Integrand size = 24, antiderivative size = 80 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {4 i a^{13}}{5 d (a-i a \tan (c+d x))^5}+\frac {i a^{12}}{d (a-i a \tan (c+d x))^4}-\frac {i a^{11}}{3 d (a-i a \tan (c+d x))^3} \] Output:
-4/5*I*a^13/d/(a-I*a*tan(d*x+c))^5+I*a^12/d/(a-I*a*tan(d*x+c))^4-1/3*I*a^1 1/d/(a-I*a*tan(d*x+c))^3
Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.55 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {a^8 \left (-2-5 i \tan (c+d x)+5 \tan ^2(c+d x)\right )}{15 d (i+\tan (c+d x))^5} \] Input:
Integrate[Cos[c + d*x]^10*(a + I*a*Tan[c + d*x])^8,x]
Output:
-1/15*(a^8*(-2 - (5*I)*Tan[c + d*x] + 5*Tan[c + d*x]^2))/(d*(I + Tan[c + d *x])^5)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 53, 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.
\(\displaystyle \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^8}{\sec (c+d x)^{10}}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i a^{11} \int \frac {(i \tan (c+d x) a+a)^2}{(a-i a \tan (c+d x))^6}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {i a^{11} \int \left (\frac {4 a^2}{(a-i a \tan (c+d x))^6}-\frac {4 a}{(a-i a \tan (c+d x))^5}+\frac {1}{(a-i a \tan (c+d x))^4}\right )d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i a^{11} \left (\frac {4 a^2}{5 (a-i a \tan (c+d x))^5}-\frac {a}{(a-i a \tan (c+d x))^4}+\frac {1}{3 (a-i a \tan (c+d x))^3}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^10*(a + I*a*Tan[c + d*x])^8,x]
Output:
((-I)*a^11*((4*a^2)/(5*(a - I*a*Tan[c + d*x])^5) - a/(a - I*a*Tan[c + d*x] )^4 + 1/(3*(a - I*a*Tan[c + d*x])^3)))/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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 587 vs. \(2 (70 ) = 140\).
Time = 1.28 (sec) , antiderivative size = 588, normalized size of antiderivative = 7.35
\[\frac {a^{8} \left (-\frac {\sin \left (d x +c \right )^{7} \cos \left (d x +c \right )^{3}}{10}-\frac {7 \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{3}}{80}-\frac {7 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}}{96}-\frac {7 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{128}+\frac {7 \cos \left (d x +c \right ) \sin \left (d x +c \right )}{256}+\frac {7 d x}{256}+\frac {7 c}{256}\right )-8 i a^{8} \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{6}}{10}-\frac {3 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{4}}{40}-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2}}{20}-\frac {\cos \left (d x +c \right )^{4}}{40}\right )-28 a^{8} \left (-\frac {\sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{5}}{10}-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{5}}{16}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{32}+\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 )}{128}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+56 i a^{8} \left (-\frac {\cos \left (d x +c \right )^{6} \sin \left (d x +c \right )^{4}}{10}-\frac {\cos \left (d x +c \right )^{6} \sin \left (d x +c \right )^{2}}{20}-\frac {\cos \left (d x +c \right )^{6}}{60}\right )+70 a^{8} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{10}-\frac {3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{7}}{80}+\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 )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )-56 i a^{8} \left (-\frac {\cos \left (d x +c \right )^{8} \sin \left (d x +c \right )^{2}}{10}-\frac {\cos \left (d x +c \right )^{8}}{40}\right )-28 a^{8} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{9}}{10}+\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 )}{80}+\frac {7 d x}{256}+\frac {7 c}{256}\right )-\frac {4 i a^{8} \cos \left (d x +c \right )^{10}}{5}+a^{8} \left (\frac {\left (\cos \left (d x +c \right )^{9}+\frac {9 \cos \left (d x +c \right )^{7}}{8}+\frac {21 \cos \left (d x +c \right )^{5}}{16}+\frac {105 \cos \left (d x +c \right )^{3}}{64}+\frac {315 \cos \left (d x +c \right )}{128}\right ) \sin \left (d x +c \right )}{10}+\frac {63 d x}{256}+\frac {63 c}{256}\right )}{d}\]
Input:
int(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^8,x)
Output:
1/d*(a^8*(-1/10*sin(d*x+c)^7*cos(d*x+c)^3-7/80*sin(d*x+c)^5*cos(d*x+c)^3-7 /96*sin(d*x+c)^3*cos(d*x+c)^3-7/128*sin(d*x+c)*cos(d*x+c)^3+7/256*cos(d*x+ c)*sin(d*x+c)+7/256*d*x+7/256*c)-8*I*a^8*(-1/10*cos(d*x+c)^4*sin(d*x+c)^6- 3/40*cos(d*x+c)^4*sin(d*x+c)^4-1/20*cos(d*x+c)^4*sin(d*x+c)^2-1/40*cos(d*x +c)^4)-28*a^8*(-1/10*sin(d*x+c)^5*cos(d*x+c)^5-1/16*sin(d*x+c)^3*cos(d*x+c )^5-1/32*sin(d*x+c)*cos(d*x+c)^5+1/128*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d *x+c)+3/256*d*x+3/256*c)+56*I*a^8*(-1/10*cos(d*x+c)^6*sin(d*x+c)^4-1/20*co s(d*x+c)^6*sin(d*x+c)^2-1/60*cos(d*x+c)^6)+70*a^8*(-1/10*sin(d*x+c)^3*cos( d*x+c)^7-3/80*sin(d*x+c)*cos(d*x+c)^7+1/160*(cos(d*x+c)^5+5/4*cos(d*x+c)^3 +15/8*cos(d*x+c))*sin(d*x+c)+3/256*d*x+3/256*c)-56*I*a^8*(-1/10*cos(d*x+c) ^8*sin(d*x+c)^2-1/40*cos(d*x+c)^8)-28*a^8*(-1/10*sin(d*x+c)*cos(d*x+c)^9+1 /80*(cos(d*x+c)^7+7/6*cos(d*x+c)^5+35/24*cos(d*x+c)^3+35/16*cos(d*x+c))*si n(d*x+c)+7/256*d*x+7/256*c)-4/5*I*a^8*cos(d*x+c)^10+a^8*(1/10*(cos(d*x+c)^ 9+9/8*cos(d*x+c)^7+21/16*cos(d*x+c)^5+105/64*cos(d*x+c)^3+315/128*cos(d*x+ c))*sin(d*x+c)+63/256*d*x+63/256*c))
