i d x} - 188 i a^{8}}{3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} + 3 d} + \begin {cases} \frac {- 4 i a^{8} d e^{4 i c} e^{4 i d x} + 32 i a^{8} d e^{2 i c} e^{2 i d x}}{d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (16 a^{8} e^{4 i c} - 64 a^{8} e^{2 i c}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**8,x)
Output:
-80*I*a**8*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-240*I*a**8*exp(4*I*c)*exp (4*I*d*x) - 420*I*a**8*exp(2*I*c)*exp(2*I*d*x) - 188*I*a**8)/(3*d*exp(6*I* c)*exp(6*I*d*x) + 9*d*exp(4*I*c)*exp(4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x ) + 3*d) + Piecewise(((-4*I*a**8*d*exp(4*I*c)*exp(4*I*d*x) + 32*I*a**8*d*e xp(2*I*c)*exp(2*I*d*x))/d**2, Ne(d**2, 0)), (x*(16*a**8*exp(4*I*c) - 64*a* *8*exp(2*I*c)), True))
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^{8} \tan \left (d x + c\right )^{3} - 12 i \, a^{8} \tan \left (d x + c\right )^{2} + 240 \, {\left (d x + c\right )} a^{8} + 120 i \, a^{8} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 93 \, a^{8} \tan \left (d x + c\right ) - \frac {48 \, {\left (5 \, a^{8} \tan \left (d x + c\right )^{3} - 6 i \, a^{8} \tan \left (d x + c\right )^{2} + 3 \, a^{8} \tan \left (d x + c\right ) - 4 i \, a^{8}\right )}}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{3 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")
Output:
1/3*(a^8*tan(d*x + c)^3 - 12*I*a^8*tan(d*x + c)^2 + 240*(d*x + c)*a^8 + 12 0*I*a^8*log(tan(d*x + c)^2 + 1) - 93*a^8*tan(d*x + c) - 48*(5*a^8*tan(d*x + c)^3 - 6*I*a^8*tan(d*x + c)^2 + 3*a^8*tan(d*x + c) - 4*I*a^8)/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/d
Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {80 i \, a^{8} \log \left (\tan \left (d x + c\right ) + i\right )}{d} + \frac {a^{8} d^{2} \tan \left (d x + c\right )^{3} - 12 i \, a^{8} d^{2} \tan \left (d x + c\right )^{2} - 93 \, a^{8} d^{2} \tan \left (d x + c\right )}{3 \, d^{3}} - \frac {16 \, {\left (5 \, a^{8} \tan \left (d x + c\right ) + 4 i \, a^{8}\right )}}{d {\left (\tan \left (d x + c\right ) + i\right )}^{2}} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")
Output:
80*I*a^8*log(tan(d*x + c) + I)/d + 1/3*(a^8*d^2*tan(d*x + c)^3 - 12*I*a^8* d^2*tan(d*x + c)^2 - 93*a^8*d^2*tan(d*x + c))/d^3 - 16*(5*a^8*tan(d*x + c) + 4*I*a^8)/(d*(tan(d*x + c) + I)^2)
Time = 0.41 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^8\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}-\frac {80\,a^8\,\mathrm {tan}\left (c+d\,x\right )+a^8\,64{}\mathrm {i}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}-1\right )}-\frac {31\,a^8\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {a^8\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,80{}\mathrm {i}}{d}-\frac {a^8\,{\mathrm {tan}\left (c+d\,x\right )}^2\,4{}\mathrm {i}}{d} \] Input:
int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^8,x)
Output:
(a^8*log(tan(c + d*x) + 1i)*80i)/d - (31*a^8*tan(c + d*x))/d - (80*a^8*tan (c + d*x) + a^8*64i)/(d*(tan(c + d*x)*2i + tan(c + d*x)^2 - 1)) - (a^8*tan (c + d*x)^2*4i)/d + (a^8*tan(c + d*x)^3)/(3*d)
Time = 0.16 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.49 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^{8} \left (240 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} i -240 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) i -240 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} i +240 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) i -240 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} i +240 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) i -96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} i +240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c +240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x +108 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -240 \cos \left (d x +c \right ) c -240 \cos \left (d x +c \right ) d x +96 \sin \left (d x +c \right )^{7}-48 \sin \left (d x +c \right )^{5}-286 \sin \left (d x +c \right )^{3}+237 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x)
Output:
(a**8*(240*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*i - 2 40*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*i - 240*cos(c + d*x)*log(tan( (c + d*x)/2) - 1)*sin(c + d*x)**2*i + 240*cos(c + d*x)*log(tan((c + d*x)/2 ) - 1)*i - 240*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*i + 240*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*i - 96*cos(c + d*x)*sin(c + d*x )**6*i + 240*cos(c + d*x)*sin(c + d*x)**2*c + 240*cos(c + d*x)*sin(c + d*x )**2*d*x + 108*cos(c + d*x)*sin(c + d*x)**2*i - 240*cos(c + d*x)*c - 240*c os(c + d*x)*d*x + 96*sin(c + d*x)**7 - 48*sin(c + d*x)**5 - 286*sin(c + d* x)**3 + 237*sin(c + d*x)))/(3*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))