\(\int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx\) [68]

Optimal result

Integrand size = 24, antiderivative size = 144 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 x}{32}-\frac {i a^{10}}{10 d (a-i a \tan (c+d x))^5}-\frac {i a^9}{16 d (a-i a \tan (c+d x))^4}-\frac {i a^8}{24 d (a-i a \tan (c+d x))^3}-\frac {i a^7}{32 d (a-i a \tan (c+d x))^2}-\frac {i a^6}{32 d (a-i a \tan (c+d x))} \]

[Out]

1/32*a^5*x-1/10*I*a^10/d/(a-I*a*tan(d*x+c))^5-1/16*I*a^9/d/(a-I*a*tan(d*x+c))^4-1/24*I*a^8/d/(a-I*a*tan(d*x+c)
)^3-1/32*I*a^7/d/(a-I*a*tan(d*x+c))^2-1/32*I*a^6/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3568, 46, 212} \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i a^{10}}{10 d (a-i a \tan (c+d x))^5}-\frac {i a^9}{16 d (a-i a \tan (c+d x))^4}-\frac {i a^8}{24 d (a-i a \tan (c+d x))^3}-\frac {i a^7}{32 d (a-i a \tan (c+d x))^2}-\frac {i a^6}{32 d (a-i a \tan (c+d x))}+\frac {a^5 x}{32} \]

[In]

Int[Cos[c + d*x]^10*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*x)/32 - ((I/10)*a^10)/(d*(a - I*a*Tan[c + d*x])^5) - ((I/16)*a^9)/(d*(a - I*a*Tan[c + d*x])^4) - ((I/24)*
a^8)/(d*(a - I*a*Tan[c + d*x])^3) - ((I/32)*a^7)/(d*(a - I*a*Tan[c + d*x])^2) - ((I/32)*a^6)/(d*(a - I*a*Tan[c
+ d*x]))

Rule 46

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^{11}\right ) \text {Subst}\left (\int \frac {1}{(a-x)^6 (a+x)} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^{11}\right ) \text {Subst}\left (\int \left (\frac {1}{2 a (a-x)^6}+\frac {1}{4 a^2 (a-x)^5}+\frac {1}{8 a^3 (a-x)^4}+\frac {1}{16 a^4 (a-x)^3}+\frac {1}{32 a^5 (a-x)^2}+\frac {1}{32 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^{10}}{10 d (a-i a \tan (c+d x))^5}-\frac {i a^9}{16 d (a-i a \tan (c+d x))^4}-\frac {i a^8}{24 d (a-i a \tan (c+d x))^3}-\frac {i a^7}{32 d (a-i a \tan (c+d x))^2}-\frac {i a^6}{32 d (a-i a \tan (c+d x))}-\frac {\left (i a^6\right ) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,i a \tan (c+d x)\right )}{32 d} \\ & = \frac {a^5 x}{32}-\frac {i a^{10}}{10 d (a-i a \tan (c+d x))^5}-\frac {i a^9}{16 d (a-i a \tan (c+d x))^4}-\frac {i a^8}{24 d (a-i a \tan (c+d x))^3}-\frac {i a^7}{32 d (a-i a \tan (c+d x))^2}-\frac {i a^6}{32 d (a-i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.85 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 \sec ^5(c+d x) (500 \cos (c+d x)+375 \cos (3 (c+d x))+149 \cos (5 (c+d x))-100 i \sin (c+d x)-225 i \sin (3 (c+d x))-125 i \sin (5 (c+d x))+120 \arctan (\tan (c+d x)) (i \cos (5 (c+d x))+\sin (5 (c+d x))))}{3840 d (i+\tan (c+d x))^5} \]

[In]

Integrate[Cos[c + d*x]^10*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Sec[c + d*x]^5*(500*Cos[c + d*x] + 375*Cos[3*(c + d*x)] + 149*Cos[5*(c + d*x)] - (100*I)*Sin[c + d*x] - (
225*I)*Sin[3*(c + d*x)] - (125*I)*Sin[5*(c + d*x)] + 120*ArcTan[Tan[c + d*x]]*(I*Cos[5*(c + d*x)] + Sin[5*(c +
d*x)])))/(3840*d*(I + Tan[c + d*x])^5)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (122 ) = 244\).