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.60 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {-6 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} - 15 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 10 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )}}{240 \, d} \] Input:
integrate(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")
Output:
1/240*(-6*I*a^8*e^(10*I*d*x + 10*I*c) - 15*I*a^8*e^(8*I*d*x + 8*I*c) - 10* I*a^8*e^(6*I*d*x + 6*I*c))/d
Time = 0.45 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.51 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=\begin {cases} \frac {- 384 i a^{8} d^{2} e^{10 i c} e^{10 i d x} - 960 i a^{8} d^{2} e^{8 i c} e^{8 i d x} - 640 i a^{8} d^{2} e^{6 i c} e^{6 i d x}}{15360 d^{3}} & \text {for}\: d^{3} \neq 0 \\x \left (\frac {a^{8} e^{10 i c}}{4} + \frac {a^{8} e^{8 i c}}{2} + \frac {a^{8} e^{6 i c}}{4}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**10*(a+I*a*tan(d*x+c))**8,x)
Output:
Piecewise(((-384*I*a**8*d**2*exp(10*I*c)*exp(10*I*d*x) - 960*I*a**8*d**2*e xp(8*I*c)*exp(8*I*d*x) - 640*I*a**8*d**2*exp(6*I*c)*exp(6*I*d*x))/(15360*d **3), Ne(d**3, 0)), (x*(a**8*exp(10*I*c)/4 + a**8*exp(8*I*c)/2 + a**8*exp( 6*I*c)/4), True))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (64) = 128\).
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.90 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {5 \, a^{8} \tan \left (d x + c\right )^{7} - 30 i \, a^{8} \tan \left (d x + c\right )^{6} - 77 \, a^{8} \tan \left (d x + c\right )^{5} + 110 i \, a^{8} \tan \left (d x + c\right )^{4} + 95 \, a^{8} \tan \left (d x + c\right )^{3} - 50 i \, a^{8} \tan \left (d x + c\right )^{2} - 15 \, a^{8} \tan \left (d x + c\right ) + 2 i \, a^{8}}{15 \, {\left (\tan \left (d x + c\right )^{10} + 5 \, \tan \left (d x + c\right )^{8} + 10 \, \tan \left (d x + c\right )^{6} + 10 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1\right )} d} \] Input:
integrate(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")
Output:
-1/15*(5*a^8*tan(d*x + c)^7 - 30*I*a^8*tan(d*x + c)^6 - 77*a^8*tan(d*x + c )^5 + 110*I*a^8*tan(d*x + c)^4 + 95*a^8*tan(d*x + c)^3 - 50*I*a^8*tan(d*x + c)^2 - 15*a^8*tan(d*x + c) + 2*I*a^8)/((tan(d*x + c)^10 + 5*tan(d*x + c) ^8 + 10*tan(d*x + c)^6 + 10*tan(d*x + c)^4 + 5*tan(d*x + c)^2 + 1)*d)
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.56 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {5 \, a^{8} \tan \left (d x + c\right )^{2} - 5 i \, a^{8} \tan \left (d x + c\right ) - 2 \, a^{8}}{15 \, d {\left (\tan \left (d x + c\right ) + i\right )}^{5}} \] Input:
integrate(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")
Output:
-1/15*(5*a^8*tan(d*x + c)^2 - 5*I*a^8*tan(d*x + c) - 2*a^8)/(d*(tan(d*x + c) + I)^5)
Time = 0.45 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^8\,\left (-5\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}+2\right )}{15\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^5+{\mathrm {tan}\left (c+d\,x\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \] Input:
int(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^8,x)
Output:
(a^8*(tan(c + d*x)*5i - 5*tan(c + d*x)^2 + 2))/(15*d*(5*tan(c + d*x) - tan (c + d*x)^2*10i - 10*tan(c + d*x)^3 + tan(c + d*x)^4*5i + tan(c + d*x)^5 + 1i))
Time = 0.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.75 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {\sin \left (d x +c \right ) a^{8} \left (192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}-504 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+452 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-155 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+15 \cos \left (d x +c \right )+192 \sin \left (d x +c \right )^{9} i -600 \sin \left (d x +c \right )^{7} i +680 \sin \left (d x +c \right )^{5} i -330 \sin \left (d x +c \right )^{3} i +60 \sin \left (d x +c \right ) i \right )}{15 d} \] Input:
int(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^8,x)
Output:
(sin(c + d*x)*a**8*(192*cos(c + d*x)*sin(c + d*x)**8 - 504*cos(c + d*x)*si n(c + d*x)**6 + 452*cos(c + d*x)*sin(c + d*x)**4 - 155*cos(c + d*x)*sin(c + d*x)**2 + 15*cos(c + d*x) + 192*sin(c + d*x)**9*i - 600*sin(c + d*x)**7* i + 680*sin(c + d*x)**5*i - 330*sin(c + d*x)**3*i + 60*sin(c + d*x)*i))/(1 5*d)