Time = 0.70 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.30

\[\frac {i a^{5} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{10}-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{20}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{60}\right )+5 a^{5} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{10}-\frac {3 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{80}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )-10 i a^{5} \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{10}-\frac {\left (\cos ^{8}\left (d x +c \right )\right )}{40}\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{9}\left (d x +c \right )\right )}{10}+\frac {\left (\cos ^{7}\left (d x +c \right )+\frac {7 \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\cos ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \cos \left (d x +c \right )}{16}\right ) \sin \left (d x +c \right )}{80}+\frac {7 d x}{256}+\frac {7 c}{256}\right )-\frac {i a^{5} \left (\cos ^{10}\left (d x +c \right )\right )}{2}+a^{5} \left (\frac {\left (\cos ^{9}\left (d x +c \right )+\frac {9 \left (\cos ^{7}\left (d x +c \right )\right )}{8}+\frac {21 \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {105 \left (\cos ^{3}\left (d x +c \right )\right )}{64}+\frac {315 \cos \left (d x +c \right )}{128}\right ) \sin \left (d x +c \right )}{10}+\frac {63 d x}{256}+\frac {63 c}{256}\right )}{d}\]

[In]

int(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^5,x)

[Out]

1/d*(I*a^5*(-1/10*sin(d*x+c)^4*cos(d*x+c)^6-1/20*cos(d*x+c)^6*sin(d*x+c)^2-1/60*cos(d*x+c)^6)+5*a^5*(-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)-10*I*a^5*(-1/10*cos(d*x+c)^8*sin(d*x+c)^2-1/40*cos(d*x+c)^8)-10*a^5*(-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))*sin(d*x+c)+7/256*d*x+7
/256*c)-1/2*I*a^5*cos(d*x+c)^10+a^5*(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))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.58 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {120 \, a^{5} d x - 12 i \, a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} - 75 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} - 200 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 300 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 300 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )}}{3840 \, d} \]

[In]

integrate(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

1/3840*(120*a^5*d*x - 12*I*a^5*e^(10*I*d*x + 10*I*c) - 75*I*a^5*e^(8*I*d*x + 8*I*c) - 200*I*a^5*e^(6*I*d*x + 6
*I*c) - 300*I*a^5*e^(4*I*d*x + 4*I*c) - 300*I*a^5*e^(2*I*d*x + 2*I*c))/d

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.45 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^{5} x}{32} + \begin {cases} \frac {- 100663296 i a^{5} d^{4} e^{10 i c} e^{10 i d x} - 629145600 i a^{5} d^{4} e^{8 i c} e^{8 i d x} - 1677721600 i a^{5} d^{4} e^{6 i c} e^{6 i d x} - 2516582400 i a^{5} d^{4} e^{4 i c} e^{4 i d x} - 2516582400 i a^{5} d^{4} e^{2 i c} e^{2 i d x}}{32212254720 d^{5}} & \text {for}\: d^{5} \neq 0 \\x \left (\frac {a^{5} e^{10 i c}}{32} + \frac {5 a^{5} e^{8 i c}}{32} + \frac {5 a^{5} e^{6 i c}}{16} + \frac {5 a^{5} e^{4 i c}}{16} + \frac {5 a^{5} e^{2 i c}}{32}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**10*(a+I*a*tan(d*x+c))**5,x)

[Out]

a**5*x/32 + Piecewise(((-100663296*I*a**5*d**4*exp(10*I*c)*exp(10*I*d*x) - 629145600*I*a**5*d**4*exp(8*I*c)*ex
p(8*I*d*x) - 1677721600*I*a**5*d**4*exp(6*I*c)*exp(6*I*d*x) - 2516582400*I*a**5*d**4*exp(4*I*c)*exp(4*I*d*x) -
2516582400*I*a**5*d**4*exp(2*I*c)*exp(2*I*d*x))/(32212254720*d**5), Ne(d**5, 0)), (x*(a**5*exp(10*I*c)/32 + 5
*a**5*exp(8*I*c)/32 + 5*a**5*exp(6*I*c)/16 + 5*a**5*exp(4*I*c)/16 + 5*a**5*exp(2*I*c)/32), True))

Maxima [A] (verification not implemented)

none

Time = 1.33 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.14 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {15 \, {\left (d x + c\right )} a^{5} + \frac {15 \, a^{5} \tan \left (d x + c\right )^{9} + 70 \, a^{5} \tan \left (d x + c\right )^{7} + 128 \, a^{5} \tan \left (d x + c\right )^{5} - 80 i \, a^{5} \tan \left (d x + c\right )^{4} - 230 \, a^{5} \tan \left (d x + c\right )^{3} + 560 i \, a^{5} \tan \left (d x + c\right )^{2} + 465 \, a^{5} \tan \left (d x + c\right ) - 128 i \, a^{5}}{\tan \left (d x + c\right )^{10} + 5 \, \tan \left (d x + c\right )^{8} + 10 \, \tan \left (d x + c\right )^{6} + 10 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1}}{480 \, d} \]

[In]

integrate(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

1/480*(15*(d*x + c)*a^5 + (15*a^5*tan(d*x + c)^9 + 70*a^5*tan(d*x + c)^7 + 128*a^5*tan(d*x + c)^5 - 80*I*a^5*t
an(d*x + c)^4 - 230*a^5*tan(d*x + c)^3 + 560*I*a^5*tan(d*x + c)^2 + 465*a^5*tan(d*x + c) - 128*I*a^5)/(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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 857 vs. \(2 (112) = 224\).

Time = 0.90 (sec) , antiderivative size = 857, normalized size of antiderivative = 5.95 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^10*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

1/15360*(480*a^5*d*x*e^(16*I*d*x + 8*I*c) + 3840*a^5*d*x*e^(14*I*d*x + 6*I*c) + 13440*a^5*d*x*e^(12*I*d*x + 4*
I*c) + 26880*a^5*d*x*e^(10*I*d*x + 2*I*c) + 26880*a^5*d*x*e^(6*I*d*x - 2*I*c) + 13440*a^5*d*x*e^(4*I*d*x - 4*I
*c) + 3840*a^5*d*x*e^(2*I*d*x - 6*I*c) + 33600*a^5*d*x*e^(8*I*d*x) + 480*a^5*d*x*e^(-8*I*c) - 195*I*a^5*e^(16*
I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 1560*I*a^5*e^(14*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 5
460*I*a^5*e^(12*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 10920*I*a^5*e^(10*I*d*x + 2*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) - 10920*I*a^5*e^(6*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 5460*I*a^5*e^(4*I*d*x - 4*I*c)*
log(e^(2*I*d*x + 2*I*c) + 1) - 1560*I*a^5*e^(2*I*d*x - 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 13650*I*a^5*e^(8*
I*d*x)*log(e^(2*I*d*x + 2*I*c) + 1) - 195*I*a^5*e^(-8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 195*I*a^5*e^(16*I*d*
x + 8*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 1560*I*a^5*e^(14*I*d*x + 6*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 546
0*I*a^5*e^(12*I*d*x + 4*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 10920*I*a^5*e^(10*I*d*x + 2*I*c)*log(e^(2*I*d*x)
+ e^(-2*I*c)) + 10920*I*a^5*e^(6*I*d*x - 2*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 5460*I*a^5*e^(4*I*d*x - 4*I*c)
*log(e^(2*I*d*x) + e^(-2*I*c)) + 1560*I*a^5*e^(2*I*d*x - 6*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 13650*I*a^5*e^
(8*I*d*x)*log(e^(2*I*d*x) + e^(-2*I*c)) + 195*I*a^5*e^(-8*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) - 48*I*a^5*e^(26*
I*d*x + 18*I*c) - 684*I*a^5*e^(24*I*d*x + 16*I*c) - 4544*I*a^5*e^(22*I*d*x + 14*I*c) - 18688*I*a^5*e^(20*I*d*x
+ 12*I*c) - 53360*I*a^5*e^(18*I*d*x + 10*I*c) - 111688*I*a^5*e^(16*I*d*x + 8*I*c) - 174944*I*a^5*e^(14*I*d*x
+ 6*I*c) - 204784*I*a^5*e^(12*I*d*x + 4*I*c) - 176048*I*a^5*e^(10*I*d*x + 2*I*c) - 44000*I*a^5*e^(6*I*d*x - 2*
I*c) - 10800*I*a^5*e^(4*I*d*x - 4*I*c) - 1200*I*a^5*e^(2*I*d*x - 6*I*c) - 107500*I*a^5*e^(8*I*d*x))/(d*e^(16*I
*d*x + 8*I*c) + 8*d*e^(14*I*d*x + 6*I*c) + 28*d*e^(12*I*d*x + 4*I*c) + 56*d*e^(10*I*d*x + 2*I*c) + 56*d*e^(6*I
*d*x - 2*I*c) + 28*d*e^(4*I*d*x - 4*I*c) + 8*d*e^(2*I*d*x - 6*I*c) + 70*d*e^(8*I*d*x) + d*e^(-8*I*c))

Mupad [B] (verification not implemented)

Time = 4.74 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.85 \[ \int \cos ^{10}(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5\,x}{32}+\frac {\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^4}{32}+\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^3\,5{}\mathrm {i}}{32}-\frac {31\,a^5\,{\mathrm {tan}\left (c+d\,x\right )}^2}{96}-\frac {a^5\,\mathrm {tan}\left (c+d\,x\right )\,35{}\mathrm {i}}{96}+\frac {4\,a^5}{15}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^5+{\mathrm {tan}\left (c+d\,x\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^5,x)

[Out]

(a^5*x)/32 + ((4*a^5)/15 - (a^5*tan(c + d*x)*35i)/96 - (31*a^5*tan(c + d*x)^2)/96 + (a^5*tan(c + d*x)^3*5i)/32
+ (a^5*tan(c + d*x)^4)/32)/(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))